["Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.55 \t\t Motion can also be defined as the change in the posi- \t\tNow, tion of a body with respect to a given object. The posi- tion of a point P at any time t is expressed in terms of a = dv = dv \u00d7 dx = dv \u00d7v the distance x from a fixed origin O on the reference dt dx dt dx x-axis, y-axis or z-axis and can be taken as positive or negative as per the usual sign convention. Acceleration is positive when velocity is increasing. A posi- tive acceleration means that the particle is either moving X2 X further in a positive direction or is slowing down in the negative direction. X1 Retardation or deceleration of a body in motion is \u2212x \u2022 O\u2022 \u2022 P \u2022 x the negative acceleration, i.e., retarding acceleration. P2 P1 + Acceleration is the rate of increase in the velocity and deceleration is the rate of decrease in the velocity. \t 2.\t Average velocity: The average velocity vav of a point Uniform Motion P, in the time interval between t + Dt and t, i.e., in the time interval Dt, during which its position changes When a particle moves with a constant velocity so that its accel- eration is zero, then the motion is termed as uniform motion. from to is defined by\u2009vav Dx . x x + Dx = Dt Uniformly Accelerated Motion t t + Dt When a particle moves with a constant acceleration, then the motion is termed as a uniformly accelerated motion. x Dt O\u2022 P\u2022 P\u20221 Motion at a Uniform Acceleration \t 3.\tInstantaneous velocity and speed: The instantane- Let the uniform acceleration be \u2018a\u2019. Then ous velocity v of a point P at time t is the limiting value of the average velocity as the increment of time v = u + at approaches zero as a limit. Mathematically it can be expressed as: And, v2 = u2 + 2as s = ut + 1 at2 2 v = Limit Dx = dx sn = u + a \u239b\u239c\u239d n - 12\u23a0\u239e\u239f Dt dt Dt\u2192 0 \t\tThe velocity v is positive if the displacement x is Where, increasing and the particle is moving in a positive direc- v \u2013 Velocity at any time instant t (secs) tion. The unit of velocity is metre per second (m\/s). u \u2013 Initial velocity s \u2013 Distance travelled during the time t (secs) \t\t\t If s is the distance covered by a moving particle at sn \u2013 Distance travelled at the nth second time t, then speed\u2009= ds . \u2009The unit of speed is the same dt as that of the velocity. NOTE \t 4.\t Average acceleration: The average acceleration a\u00adav of For motion under constant retardation or deceleration, a point P, in the time interval between t + Dt and t, assign negative sign for acceleration (a). i.e., in the time interval Dt, during which its velocity changes from v to v + Dv is defined by\u2009 aav = Dv . Vertical Motion Under Gravity Dt A body in motion above the ground will be under influence \t 5.\t Instantaneous acceleration: The instantaneous accel- eration of a point P is the limiting value of the average of the gravitational force of attraction (g). If the body moves acceleration as the increment of time approaches zero. Mathematically it can be expressed as: upwards then it is subjected to gravitational retardation, i.e., a = -g. Then, the equations for the upward motion of a body under gravity will be a = Limit Dv = dv v = u - gt Dt dt Dt\u2192 0 v2 = u2 \u2013 2gs \t\tThen, s = ut - 1 \u2009gt2 2 d2x a = dv = dt 2 sn = u \u2013 g \u239c\u239b\u239d n - 1 \u23a0\u239e\u239f dt 2","3.56\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics If the body moves downwards then it is subjected to gravi- Dot product zero means the new resultant between 2v1 and v2 is at right angles to v2. tational attraction and hence an acceleration, i.e., a = g. Example 2:\u2002 If the two ends of a train, moving with a Then, the equations for the downward motion of a body constant acceleration, pass a certain point with velocities u and v respectively, the velocity with which the middle point under gravity will be of the train passes through the same point is \u2009v = u + gt \u2009v2 = u2 + 2gs s = ut + 1 gt2 (A)\t u + v \t(B)\tu2 + v2 2 2 u+v sn = u + g(n \u2013 1) 2 (C)\t u - v\t(D)\tu2 + v2 NOTES 2 1.\u2002 For a body that is just dropped, a = g and u = 0. Solution: 2.\u2002T\u0007he final vertical velocity of a body thrown upwards as We have the relation v2 = u2 + 2as\b(1) it reaches the maximum height, will be zero, i.e., v = 0. If v is the velocity with which the mid point of the train Motion Curves crosses the point, we have These are the graphical representation of displacement, \t\tv2 = u2 + 2 a s \b(2) velocity and acceleration against time. 2 a Eliminating s from (1) and (2), We have, a v2 - u2 = as v dv And, v dt s v2 - u2 = 2as s ds dt Now, t v2 - u2 = 1 v2 - u2 2 Considering the general case of acceleration not being a constant, the above graphical representation is made. Or, \u2234 2v2 - 2u2 = v2 - u2 Now, 2v2 = v2 + u2 The slope of the displacement-time curve \u2192 Velocity The slope of the velocity-time curve \u2192 Acceleration v2 = v2 + u2 The area under the velocity-time curve \u2192 Displacement 2 The area under the acceleration-time curve \u2192 Velocity Solved Examples \u2234 v = v2 + u2 . 2 Example 1:\u2002 A particle has two velocities v1 and v2. Its Direction for examples 3 and 4:\u2002 The motion of a particle resultant is v1 in magnitude. When the velocity v1 is doubled, is defined as s = 2t3 \u2013 6t2 + 15, where s is in metres and t is the new resultant is in seconds. (A)\t Perpendicular to v2\t (B)\t Parallel to v2 Example 3:\u2002 The acceleration when the velocity is zero is (C)\t Equal to v2\t (D)\t Equal to 2 v2 (A)\t 2 m\/s2 \t(B)\t8 m\/s2 Solution: (C)\t 6 m\/s2 \t(D)\t4 m\/s2 Applying the principle of vector, the magnitude of the resultant between\u2009 v1 + v2 Solution: Given that v1 + v2 = v1 s = 2t3 \u2013 6t2 + 15 Squring both side, v1 + v2 2 = v12 \u2234 (v1 + v2 ) \u22c5 (v1 + v2 ) = v1.v1 ds = 6t 2 \u2013 12t dt v1 \u22c5 v1 + 2v1 \u22c5 v2 + v2 \u22c5 v2 = v1 \u22c5 v1 2v1 \u22c5 v2 + v2 \u22c5 v2 = 0 a= ds2 = 12t - 12 dt 2 (2v1 + v2 ) \u22c5 v2 = 0.","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.57 When velocity is zero, 6t2 \u2013 12t = 0, When t = 0, v = 0 \u2234 t = 2 sec \\\\C=0 Then acceleration is, a = 12 \u00d7 2 \u2013 12 =\u200912 m\/s2 \\\\ v = 50t \u2013 12t3 Example 4:\u2002 The minimum velocity is ds = 50t - 12t 3 (A)\t -2 m\/s \t(B)\t6 m\/s dt (C)\t -6 m\/s \t(D)\t2 m\/s Integrating, s = 50 t2 - 12 t4 + C1 2 4 Solution: = 25t2 \u2013 3t4 + C1 When t = 0, s = 0 Velocity is minimum when dv = 0, \u2009i.e., when 12t \u2013 12 = 0, s\\\\=C215=t2 0 3t4 dt \u2013 \u2234 t = 1 sec Here we can find the time when s = 52 m. \\\\ 25t2 \u2013 3t4 = 52 Let t2 = u, then 25u \u2013 3u2 = 52 (Velocity)min\u2009= 6t 2 \u2013 12t = 6 \u2013 12 = -6 m\/s 3u2 - 25u + 52 = 0 Example 5:\u2002 The velocity of a particle along the x-axis is given by v = 5x3\/2 where x is in metres and v is in\u2009m\/s. u = 25 \u00b1 625 - 624 6 The acceleration when x = 2m is u = 25 \u00b11 = 26 or 24 (A)\t 300 m\/s2 \t(B)\t200 m\/s2 6 6 6 (C)\t 180 m\/s2 \t(D)\t150 m\/s2 24 6 Solution: Case 1: when t2 = = 4 Given v = 5x3\/2, differentiating with respect to t, we have \\\\ t = 2 sec v = 50t \u2013 12t3 dv = 5\u00d7 3 x 3\/ 2 -1 \u239b\u239d\u239c dx \u23a0\u239f\u239e = 50 \u00d7 2 - 12 \u00d7 8 dt 2 dt = 100 - 96 = 4 m\/s = 15 x1\/ 2 dx , but dx = v Case 2: when t2 = 26 = 4.333 2 dt dt 6 \u2234 a = 15 x1\/ 2 \u00d7 5x3\/2 = 75 x2 \\\\ t = 2.08 sec 2 2 The value of the velocity calculated with this t value is not 75 m\/s2 . available in the options provided. 2 When x = 2, a = \u00d7 4 = 150 Example 7:\u2002 A body dropped from a certain height covers Example 6:\u2002 A particle is moving in a straight line starting 5 th of the total height in the last second, the height from from rest. Its acceleration is given by the expression a = 9 50 \u2013 36t2, where t is in seconds. The velocity of the particle which the body is dropped is when it has travelled 52 m can be (A)\t 2.3 m\/s \t(B)\t4 m\/s (A)\t 36.8 m\t (B)\t 40.3 m (C)\t 6.7 m\/s \t(D)\t8 m\/s (C)\t 44.1 m\t (D)\t 50.6 m Solution: Let \u2018h\u2019 be the height and let \u2018n\u2019 be the time taken for the Solution: fall. Then, Given, s = u + a \u239b\u239c\u239d n - 12\u23a0\u239f\u239e a = 50 \u2013 36t2 5 h = 0 + g \u239d\u239c\u239b n - 1 \u239f\u23a0\u239e 9 2 So, dv \u2009= 50 \u2013 36t2 dt \t\t 5 h= g \u239c\u239d\u239b n - 12\u23a0\u239e\u239f \b(1) 9 Or, dv = 50dt \u2013 36t2 dt Integrating the above equation, we have Also, h = un + 1 an2 2 t3 1 v = 50t - 36 3 + C = 50t - 12t 3 +C \t\th = 0 + 2 gn2\b(2)","3.58\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Putting (2) in (1), For the second ball, 5 \u00d7 1 gn2 = g \u239c\u239b\u239d n - 1 \u239f\u239e\u23a0 h = 49 \u00d7 (T - 2) \u2013 1 g(T - 2)2 9 2 2 2 \\\\ 5n2 \u2013 18n + 9 = 0 Equating, 49T \u2013 1 gT2 = 49 (T - 2) \u2013 1 g(T - 2)2 5n2 \u2013 15n \u2013 3n + 9 = 0 2 2 5n(n - 3) \u2013 3(n - 3) = 0 \u2234 T = 11.99 sec \u2234 t = T \u2013 2 = 9.99 sec \u2248 10 sec \\\\ (5n \u2013 3)(n \u2013 3) = 0 \u2234n = 3 or n = 3, but n >1 Example 10:\u2002 Two bodies are moving uniformly towards 5 each other. The distance between them decreases at a rate \u2234n=3 of 6 m\/s. If both the bodies move in the same direction with the same speeds, then the distance between them increases \u2234h= 1 gn2 = 1 \u00d7 9.81 \u00d7 9 = 44.1 m. at a rate of 4 m\/s. The respective speeds of the bodies are 2 2 (A)\t 3 m\/s and 1 m\/s\t (B)\t 5 m\/s and 1 m\/s (C)\t 4 m\/s and 2 m\/s\t (D)\t 3 m\/s and 5 m\/s Example 8:\u2002 A stone falls past a window 2 m high in a time of 0.2 seconds. The height above the window from where the stone has been dropped is (A)\t 4.15 m\t (B)\t 5.23 m Solution: Let u and v be the velocities of the bodies. From the state- (C)\t 5.87 m\t (D)\t 6.32 m ment of the problem, Solution: u+v=6 u\u2013v=4 A h \u2234 u = 5 m\/s and v = 1 m\/s. window Example 11:\u2002 Two cars are moving in the same direction The stone is dropped from A. Let the body reach the top of each with a speed of 45 km\/h. The distance separating the window with a velocity of u m\/s. Then, them is 10 km. Another vehicle coming from the opposite u2 = 02 + 2gh direction meets these two cars in an interval of 6 minutes. u2 = 2gh\b(1) The speed of the vehicle is (A)\t 45 km\/h\t (B)\t 50 km\/h Falling with an initial velocity u, it covers the window 2 m (C)\t 55 km\/h\t (D)\t 60 km\/h high in 0.5 seconds. Solution: The distance between the cars moves with a velocity of 45 s = ut + 1 at2 2 km\/h. If the speed of the vehicle is u, then its velocity rela- 1 2 = u \u00d7 0.2 + 2 \u00d7 9.81 \u00d7 0.22 tive to the moving distance is 45 + u m\/s. It takes 6 minutes to cover the distance of 10 km. 1 2 = 0.2u + 2 \u00d7 9.81 \u00d7 0.04 2 = 0.2u + 9.81 \u00d7 0.02 \u2234 (45 + u) \u00d7 6 = 10 60 u = 9.019 m\/s \u2234 45 + u = 100 u = 55 km\/h. From (1), u2 = 2gh, \u2234 h 9.0192 = 4.145 m. Motion under Variable Acceleration 2 \u00d7 9.81 In practical conditions a body may very often move with Example 9:\u2002 A ball is projected vertically upwards with a variable acceleration. The rate of change of velocity will not remain constant. We know that acceleration, velocity of 49 m\/s. If another ball is projected in the same manner after 2 seconds, and if both meet t seconds after the dv dv ds dt ds dt second ball is projected, then t is equal to a = = \u22c5 (A)\t 3 s\t (B)\t 10 s (C)\t 5 s\t (D)\t 6 s Or\u2003\u2003\u2003\u2003\u2003\u2003\u2003 a= v \u22c5 dv ds Solution: Let both the balls meet T seconds after the first ball is pro- Also when displacement can be expressed as a third degree jected. Therefore when the balls meet, for the first ball, or higher degree equation in time, the acceleration becomes a variable with respect to time. h = 49 \u00d7 T \u2013 1 gT2 2","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.59 For example, if s = 4t3 + 3t2 + 5t + 1 Solution: ds 12t 2 During the first 6 seconds, the average acceleration dt = + 6t + 5 = 12 +10 = 11 m\/s2 2 d2s dt 2 = 24t + 6 \u2234 Increase in velocity during this interval of 6 seconds The velocity and displacement are evaluated by integration. = average acceleration \u00d7 6 = 66 m\/s Example 12:\u2002 A body is starting from rest and moving \u2234 Velocity at the end of 6 second =\u2009 66 m\/s. along a straight line whose acceleration is given by f = 10 \u2013 0.006x2 where x is the displacement in m and f is the Example 14:\u2002 The distance travelled during these six acceleration in\u2009m\/s2. The distance travelled by it when it comes to rest is seconds is (A)\t 242 m\t (B)\t 218 m (C)\t 198 m\t (D)\t 124 m (A)\t 70.7 m\t (B)\t 68.3 m Solution: (C)\t 62.6 m\t (D)\t 58.5 m Average velocity during this interval = 0 + 66 = 33 m\/s 2 Solution: Given that f = 10 \u2013 0.006x2 \u2234 Distance travelled during this interval = 33 \u00d7 6 = 198 m. dv = 10 - 0.006 x 2 Example 15:\u2002 At any instant, the acceleration of a train dt starting from rest is given by f = 10 \u2009where u is the u +1 dv dx velocity of the train in m\/s. The distance at which the train dx \u22c5 dt = 10 - 0.006 x 2 will attain a velocity of 54 km\/h is dv (A)\t 123.7 m\t (B)\t 185.4 m dx v\u22c5 = 10 - 0.066 x 2 (C)\t 214.4 m \t (D)\t 228.2 m Solution: vdv = (10 \u2013 0.006x2)dx integrating It is given f = 10 u +1 v2 x3 v = 10x - 0.006 3 +C u\u22c5 du = 10 dx u +1 when x = 0, v = 0 u(u + 1)du = 10dx \u2234C=0 Integrating we have, u3 u2 = 10x + c v2 x3 3 +2 2 = 10x - 0.006 3 when x = 0, u = 0. v2 = 20x \u2013 0.004x3 \u2234c=0 when v = 0; 20x \u2013 0.004x3 = 0 u3 + u2 = 10x \u2234 0.004x2 = 20 (note that the solution of x = 0 is also pos- 3 2 sible for the above equation, but the value of x > 0 is sought for) \u2234 x = 70.7 m. when u = 54 km\/h = 54 \u00d7 5\/18 = 15 m\/s Direction for examples 13 and 14: An electric train starting 153 152 = 10x from rest has an acceleration f in m\/s2. which vary with time 3 +2 as shown in the table below: 1125 + 112.5 = 10x \t\t \u2234 x = 123.7 m. t(secs) 0 6 12 18 Example 16:\u2002 The motion of a particle is given by the f\u2009\uf8ec\uf8eb\uf8ed m \uf8f6 equation a = t3 \u2013 3t2 + 5, where \u2018a\u2019 is acceleration in m\/ s2 \uf8f7\uf8f8 s2 and t is time in seconds. It is seen that the velocity and 12 10 9.5 8 displacement of the particle at \u2018t\u2019 = 1 sec are 6.25 m\/s and Example 13:\u2002 The velocity at the end of the first 6 seconds is 8.3 m respectively. Then the displacement at time t = 2 sec is (A)\t 18 m\/s \t(B)\t27 m\/s (C)\t 43 m\/s \t(D)\t66 m\/s (A)\t 17.3 m\t (B)\t 15.6 m (C)\t 14.8 m\t (D)\t 12.6 m","3.60\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Solution: Solution: Given a = t3 \u2013 3t2 + 5 Let AB = l, AC = l cosa dv = t3 \u2013 3t 2 + 5 Consider sliding along AC, acceleration is gcosa dt We have, Integrating, v t4 3 t3 5t c at t = 1 sec, v = 6.25 m\/s s = ut + 1 at 2 i.e., 4 3 2 = - + + Now, 6.25 = 1 -1+ 5 + c l cos \u03b1 = 0 + 1 g cos \u03b1t22 4 2 = 4.25 + c \u2234 t22 = 2l or t2 = 2l \u2234c=2 g g \u2234 v = t4 - t3 + 5t + 2 Consider sliding along AB, 4 I = 0 + 1 gt12 ds t4 2 dt = 4 - t3 + 5t + 2 t5 t4 t2 t1 = 2l 20 - 4 2 g Integrating, s= + 5 \u22c5 + 2t + c, at t = 1, s = 8.3 m \u2234 t1:t2 = 1:1. 8.3 = 1 - 1 + 5 + 2 + c, Relative Velocity 20 4 2 The motion of one body with respect to another moving 8.3 = 1 + 4.25 + c, body is known as relative motion. 20 Take the case of two bodies P and Q moving along the c = 8.3 \u2013 4.25 \u2013 0.05 = 4.05 \u2013 0.05 = 4 same straight line. The position of the bodies is specified with reference to an origin O. s = t5 - t4 + 5 \u22c5 t2 + 2t + 4 20 4 2 xP and xQ are measured from the origin O. The difference xQ - xP defines the relative position of Q with respect to P. It is denoted as s at t = 2 sec is xQ\/P = xQ - xP \u2234 xQ = xP + xQ\/P 32 16 s = 20 -4 + 10 + 4 + 4 Consider the rate of change of displacement, then 32 Q P Q 20 x xP x x = + 14 = 15.6 m. xQ Example 17:\u2002 In the figure shown, AB is the diameter \u2018d\u2019 of vQ\/P = vQ - vP the circle and AC is the chord of the same circle? \u2234 vQ = vP + vQ\/P Similar relations hold good for acceleration also, i.e., A \u2234 aQ = aP + aQ\/P Working rule: Let two particles A and B move with veloci- a ties v1 m\/s and v2 m\/s respectively in directions as shown in the following figure. C vA = v1 B Making an angle \u03b1 with AB. Two particles are dropped from vB = v2 m\/sec rest one along AB and the other along AC. If t1 is the time If we want to find out the velocity of A relative to B, the taken by the particle to slide along AB and t2 is the time velocity of B is to be made zero. For that we provide veloc- taken to slide along AC, then t1: t2 is ity v2 in the reverse direction of OB and find the vector sum with v1 = OA. (A)\t1:cosa \t (B)\t 1:seca (C)\t1:1\t (D)\t1:15","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.61 D A Method II: v1 1 v1 C C O B AN B\u2032 V v 30\u00b0 The vector\u2009OD \u2009gives both the magnitude and direction of 30 km\/h E the velocity of A relative to B. WO Another method is to resolve their velocities into their components with sign. Then evaluate the relative velocity 45\u00b0 in the x-direction and in the y-direction. Find their result- B ant vector. This vector will be the relative velocity, both in 40 km\/h magnitude and in direction. The vector\u2009OC \u2009is the resultant velocity vector. Velocity of B is reversed and considered. Therefore the resultant is the velocity of A relative to B. Example 18:\u2002 Two boats start from a point at the same OC = 402 + 302 + 2 \u00d7 40 \u00d7 30 \u00d7 cos 75\u00b0 time. The boat A has a velocity of 30 km\/h and move in the direction N 30\u00b0 W. The boat moves in the south west = 1600 + 900 + 40 \u00d7 60 \u00d7 0.258 = 55.86 m\/s direction with a velocity of 40 km\/h. The distance between Relative distance after half and hour the boats after half an hour is (A)\t 27.9 km\t (B)\t 32.3 km = 55.86 \u00d7 1 = 27.9 km. 2 (C)\t 36.7 km\t (D)\t 42.3 km Solution: Example 19:\u2002 A vessel which can steam in still water with Method 1: a velocity of 48 km\/h is steaming with its bow pointing due 30 km\/h N east. It is carried by a current which flows northward with a 30\u00b0 speed of 14 km\/h. The distance it would travel in 12 minutes is (A)\t 14 km\t (B)\t 12 km (C)\t 10 km\t (D)\t 8 km WO E Solution: 45\u00b0 N 40 km\/h 14 km\/h 48 km\/h S E Resolving along the x-axis, 14 km\/h (vA )x = -(30 sin 30\u00b0)i for A and To find the velocity of the steamer relative to the flow, the (vB )x = -(40 cos 45\u00b0)i for B,\u2009where\u2009 i is a \u2009unit vector along flow velocity is reversed and vector sum is found. the x-axis. Relative velocity = 482 + 142 = 50 km\/h (vA\/B)x = (vA)x - (vB)x = \u2013(30 sin 30\u00b0)i \u2013 (\u201340 cos 45\u00b0)i = \u239b 40 - 15\u23a0\u239f\u239e i Distance after 12 minutes = 50 \u00d7 12 = 10 km. \u239c\u239d 2 60 = 13.28i km\/h Example 20:\u2002 A man keeps his boat at right angles to the Similarly, (vA\/B)y = (vA)y - (vB)y current and rows across a stream 0.25 km broad. He reaches = (30 cos30\u00b0) j - (- 40sin45\u00b0) j, the opposite bank 0.125 km below the point opposite to the Where\u2009 j \u2009is a unit vector along the y-axis. starting point. If the speed of the boat in rowing alone is 6\u00a0km\/ph, the speed of the current is \u239b 40 + 30 3\u239e j = 54.26 j km\/h (A)\t 5 km\/h\t (B)\t 4 km\/h = \u239c\u239d 2 2 \u239f\u23a0 (C)\t 3 km\/h\t (D)\t 2 km\/h vA\/ B = 54.262 + 13.282 = 55.86 km\/h Solution: The speed responsible for reaching the opposite side is the The relative distance after half an hour = 55.86 \u00d7 1 = 27.9 km. rowing velocity 6 km\/h. Due to the velocity of the current 2 by the time the boat can cross the stream with its absolute","3.62\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics velocity, the boat flows down 0.125 km due to the speed of The swimmer must swim in the direction OA with velocity the current. 2.5 m\/s so that he can cross the stream at right angles. Time for crossing = 0.25 = 0.04166 hr From geometry 2.5 sin\u03b8 = 1.5 6 \u2234 sin \u03b8 = 1.5 = 0.6 Let the stream velocity be v m\/s. 2.5 \u2234 Resultant speed = v2 + 62 \u03b8 = 36.8\u00b0. A 0.125 km Example 23:\u2002 An aeroplane is flying in a horizontal direction with a velocity of 1800 km\/h. At a height of 1960 metres, when it is above a point A on the ground, a body is 0.25 km dropped from it. If the body strikes the ground at point B, then the distance AB is O (A)\t 18 km\t (B)\t 15 km The distance covered by the boat within this time is (C)\t 10 km\t (D)\t 8 km OA = 0.252 + 0.1252 Solution: The time taken by the body to fall down the distance \u2234 0.04166 \u00d7 v2 + 62 = 0.252 + 0.1252 1960 m is \u2234 v = 3 km\/h. h = 1 gt 2 2 Example 21:\u2002 A boat of weight 45 kg is initially at rest. A 1960 = 1 9.8 t2 2 boy of 32 kg is standing on it. If he jumps horizontally with a speed of 2 m\/s relative to the boat, the speed of the boat is 2 \u00d7 1960 = t2 9.8 (A)\t 2 m\/s \t (B)\t 3.42 m\/s (C)\t 4.92 m\/s\t (D)\t 5.36 m\/s 400 = t2; t = 20 sec Solution: AB = v \u00d7t = 1800 \u00d7 20 = 10 km Given vA\/B = 2 m\/s 60 \u00d7 60 It is relative velocity of the boy with respect to the boat. Example 24:\u2002 Two ships leave the port at the same time. The vA\/B = vA - vB 2 = vA - vB first ship A steams north-west at 32 km\/h and the second \u2234 vA = 2 + vB ship B 40\u00b0 south of west at 24 km\/h. The time after which By conservation of momentum, they will be 160 km apart is 0 = 32 (2 + vB) \u2013 45 vB = 64 \u2013 13 vB (A)\t 2.15 hrs\t (B)\t 2.86 hrs \u2234 vB = 4.92 m\/s. (C)\t 3.46 hrs\t (D)\t 4.19 hrs Example 22:\u2002 A stream of water flows with velocity of 1.5 km\/h. A swimmer swims in still water with a velocity Solution: of 2.5 km\/h. If the breadth of the stream is 0.5 km, the Let us find the velocity of the second ship relative to the first. direction in which the swimmer should swim so that he can For that consider the velocity of the first ship in the reverse cross the stream perpendicularly is direction and evaluate the vector sum of the velocities. (A)\t26\u00b0 with the vertical (B)\t29.4\u00b0 with the vertical Resultant or velocity of B relative to A is (C)\t32.5\u00b0 with the vertical (D)\t36.8\u00b0 with the vertical = 242 + 322 + 2 \u00d7 32 \u00d7 24 cos 95\u00b0 = 1466 = 38.3 km\/h 32 km\/h N Solution: A 45\u00b0 A WO E 2.5 km\/h 40\u00b0 1.5 km\/h B 95\u00b0 S 24 km\/h 0.5 km Time for two ships to be 160 km apart = 160 = 4.19 hrs. q 38.3 1.5 km\/h O","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.63 Example 25:\u2002 A particle is accelerated from (1, 2, 3), where Example 27:\u2002 The acceleration due to gravity on a planet is it is at rest, according to the equation a = 6ti \u2013 24 t 2 j + 10 km\/s2, \u2009where\u2009 i , j and k \u2009are unit vectors along the x, y 200 cm\/s2. If it is safe to jump from a height of 2 m on earth, and z axes. The magnitude of the displacement after the then the corresponding safe height on the planet is lapse of 1 second is (A)\t 2 m\t (B)\t 9.8 m (C)\t 10 m\t (D)\t 8 m (A)\t 5 m\t (B)\t 30 m Solution: (C)\t 6 m\t (D)\t 47 m Let hse and hsp denote the safe heights on the earth and the planet. Solution: On the earth, v2 = 2ghse = 2 \u00d7 9.8 \u00d7 2 It is given that a = 6ti - 24t 2 j +10k = 39.2 m2\/s2 \u2234 v = 3t 2i \u2013 8t3 j + 10t k + c On the planet, v2 = 2 \u00d7 2 \u00d7 hsp For a safe jump the final velocity (v) should be same on when t = 0, v = 0 \u2234c=0 earth and the planet, hence, 2 \u00d7 2 \u00d7 hsp = 39.2 \u2234 hsp = 9.8 m. \u2234 v = 3t 2i \u2013 8t3 j + 10tk Example 28:\u2002 A ball weighing 500 gm is thrown vertically dx = 3t 2i - 8t3 j + 10tk upwards with a velocity of 980 cm\/s. The time that the ball dt will take to return back to earth would be x 3 t3 i 8 t4 j + 10 t2 k C (A)\t 1 s\t (B)\t 2 s 3 4 2 = - + (C)\t 3 s\t (D)\t 4 s x = t3i - 2t 4 j + 5t 2k + C Solution: when t = 0, position of the particle is at (1, 2, 3) i.e., at t = 0, For the upward journey, u = u0 \u2013 gt x = 1i + 2 j + 3k 0 = 980 \u00d7 10-2 - 9.8 t \u2234C = 1i + 2 j + 3k \u21d2\u2003 t = 1 s \u2234 x = t3i - 2t 4 j + 5t 2k + 1i + 2 j + 3k v2 - u2 = 2gs\u2003 \u21d2\u2003 0 \u2013 9.82 = -2 \u00d7 9.8 s = (t3 + 1)i - (2 - 2t 4 ) j + (3 + 5t 2 )k s = 4.9 m For the downward journey, When t = 1, s = ut + 1 gt 2 x = 2i + 8k 2 \\\\ Displacement vector 4.9 = 0 + 1 \u00d7 9.8t 2 2 = 2i + 8k - (1i + 2 j + 3k ) = 1i - 2 j + 5k t=1s Total time taken to return back to earth = 1 + 1 = 2 s. Magnitude of the displacement vector Kinetics of a Particle = 1+ 4 + 25 = 30 m. Kinetics can be used to predict a particle\u2019s motion, given a set of forces (acting upon the particle) or to determine Example 26:\u2002 If a particle, moving with uniform accelera\u00ad the forces necessary to produce a particular motion of the particle. Kinetics of the rectilinear motion of a particle are tion, travels the distances of 8 and 9 cms in the 5th sec and governed mainly by Newton\u2019s three laws of motion. 9th sec respectively, then its acceleration will be Newton\u2019s first law: Every body continues in its state of rest or of uniform motion in a straight line, unless it is com- (A)\t 1 cm\/s2\t (B)\t 5 cm\/s2 pelled to change that state by forces impressed upon it. This law is sometimes called as the law of inertia. (C)\t 25 cm\/s2\t (D)\t 0.5 cm\/s2 From Newton\u2019s first law, it follows that any change in the Solution: velocity of a particle is the result of a force. The question, of the relationship between this change in the velocity of the s in the nth sec = u+ a (2n - 1) particle and the force that produces it, is answered by the 2 second law of motion which is as follows. 8 = u + a (2 \u00d7 5 - 1) = u + 4.5a \b(1) Newton\u2019s second law: The acceleration of a given particle 2 is proportional to the force applied to it and takes place in the direction of the straight line in which the force acts. 9 = u + a (2 \u00d7 9 - 1) = u + 8.5a \b(2) 2 Subtracting Eq. (1) from Eq. (2), 1 = 4a\u2003or\u2003a = 0.25 cm\/s2.","3.64\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Newton\u2019s third law: Every action there is always an equal force subjected to a constant acceleration. Let us consider and opposite reaction, or the forces of two bodies on each other are always equal and directed in opposite directions. a particle moving along the x-axis (see figure) where the initial (at t = 0) displacement and velocity of the particle is x0 and\u2009 x0 \u2009respectively. General Equation of Motion for a Particle X0 CF D From Newton\u2019s second law, the relationship between the OX X acceleration \u2018a\u2019 produced in a body of mass \u2018m\u2019 (mass is always assumed to be invariant with time) by a resultant, If F is the magnitude of the constant force acting on the \u2018F\u2019, of all the forces acting on the body can be derived as follows: F = ma, which is the general equation of motion particle, then from the differential equation of rectilinear for a particle. motion,\u2009 x = F = a, \u2009where a is the constant acceleration pro- For a stationary body lying on a surface (body with no m motion), there is a force (F) exerted by the body on the surface which is equal to the weight of the body (W), i.e., duced in the particle due to the constant force. The equation F = W = mg, where \u2018m\u2019 is the mass of the body and \u2018g\u2019 is the acceleration due to gravity. There is an equal and oppo- x = a \u2009can be written as d(x) = a. site force exerted by the surface on the body (consequence dt of Newton\u2019s third law). Note that the weight of a body is Integration of the above equation with the initial value obtained by multiplying the mass of the body by the accel- eration due to gravity. condition, at t = 0 x = x0 ,\u2009gives: x = x0 + at \b Which is the general velocity-time equation for the recti- linear motion of a particle under the action of a constant Differential Equation of Rectilinear Motion force F producing the constant acceleration a in the parti- The general equation of motion for a particle can be applied cle. With x = dx , equation (1) can be rewritten as follows: dt directly to the case of the rectilinear translation of a rigid dx dt = x0 + at. body, since all the particles of the rigid body have the same velocity and acceleration (same motion) where the particles Integration of the above equation with the initial value move in parallel straight lines. Here, the rigid body is con- condition, at t = 0, x = x0, gives:\u2009 x = x0 + x0t + 1 at 2 ,\u2009which 2 sidered as a particle concentrated at the center of gravity of the rigid body. is the general displacement-time equation for the rectilinear Whenever such a body or particle moves under the action motion of a particle under the action of a constant force F of a force applied at its center of gravity and having a fixed producing the constant acceleration a in the particle. line of action, acceleration of the body is produced in the Freely falling body same direction, and if any initial velocity of the body is also The force acting on a freely falling body is the weight of the body (assuming no friction in the motion) and therefore the directed along this line, then the motion corresponding to acceleration produced in the body is the acceleration due to gravity, i.e., F = W = mg and \\\\ a = g. Hence, the velocity- this case is known as rectilinear translation. time and displacement-time equations for a freely falling body are respectively as follows: If the line of motion of a particle is taken to be along the x-axis (i.e., displacement at a time t is denoted by x), x = d2x \u2009represents the acceleration and F represents the dt 2 resultant force acting, then the differential equation of the x = x0 + gt rectilinear motion of the particle is given by\u2009F = mx. Two types of problems that can be solved by the above 1 gt 2 2 equation are: (a) Determination of the force necessary to x = x0 + x0t + produce a given motion of the particle where the displace- If the freely falling body starts to fall from a resting position, i.e., it has an initial velocity of zero (x(0) = 0) \u2009and if the ori- ment x is given as a function of time t, (b) Determination of gin of displacement of the body is taken to coincide with the initial position of the body, i.e., it has an initial displacement the motion of a particle given a force F acting on the parti- of zero (x0 = 0), then the above equations reduce to: cle, i.e., to determine a function relating x and t such that the above equation is satisfied. Motion of a Particle Acted upon x = gt by a Constant Force x = 1 gt 2 A particle, acted upon by a force of constant magnitude 2 and direction, will move rectilinearly in the direction of the","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.65 Force as a Function of Time a downward acceleration\u2009x \u2009of the weight W1 is obtained. The inertia forces acting on the corresponding weights are If the force acting on the particle is a function of time t, i.e., shown in the above figure. the acting force = F(t), then the acceleration a(t), velocity By adding the inertia forces to the real forces such as x(t) and displacement x(t) of the particle at time t (with ini- (W1 and W2 and the string reactions S), a system of forces in equilibrium is obtained for each particle. Hence, the entire tial time, t = 0) is given by the following respective equations. system of forces can be concluded to be in equilibrium. a(t ) = F (t ) An equation of equilibrium can be written for the entire m system (instead of separate equilibrium equations for the individual weights) by equating to zero the algebraic sum of t moments of all the forces (including the inertia forces) with respect to the axis of the pulley or by using the principle x(t) = \u222b a(t) dt of virtual work. In either case, the internal forces \u2018S\u2019 of the 0 system need not be considered and the following equation t of equilibrium can be obtained for the entire system. x (t) = \u222b x(t) dt 0 Dynamics of a Particle W2 + m2 x = W1 - m1x or x = \u239b W1 - W2 \u239e g \u239c\u239d W1 + W2 \u239f\u23a0 D\u2019Alembert\u2019s Principle Momentum and Impulse Let \u03a3Fi, where Fi denotes the ith force, be the resultant of a set of forces acting on a particle in the x-axis direction. The differential equation of the rectilinear motion of a par- From the 0 differential equation of the rectilinear motion of ticle may be written as a particle, we have m dx = F or d(mx) = Fdt \b(1) \u2211 Fi - mx = 0 or dt \u2211 Fi + (-mx) = 0 X From the above equation, it can be seen that if a fictitious force\u2009(-mx) \u2009is added to the system of forces acting on X the particle, then an equation resembling equilibrium is BC obtained. The force\u2009(-mx) \u2009which has the same magnitude as\u2009mx \u2009but opposite in direction is called as the inertia force. t Hence, it can be seen that if an inertia force is added to the system of forces acting on a particle, then the particle is 0 dt D t brought into an equilibrium state called as dynamic equi- librium. This is called as D\u2019Alembert\u2019s principle. The above It is assumed that the force \u2018F\u2019 is known as a function of equation thus represents the equation of dynamic equilib- rium for the rectilinear translation of a rigid body. time and is given by the force-time diagram as shown in the Let us consider, now any system of particles connected above figure. The right hand side of equation (1) is then rep- between them and so constrained that each particle can have only a rectilinear motion. To exemplify such a system, resented by the area of the shaded elemental strip of height the case of two weights W1 and W2 attached to the ends of a flexible but inextensible string overhanging a pulley, as F and width dt in the force-time diagram. This quantity is shown in the figure below, is considered. called as the impulse of the force F in the time interval dt. The expression\u2009mx \u2009on the left hand side of the equation is called as the momentum of the particle. The equation states that the differential change of the momentum of the par- S S ticle during the time interval dt is equal to the impulse of m2 m1 the acting force during the same time interval. Impulse and momentum have the same dimensions of the product of W1 mass and velocity. t m1 X W2 Integrating equation (1), we get mx - mx0 = \u222b Fdt , m2 X 0 The inertia of the pulley and the friction on its axle are where\u2009 x0 \u2009is the velocity of the particle at time t = 0. assumed to be negligible. If the motion of the system is Thus the total change in the momentum of a particle assumed to be in the direction as shown by the arrow on the pulley, an upward acceleration\u2009x \u2009of the weight W2 and during a finite time interval is equal to the impulse of the acting force during the same time interval. This impulse is","3.66\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics represented by the area OBCD of the force-time diagram. create their own moment of force. The net moment of the The equation of momentum-impulse is particularly useful couple is independent of the location of the point considered. when dealing with a system of particles, since in such cases the calculation of the impulse can often be eliminated. As Moment of couple = Force \u00d7 Perpendicular distance a specific example, consider the case of a gun and shell as between the forces. shown in the figure below, which may be considered F Fx F\u2003\u2003F V2 Moment of couple = F\u2009.\u2009x V1 \u2022\u2022 Moment is the measure of the turning effect produced by a as a system of two particles. During the extremely short force about a point. Couple consists of two forces, equal and interval of explosion, the forces \u2018F\u2019 acting on the shell and opposite, acting in two different but parallel lines of action. gun and representing the gas pressure in the barrel are vary- ing in an unknown manner and a calculation of the impulses \u2022\u2022 Moment of a couple is independent of the location of the of these forces would be extremely difficult. pivot or point considered. However, the relation between the velocity of the shell Work and Energy and velocity of recoil of the gun can be obtained without calculation of the impulse. Since the forces \u2018F\u2019 are in the The differential equation of the rectilinear motion of a par- nature of action and reaction between the shell and gun, ticle can be written in the following form: they must at all times be equal and opposite, and hence their impulses for the interval of explosion are equal and oppo- m dx = F site since the forces act exactly the same time \u2018t\u2019. dt Let m1 and m2 be the masses of the shell and gun respec- Multiplying both sides of the above equation by\u2009 x \u2009and with tively. If the initial velocities of the shell and gun are assumed suitable modifications, the above equation can be written as to be zero and if the external forces are neglected, then follows: m1v1 m2v2 , i.e., v2 m1 d \u239b mx2 \u239e = Fdx \b(2) v1 m2 \u239d\u239c 2 \u23a0\u239f = = The velocities of the shell and gun after discharge are in It is assumed that the force F is known as a function of opposite directions and inversely proportional to the cor- the displacement x of the particle and is represented by the responding masses. Internal forces in a system of parti- below force-displacement diagram. cles always appear as pairs of equal and opposite forces and need not be considered when applying the equation of BC momentum and impulse. Thus it may be stated that, for a system of particles on which no external forces are applied, h the momentum of the system remains unchanged, since the total impulse is zero. This is sometimes called as the princi- A DX ple of conservation of momentum. X0 Moment and Couple dx Moment or moment of a force is the turning effect caused by The right hand side of equation (2) is represented by the the force. It is the force acting at a perpendicular distance d area of the elemental strip of the height h and width dx in the above figure. This quantity represents the work done Moment of a force = Force \u00d7 Perpendicular distance. by the force F on the infinitesimal displacement dx, and the expression in the parenthesis on the left hand side of x equation (2) is called the as kinetic energy of the particle. F Equation (2) thus states that the differential change in the kinetic energy of a moving particle is equal to the work \u2022 done by the acting force on the corresponding infinitesimal displacement dx. Work and kinetic energy have the same Moment = F \u00d7 x. dimensions of the product of force and length. They are usu- ally expressed in the unit of Joules (J). Couple Tow equal and opposite forces with separate lines of action present in a system of forces constitute a couple. Both forces","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.67 Integrating equation (2), with the assumption that the In such a case, the resultant acting force, in the direction velocity of the particle is\u2009xo when the displacement is x0, we have of motion F = Wsina - mWcosa. Then through the displacement\u2009\u239c\u239b\u239d sinh \u03b1 \u239e \u2009between the \u239f\u23a0 \u222bmx2 mx02 x - 2 = \b(3) points A and B, the work done is\u2009 = Wh - mWh cot \u03b1 Equation 2 Fdx (3) would then yield x0 The definite integral on the right hand side of the above v = 2gh(1- mcot (\u03b1)) equation is represented by the area ABCD of the force-dis- placement diagram. This is the total work of the force F When \u03b1= \u03c0 , the above equation agrees with the velocity on the finite displacement of the particle from x0 to x. The 2 work of a force is considered positive if the force acts in the equation derived for a freely falling body and when m\u00a0= 0, direction of the displacement and negative if it acts in the opposite direction. The total change in the kinetic energy of the above equation agrees with the velocity equation derived a particle during a displacement from x0 to x is equal to the work of the acting force on the displacement. for the inclined plane motion of the body with no friction. The equation of work and energy is especially useful Also from the above equation, it can be noted that to obtain in cases where the acting force is a function of displace- ment and where the velocity of the particle as a function of a real value for the velocity, m < tan \u03b1, \u2009otherwise the block displacement is of interest. For example, the velocity with would not slide down. which a weight W falling from a height h strikes the ground is to be determined. In this case, the acting force F = W and Work done by Torque the total work is Wh. Thus if the body starts from rest, the initial velocity x0 = 0\u2009and hence equation (3) becomes Consider a light rod of length l pin joined at one end and is turned by an angle q by the force F from the position A to mx2 = Wh \b(4) B. Work done by the constant torque is the product of the 2 torque and the angle turned by the rod. \u2234 Work done = F\u2009.\u2009s. = F\u2009.\u2009r\u2009.\u2009q = T\u2009.\u2009q B S Which yields\u2009 x = v = 2gh. F Let the same body slide, without friction, along an ql A inclined plane AB starting from an elevation h above point B as shown in the figure below. O A W sina Work Energy Formulations h \u2022\u2022 Kinetic energy of a body\/particle in translation = 1 mv 2. 2 Wa \u2022\u2022 Kinetic energy of a body\/particle in rotation and rotating B 1 2 The equation of work and energy can be used to determine about a point = IW 2. the velocity of the body when it reaches point B. Here only \u2022\u2022 Work Energy principle for a body\/particle in translation. the component W sin a of the gravity force does work on Work done on body\/particle between points 1 and 2 is x2 the displacement and the component perpendicular to the \u222b \u2211W1-2 = Fx dx. inclined plane is at all times balanced by the reaction of x1 the plane. In short, the resultant of all the forces acting Y on the body is F = Wsina in the direction of motion, and F2 this force acts through the distance \u239bh \u03b1 \u239e . The work F1 t1 V1 t2 of the force acting on the body is = \u239d\u239c sin \u23a0\u239f = Wh 1W h \u03bcR 2 W sin \u03b1 \u00d7 sin \u03b1 \u2022 X1 \u2022 V2 and hence velocity at the point B (derived from equation (0,0) R \u2022X X2 (4)), v = 2gh. \u2009Hence, the velocity is the same as that gained in a free fall through the height h. Change in kinetic energy from the positions 1 to 2 is If (is the coefficient of friction between the block and the 1 2 inclined plane, then the work of friction has to be consid- (DK \u22c5 E )1- 2 = m (v22 - v12 ). ered in equation (3).","3.68\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics \u222b \u2211 ( )x2 1 1.5 252 468.75 Nm 2 2 \u2234W1\u20132 = x1 Fxdx = m v22 \u2013 v12 . = \u00d7 = Work energy principle for a body\/particle in rotation. Total work done = Work done to pull the bucket + work done to pull the rope IO 1 O (q1 \u2212 q2) = 15 \u00d7 25 + 468.75 = 843.75 Nm. \u2022 Example 30:\u2002 A uniform chain of length 10 m and mass w1 q = q1 100 kg is lying on a smooth table such that one third of its length is hanging vertically down over the edge of the table. IO \u2022 q = q2 If g is the acceleration due to gravity, then the work required 2 to pull the hanging part of the chain is (A)\t50g\t (B)\t55.55g\t (C)\t100g\t (D)\t150g w2 Solution: Work done from 1 to 2 is given by. Work done = potential energy change in the raising of the L \u03b82 centre of mass over the distance 6 . \u222b \u2211W1-2 = MO d\u03b8. m L 100 \u00d7 g \u00d7 10 1000 g 3 6 18 18 \u03b81 = g = = = 55.55g. Change in kinetic energy from 1 to 2 is ( )KE1\u20132 1 Alternate method: = 2 IO \u03c912 - \u03c922 \u03b82 1 dx 2 \u222b \u2211 ( )\u2234 Work done W1\u20132 = MO d\u03b8 = IO \u03c912 - \u03c922 NOTES \u03b81 1.\u2002\u0007Work done by a force is zero if either displacement \u222bW = L x dx = mg \u239b L = mg \u00d7 L2 = mg \u00d7 L . is zero or the force acts normal to the displacement L \u239d\u239c L 18 18 for example, gravity force does no work when a body 3 mg x2 \u239e 3 moves horizontally. 0L 2 \u23a0\u239f 0 2.\u2002\u0007Work done by a force is positive if the direction of force and the direction of displacement are same. For Ideal Systems: Conservation of Energy example, work done by force of gravity is positive when a body moves from a higher elevation to a lower m2 x1 elevation. A position work can be said as the work m3 O m1 done by a force and negative work as the work done against a force. a 3.\u2002\u0007Work is a scalar quantity and has magnitude but no direction. 4.\u2002\u0007Work done by a force depends on the path over which the force moves except in the case of conservative forces. Forces due to gravity, spring force are conservative forc- es, where as friction force is a non-conservative force. Example 29:\u2002 If a bucket of water weighing 15 N is pulled up from a well of 25 m depth by a rope weighing 1.5 N\/m, The method of work and energy for a single particle can be extended to apply to a system of connected particles as then the work done is (B)\t 500 Nm shown in the above figure. In doing so, it is to be noted that (A)\t 843.75 Nm\t the attention is limited to ideal systems with one degree of freedom. That is, it is assumed that the system has friction- (C)\t 575 Nm\t (D)\t 600 Nm less constraints and inextensible connections and that its configuration can be completely specified by one coordi- Solution: nate such as x1 in the below figure. In the case shown in the above figure, for example, the assumptions involve a The work done to pull the rope smooth inclined plane, frictionless bearings, inextensible 25 \u222b= 1.5 \u00d7 (25 - h) dh \u2009(h is the tip of the rope from the bottom 0 of the well)","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.69 strings and neglecting entirely the rotational inertia of the NOTE pulleys. Then the system may be regarded simply as three particles, m1, m2 and m3, each of which performs a rectilin- If a particle of weight w is at an elevation x above a cho- ear motion. From kinematics, the displacements and veloci- sen datum plane, then the potential energy of the particle, ties of all the three masses can be expressed in terms of one V = mx. Similarly for a system of particles at an elevation, variable, say the coordinate x1 of the particle m1. the potential energy of the system,V = \u2211 wi xi = Wxc , During motion of the system, an infinitesimal interval Where wi and xi are the weight and elevation above a chosen of time dt is considered during which the system changes datum plane for the ith particle, W is the total weight of the its configuration slightly and each particle is displaced by a system and xc is the elevation of the center of gravity of the length of dxi,\u00ad along its line of motion. If Fi\u00ad is the resultant system above the chosen datum plane. force acting on any particle mi, then the total increment of work of all the forces during such a displacement, For the system of particles moving from the configura- tion A to the configuration B, it can be shown that TB + VB \u2211dU = Fidxi \b(5) = TA + VA. For the system of particles, it can be shown that Law of Conservation of Energy dT = dU \b(6) That is, as the system moves from one configuration to another, the total energy (kinetic + potential) remains \u2211WhereT=1(mi xi2 ), T is the total kinetic energy of the constant. Kinetic energy may be transformed into potential 2 energy and vice versa but the system as a whole can neither system of particles with the mass and velocity of the ith gain nor lose energy. This is the law of conservation of energy as it applies to a system of particles with ideal constraints. particle being mi and\u2009xi \u2009respectively. Equation (6) states Such systems are sometimes called conservative systems. that the differential change in the total kinetic energy of the Impact system when it changes its configuration slightly is equal to The impact between two moving bodies refers to the col- the corresponding increment of work of all forces. lision of the two bodies that occurs in a very small time interval and during which the bodies exert a very large force Consider any two configurations of the system denoted (active and reactive force) on each other. The magnitudes of the forces and the duration of impact depend on the shapes by the subscripts A and B, then from equation (6) we have of the bodies, their velocities, and their elastic properties. TB xB Consider the impact of two spheres of masses m1 and m2 as shown in the below figure. Let the spheres have the \u222b dT = \u222b dU or respective velocities of u\u00ad1 and u2, where u1 > u2, before impact and the respective velocities of v1 and v2 after impact. TA xA xB m1 u1 u2 \u222bTB - TA = dU \b(7) x xA Before impact This is the equation of work and energy for a system of particles. It states that the total change in the kinetic energy m1 u1 m2 u2 of the system when it moves from configuration A to con- x figuration B is equal to the corresponding work of all the forces acting upon it. In the case of an ideal system, the After impact reactive forces will produce no work and work of all the internal forces which occur in equal and opposite pairs will It is assumed that these velocities are directed along the line cancel out each other. Thus for such systems, only the work joining the centers of the two spheres and are considered to of active external forces are to be considered on the right be positive if they are in the positive direction of the x-axis. hand side of equation (7). This is called the case of direct central impact. Two equal and opposite forces, i.e., action and reaction, are produced The potential energy of a system in any configuration (A at the point of contact during impact. In accordance with or B) is defined as the work which will be done by the acting the law of conservation of momentum, such forces cannot forces if the system moves from that configuration (A or B) change the momentum of the system of two balls and hence, back to a certain base or reference configuration (O). If VA and VB are the potential energies of the system in configura- tions A and B respectively, then 00 m1u1 + m2u2 = m1v1 + m2v2 \b(8) VA = \u222b dU and VB = \u222b dU AB","3.70\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Elastic Impact e = coefficient of restitution e satisfies the condition 0 \u2264 e \u2264 1. In an elastic impact, the momentum and kinetic energy is conserved. If the kinetic energy is conserved during impact, If e = 1\u2003 \u21d2\u2003 the collision is perfectly elastic. then If e = 0\u2003 \u21d2\u2003 the collision is inelastic If 0 < e < 1\u2003 \u21d2\u2003 the collision is said to be elastic. 1 m1u12 + 1 m2u22 = 1 m1v12 + 1 m2 v22 \b(9) Energy loss due to impact: The energy lost in impact 2 2 2 2 when e \u2260 1, i.e., when the collision is not perfectly elastic Since momentum is conserved, equation (8) is also applica- is given by ble in this type of impact. From equations (8) and (9), it can be shown that Loss in kinetic energy = 1 m1m2 (u1 - u2 )2 (1 - e2 ). 2 m1 + m2 v1 - v2 = - (u1 - u2)\b (10) \u2234 When e = 1 the loss is zero. This equation represents a combination of the law of con- Coefficient of restitution: It is defined as the ratio of the servation of momentum and conservation of energy. It relative velocity of the impacting bodies after impact to states that for an elastic impact the relative velocity after their relative velocity before impact. The coefficient of res- impact has the same magnitude as that before impact but titution \u2018e\u2019 is given by the following equation. with reversed sign. e = (v2 - v1) For two bodies of equal masses undergoing an elastic (u1 - u2 ) impact, from equations (8) and (10) it can be shown that they will exchange their velocities, i.e., v1 = u2 and v2 = u1. If Example 31:\u2002 A bullet travelling with a velocity of 800 m\/s the second body was at rest before the impact, i.e., u2 = 0, and weighing 0.25 N strikes a wooden block of weight 50 then it would seem that the striking body stops, i.e., v1 = Nresting on a horizontal floor. The coefficient of friction 0, after having imparted its velocity to the other ball. This between floor and the block is 0.5. Determine the distance phenomenon can be observed in the case of a moving billiard through which the block is displaced from its initial position. ball which squarely strikes one that was at rest. Again, if the two balls were moving toward each other with equal speeds Solution:\u2002 Velocity of the bullet before impact, va = 800 m\/s before impact, an exchange of velocities will simply mean Velocity of the block before impact, vb = 0 m\/s that they rebound from one another with the same speed with which they collided. As another special case, we assume that Mass of the bullet,\u2009 ma = 0.25 kg m2 = \u221e while m1 remains finite and further u2 = 0. This will g represent the case of an elastic impact of a ball against a flat immovable obstruction, such as the dropping of a ball on a Mass of the block,\u2009 mb = 50 kg cement floor. In this case, it is obtained that v1 = -u1, i.e., the g striking ball rebounds with the same speed with which it hits the obstruction. The bullet after striking the block remains buried in the block and both move with a common velocity v. Plastic or Inelastic Impact Applying the principle of conservation of momentum, In a plastic or inelastic impact, the momentum is conserved but the kinetic energy is not (part of the kinetic energy is mava + mbvb = (ma + mb)v converted to a different form of energy). In a perfectly plas- tic impact, the colliding bodies will stick to each other after 0.25 \u00d7 800 + 50 \u00d7 0 = \u239b 0.25 + 50 \u239e v collision and will move with a common velocity. If v is the g g \u239d\u239c g g \u239f\u23a0 common velocity of two colliding bodies after a perfectly plastic impact, then from equation (8), we have v = 3.98 m\/s v = m1u1 + m2u2 To find the distance travelled by the block, apply the prini- m1 + m2 ciple of work and energy. Kinetic energy lost by the block with the bullet buried = Work done to overcome the fric- tional force If s is the distance travelled by the block, then Newton\u2019s experimental law of colliding bodies: Newton pro- 1 \u2009(ma + mb)v2 = mR s posed an experimental law that describes how the impact of 2 moving bodies was related to their velocities and found that: = mg(ma + mb) s (\u2234 R = g(ma + mb)) Speed of separation = e \u2234s = 3.982 = 1.61 m Speed of approach 2 \u00d7 9.81\u00d7 0.5","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.71 Example 32:\u2002 Three spherical balls of masses 3 kg, 9 kg e = - vc\u2032 - vb\u2032 and 12 kg are moving in the same direction with velocities vc - vb 12 m\/s, 4 m\/s and 2 m\/s respectively. If the ball of mass 3 kg impinges with the ball of mass 9 kg which in turn impinges v\u2032c - v\u2032b = e(vb \u2013 vc) with the ball of mass 12 kg. Prove that the balls of masses 3 kg and 9 kg will be brought to rest by the impacts. Assume = 1 \u00d7 (8 - 2) = 6\b (4) the balls to be perfectly elastic. Solving Eqs. (3) and (4), we get v\u2032c = 6 m\/s and v\u2032b = 0 m\/s, 2 m\/s i.e., the ball of mass 9 kg is brought to rest. 4 m\/s Direction for questions 33 and 34: The blocks 1 and 2, hav- ing a weight of 1 kg each and the respective velocities of 10 12 m\/s m\/s and 4 m\/s, undergo a perfect inelastic collision. 3 kg 9 kg 12 kg Example 33:\u2002 The final velocity of the blocks is Solution:\u2002 For perfectly elastic balls, e = 1 (A)\t 7 m\/sec\t (B)\t 6 m\/sec ma = 3 kg, mb = 9 kg, mc = 12 kg (C)\t 3 m\/sec\t (D)\t 4 m\/sec Impact of balls A and B: Conservation of momentum gives, Solution: V = M1V1 + M2V2 = 1 \u00d7 10 + 4 \u00d7 1 = 7 m\/s. m1 + m2 1+1 mava + mbvb = mav\u2032a + mbv\u2032b 3 \u00d7 12 + 9 \u00d7 4 = 3v\u2032a + 9v\u2032b\b(1) Example 34:\u2002 The energy converted into heat as a result of vb\u2032 va\u2032 e = - vb - va the collision is - (A)\t 40 J\t (B)\t 9 J v\u2032b - v\u2032a = e(va - vb) = 1 \u00d7 (12 - 4) = 8\b (2) (C)\t 50 J\t (D)\t 54 J Solving Eqs. (1) and (2), we get v\u2032b = 8 m\/s and v\u2032a = 0 m\/s, Solution: i.e., the ball of mass 3 kg is brought to rest. The original kinetic energy was, Impact of balls B and C: Consider now the impact of the K1 = 1 \u00d7 1 \u00d7 100 + 1 \u00d7 1 \u00d7 16 = 58 J ball B, of mass 9 kg and moving with the initial velocity of 2 2 8 m\/s, with the ball C, of mass 12 kg and moving with the velocity of 2 m\/s. The final kinetic energy is, Conservation of momentum gives K2 = 1 \u00d7 2\u00d7 49 = 49 J 2 mbvb + mcvc = mbv\u2032b + mcv\u2032c 9 \u00d7 8 + 12 \u00d7 2 = 9v\u2032b + 12v\u2032c\b(3) Loss of Kinetic energy = 58 - 49 = 9 J (converted to heat energy). Exercises Practice Problems 1 \t 3. \tA block is made to slide down an inclined plane which Direction for questions 1 to 10:\u2002 Select the correct alterna- is smooth. It starts sliding from rest and takes a time of tive from the given choices. t to reach the bottom of the plane. An identical body is \t1.\tA car starts with an acceleration of 2 m\/s2. Another car freely dropped from the same point. The time the body starts from the same point after 5 seconds and chases takes to reach the bottom of the plane is the first car with a uniform velocity of 20 m\/s. The time \t (A)\tt\t (B)\t t 2 at which the second car, after it starts, will overtake the \t(C)\tt \t(D)\tt first car is 34 \t (A)\t 5 sec\t (B)\t 7 sec \t (C)\t 9 sec\t (D)\t 11 sec \t4.\tA stone is dropped into a well. The sound of the splash \t2.\tA body is moving with uniform acceleration. In the 4th is heard 3.63 seconds later. Assume the velocity of the second of its travel it covers 20 m and 30 m in the 8th sound to be 331 m\/s. The depth of the surface of water second. The distance travelled at the 10th second is from the ground is \t (A)\t 24 m\t (B)\t 35 m \t (A)\t 46.38 m \t (B)\t 51.36 m \t (C)\t 43 m \t (D)\t 52 m \t (C)\t 58.39 m\t (D)\t 64.62 m","3.72\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics \t5.\tA motor cycle starts from rest from point A, 2 seconds \t8.\t A baggage truck pulls two carts A and B. If the mass of the truck is 400 kg and the carts A and B carry 800 and after a car, speeding at a constant velocity of 120 km\/h, 400 kg respectively, and the truck develops a tractive force of 2 kN. The horizontal forces between the truck passes that point. The motor cycle accelerates at a rate and the cart A and between the two carts, respectively, are of 6 m\/s2 until the motor cycle attains a maximum \t (A)\t 1200 N and 400 N\t (B)\t 1000 N and 450 N speed of 150 km\/h. The distance from the starting \t (C)\t 1500 N and 500 N\t (D)\t 500 N and 500 N point to the point at which the motor cycle overtakes the car is \t (A)\t 912 m\t (B)\t 1024 m \t (C)\t 1286 m\t (D)\t 1356 m \t9.\tA body of weight 200 N is placed on a rough horizontal \t 6.\t A rail road coal car of tare weight mo is moving at a plane. The coefficient of friction, if a horizontal force constant speed v while being loaded with coal at a con- stant rate of w per second. The force necessary to sus- of 80 N just causes the body to slide over the horizontal tain the constant speed, neglecting friction, is plane, is \t(A)\tw2v\t(B)\twv \t (A)\t0.6\t (B)\t0.1 \t(C)\tw22v \t(D)\tw2v2 \t (C)\t 0.2\t (D)\t0.4 \t10.\t A body of weight 400 N is pulled up along an inclined plane having an inclination of 30\u00b0 to the horizontal at \t7.\t A 10 kg shell is fired with a velocity of 800 m\/s at an a steady speed. The pulling force applied on the body angle of 30\u00b0 from an old 2000 kg gun. Assuming that is parallel to the inclined plane. The coefficient of fric- barrel and frame can recoil freely, the reaction of the tion between the body and the plane is 0.25. If the dis- gun, if the shell leaves the barrel 10 milliseconds after tance travelled by the body is 10 m along the plane, firing, is then the work done on the body is \t (A)\t 400 kN\t (B)\t 450 kN \t (A)\t 3412 J\t (B)\t 2866 J \t (C)\t 600 kN\t (D)\t 550 kN \t (C)\t 1002 J\t (D)\t 4956 J Practice Problems 2 \t 5.\t An aircraft is flying at an elevation of 1500 m above Direction for questions 1 to 10:\u2002 Select the correct alterna- the ground horizontally. The velocity is 100 km\/h, tive from the given choices. horizontal and uniform. The aircraft releases a bomb \t 1.\t A boat goes 30 km down the stream in 75 minutes and at this elevation. If the target on the ground was just the same distance up the stream in 90 minutes. The below the plane at the time of releasing the bomb, the speed of the stream is distance away from the target, the bomb will hit the \t (A)\t 0.8 km\/h\t (B)\t 1.2 km\/h ground is \t (C)\t 1.6 km\/h\t (D)\t 2 km\/h \t (A)\t 2.35 km\t (B)\t 3.42 km \t 2.\t The motion of a body is explained by the equation: s = \t (C)\t 4.86 km\t (D)\t 5.32 km t3 \u2013 3t2 \u2013 9t + 12, where s is the displacement in metres Direction for questions 6 and 7:\u2002 A pile of mass 400 kg is driven by a distance of d into the ground by the blow of a at any time t in seconds. The acceleration of the particle hammer of mass 800 kg through a height of h onto the top when its velocity is zero is of the pile. Assume the impact between the hammer and pile \t (A)\t 4.5 m\/s2\t (B)\t 6.2 m\/s2 to be plastic. \t (C)\t 8 m\/s2\t (D)\t 12 m\/s2 Direction for questions 3 and 4:\u2002There are three marks M1 along a straight road at a distance of 100 m. A vehicle h starting from rest and accelerating uniformly passes the 2 first mark (P) and takes 10 seconds to reach the second \u03b4 mark (Q). Further it takes 8 seconds to reach the third 3 mark\u00a0(R). \u03b4 \t 3.\t The velocity of the car at Q is Given M = 800 kg, m = 400 kg, h = 1.2 m, d = 10 cm. \t (A)\t 11.38 m\/s\t (B)\t 13.5 m\/s \t (C)\t 14.8 m\/s\t (D)\t 15.5 m\/s \t 4. \tThe distance of mark P from the starting point is \t (A)\t 218 m\t (B)\t 183 m \t (C)\t 156 m\t (D)\t 134 m","Chapter 4\u2002 \u2022\u2002 Rectilinear Motion\u2002 |\u2002 3.73 \t6.\t The work done is \t 8.\t The velocity with which the gun will recoil is \t (A)\t 5.28 kJ\t \t (C)\t 7.126 kJ\t (B)\t 6.278 kJ \t (A)\t \u20137.5 m\/s\t (B)\t \u20138.4 m\/s (D)\t 6.8 kJ \t (C)\t 9.2 m\/s\t (D)\t 10 m\/s \t 7.\t The kinetic energy of the whole system in the position \t9.\tThe uniform force required to stop the gun in 0.6 m is 3 is \t (A)\t 55310 N\t (B)\t 46875 N \t (A)\t 0 J\t (B)\t 10 J \t (C)\t 55475 N\t (D)\t 82750 N \t (C)\t 100 J\t (D)\t 20 J 1\t 0.\t A tennis ball is having a velocity of 40 m\/s at an angle of 30\u00b0 with the horizontal just after being struck by the Direction for questions 8 and 9:\u2002 A gun of mass 2000 kg player. The radius of curvature of its trajectory is fires horizontally a shell of mass 50 kg with a velocity of 300 m\/s. \t (A)\t 188.2 m\t (B)\t 198.6 m \t (C)\t 200 m\t (D)\t 168.2 m Previous Years\u2019 Questions 2s 2s (tan\u03b8 \t1.\t A 1 kg mass of clay, moving with a velocity of 10 \t(A)\t g cos\u03b8(tan\u03b8 - m) \t(B)\tg cos\u03b8 + m) m\/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1m. 2s 2s \u03b8(tan\u03b8 \u03b8(tan\u03b8 Assuming that the wheel and the ground are both \t(C)\tg sin - m) \t(D)\tg sin + m) rigid and that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after [2005] \t4.\t Match the approaches given below to perform stated the impact is approximately\b kinematics\/dynamics analysis of machine.\b [2009] 20 kg 10 m\/s + Analysis Approach 1 kg 1m (P) Continuous relative (1) D\u2019 Alembert\u2019s rotation principle (Q) Velocity and (2) Grubler\u2019s criterion acceleration \t (A)\tZero\t (B)\t 1 rad\/s (R) Mobility (3) Grashoff\u2019s law 3 (S) Dynamicstatic analysis (4) Kennedy\u2019s theorem \t(C)\t130 rad\/s \t(D)\t130 rad\/s \t(A)\tP\u20131, Q\u20132, R\u20133, S\u20134\t(B)\tP\u20133, Q\u20134, R\u20132, S\u20131 \t(C)\tP\u20132, Q\u20133, R\u20134, S\u20131\t(D)\tP\u20134, Q\u20132, R\u20131, S\u20133 \t 2.\t During inelastic collision of two particles, which one \t5.\t The coefficient of restitution of a perfectly plastic of the following is conserved?\b [2007] impact is\b [2011] \t (A)\t Total linear momentum only \t (A)\t0\t (B)\t1 \t (B)\t Total kinetic energy only \t (C)\t2\t (D)\t\u221e \t (C)\t Both linear momentum and kinetic energy \t 6.\t A truck accelerates up a 10\u00b0 incline with a crate of \t (D)\t Neither linear momentum nor kinetic energy 100 kg. Value of static coefficient of friction between \t3.\t A block of mass M is released from point P on rough the crate and the truck surface is 0.3. The maximum inclined plane with inclination angle \u03b8, shown in the value of acceleration (in m\/s2) of the truck such that figure below. The coefficient of friction is \u03bc. If \u03bc < tan \u03b8, then the time taken by the block to reach another the crate does not slide down is \b [2014] point Q on the inclined plane, where PQ = s, is \b [2007] \t7.\t A mass m1 of 100 kg travelling with a uniform veloc- ity of 5 m\/s along a line collides with a stationary P mass m2 of 1000 kg. After the collision, both the g masses travel together with the same velocity. The coefficient of restitution is\b [2014] \t (A)\t0.6\t (B)\t0.1 qQ \t (C)\t 0.01\t (D)\t0","3.74\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics \t8.\t A swimmer can swim 10 km in 2 hours when swim- \t13.\t A point mass having mass M is moving with a veloc- ming along the flow of a river. While swimming ity V at an angle \u03b8 to the wall as shown in the fig- against the flow, she takes 5 hours for the same dis- ure. The mass undergoes a perfectly elastic collision tance. Her speed in still water (in km\/h) is _____. with the smooth wall and rebounds. The total change \b [2015] (final minus initial) in the momentum of the mass is: \b [2016] \t 9.\t A ball of mass 0.1 kg, initially at rest, is dropped from y, j^ height of 1 m. Ball hits the ground and bounces off the V ground. Upon impact with the ground, the velocity \u03b8 reduces by 20%. The height (in m) to which the ball x, i^ will rise is _____.\b [2015] \t (A)\t\u20132MV cos \u03b8 \u02c6j \t (B)\t2MV sin \u03b8 \u02c6j 10.\t A small ball of mass 1 kg moving with a velocity \t (C)\t2MV cos \u03b8 \u02c6j \t (D)\t\u20132MV sin \u03b8 \u02c6j of 12 m\/s undergoes a direct central impact with a 1\t 4.\tAn inextensible mass less string goes over a fric- stationary ball of mass 2 kg. The impact is perfectly tionless pulley. Two weights of 100 N and 200 N are attached to the two ends of the string. The weights elastic. The speed (in m\/s) of 2 kg mass ball after the are released from rest, and start moving due to the gravity. The tension in the string (in N) is _______. impact will be ______.\b [2015] \b [2016] \t11.\t The initial velocity of an object is 40 m\/s. The accel- eration a of the object is given by the following expression: \t\ta = \u20130.1v, \t\twhere v is the instantaneous velocity of the object. The velocity of the object after 3 seconds will be _______.\b [2015] 1\t 2.\t A bullet spins as the shot is fired from a gun. For this purpose, two helical slots as shown in the figure are cut in the barrel. Projections A and B on the bullet engage in each of the slots. Gun Barrel A Bullet 0.5 m B \t\t Helical slots are such that one turn of helix is com- 200 N 100 N pleted over a distance of 0.5 m. If velocity of bullet when it exists the barrel is 20 m\/s, it spinning speed in rad\/s is _____.\b [2015] Answer Keys Exercises Practice Problems 1 \t 1.\u2002A\t 2.\u2002B\t 3.\u2002B\t 4.\u2002C\t 5.\u2002A\t 6.\u2002B\t 7.\u2002A\t 8.\u2002C\t 9.\u2002D\t 10.\u2002B Practice Problems 2 \t1.\u2002D\t 2.\u2002D\t 3.\u2002A\t 4.\u2002D\t 5.\u2002C\t 6.\u2002B\t 7.\u2002A\t 8.\u2002A\t 9.\u2002B\t 10.\u2002A Previous Years\u2019 Questions \t 1.\u2002B\t 2.\u2002A\t 3.\u2002A\t 4.\u2002B\t 5.\u2002A\t 6.\u2002 1 to 1.3\t 7.\u2002D\t 8.\u20023.5\t 9.\u20020.64 14.\u2002 133\u2013134 10.\u2002 7.8 to 8.2\t\t 11.\u2002 29.5 to 29.7\t 12.\u2002 251 to 252\t 13.\u2002D","Chapter 5 Curvilinear Motion CHAPTER HIGHLIGHTS \u261e\u261e Laws for Rotary Motion \u261e\u261e Angular Momentum or Moment of Momentum \u261e\u261e Kinematics of Curvilinear Translation \u261e\u261e Conservation of Angular Momentum \u261e\u261e Projectile Motion \u261e\u261e Simple Harmonic Motion and Free Vibrations \u261e\u261e Equations of the Path of Projectile \u261e\u261e Oscillation, Amplitude, Frequency and Period \u261e\u261e Motion of a Projectile on an Inclined Plane \u261e\u261e Velocity and Acceleration \u261e\u261e Kinematics of Rotation \u261e\u261e Frequency of Vibration of a Spring Mass System \u261e\u261e Angular Displacement and Angular Velocity \u261e\u261e Oscillations of a Simple Pendulum \u261e\u261e Angular Acceleration \u261e\u261e Equations of Motion Along a Circular Path \u261e\u261e C\u0007 urvilinear and Rotary Motion Kinetics of Curvilinear and Rotary Motion Kinematics of Curvilinear Its projections on the x and y co-ordinates are Translation (Vav) x = ds dx = dx Motion of a particle describing a curved path is called as dt ds dt curvilinear motion, (vav) y = ds dy = dy dt ds dt Velocity and Acceleration Now dx and dy are the average velocities of the dt dt The curvilinear motion of a body P may be imagined as the resultant of two rectilinear motions of its projections Px and p\u00ad rojections Px and Py respectively in the direction of their Py on Ox and Oy axis respectively. respective co-ordinates. Velocity If dt approaches zero, vav becomes the instantaneous velocity. Instantaneous velocity at P, v = lim ds ds Let us consider a body moving through a distance ds from dt \u21920 dt = dt and position P to P1 along a curved path in time dt. its direction will be tangential to the path at P. y Similarly vx = dx , vy = dy V V + dv dt dt xO ds dy (dv )y dv Total velocity v = vx2 + vy2 dl V (dv)x sy V + dv Acceleration O The average acceleration during the interval t is aav = dv dt The direction will be same as that of the change of veloc- Px x ity dv. Consider PP1 as a chord instead of an arc, we have The projections of aav on x and y co-ordinates will be Vav = ds dvx and dvy respectively. dt dt dt","3.76\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics When dt approaches zero, the instantaneous acceleration, qq\u2032 = pq sin dq = (v + dv) dq a = lim dv = dv (dq being small dq = dq in radians) = vdv + dvdq = vdq dt dt (dq and dv being very small, their product will be negligible) dt \u21920 From above figure OPP1 a= d ds = d2s dt dt dt 2 PP1 ds dq = r = r Similarly the components of the instantaneous accelera- tion a are, vds r ax = d2x , qq\u2032 = dt 2 Substituting qq\u2032 in equation we have d2y ay = dt 2 an = lim vds rdt We get, dt \u21920 a = ax2 + a2y an = v \u00d7 ds ,\ufffd r dt Tangential and Normal Acceleration But ds = v dt A particle moves on a curved path and from position P, moves a distance ds to position P1, in the time interval dt, \u2234 an = v2 such that at P the instantaneous velocity is v and that at P1 r it is (v + dv) Normal acceleration is also known as \u2018centripetal acceleration\u2019. P v p NOTE q ds dq During the motion of a particle along a curved path there P1 dy is a change in the direction of its velocity from instant to instant with or without any change in magnitude. When r v + dv q\u2032 both magnitude and direction of velocity change, the par- dq ticle has the tangential and normal acceleration. When there is only change in the direction of velocity, the parti- O cle has only normal acceleration. Resolving the acceleration into two components: \t 1.\t Tangential to the path at the position P. Solved Examples \t 2.\t Normal to the path at position P. Example 1:\u2002 The equation of motion of a particle moving on Let r be the radius of the curved path PP1 and dq, the a circular path, radius 400 m, is given by S = 18t + 3t2 + 2t3. angle subtended at the centre O. Where S is the total distance covered from the starting Let q be the angle included between the normals at P1 point, in metres, till the position reached at the end of t and P. seconds. (i)\u2002 The acceleration at the start is From the figure we see that Pp = instantaneous velocity (A)\t 6m\/s2 \t(B)\t5m\/s2 v at P. (C)\t 10m\/s2 \t(D)\t7m\/s2 Resolving dV into two components (pq) in the direction tangential at P and qq\u2032 in the direction normal at P as shown. (ii)\u2002T\u0007he time when the particle reaches its maximum Tangential acceleration tangential change in velocity pq velocity is dt dt a = lim = lim (A)\t 0.5 s\t (B)\t 0.6 s dt \u21920 dt \u21920 (C)\t 0.8 s\t (D)\t 0.95 s From the triangle Pqq\u2032; (iii)\u2002 The maximum velocity of the particle is pq = Pq - Pp = (v + dv) cos dq - v = v + dv - v = dv (A)\t 19.58 m\/s\t (B)\t 20.53 m\/s (dq being very small, cos dq = 1) (C)\t 18.65 m\/s\t (D)\t 13.5 m\/s Then at = lim dv = dv Solution: dt \u21920 dt dt (i)\u2002Given, s = 18t + 3t2 - 2t3 Now normal acceleration an = lim qq\u2032 v = ds = 18 + 6t - 6t 2 dt dt dt \u21920","Chapter 5\u2002 \u2022\u2002 Curvilinear Motion\u2002 |\u2002 3.77 From equation, a = d2s = 6 - 12t Projectile Motion dt 2 Definitions At the starting point when t = 0, Acceleration \t 1.\tProjectiles: A particle projected at a certain angle is called projectile. a = 6 \u2013 0 = 6 m\/s2. (ii)\u2002\u0007For determining the condition for maximum velocity, \t 2.\t Angle of Projection: Angle between the direction of projection and the horizontal plane through the point we have of projection is called as the angle of projection. It is denoted by a. d2s = 6 - 12t = 0 = 0.5 secs dt 2 \t 3.\tTrajectory: The path traced out by the projectile is called the trajectory of the projectile. (iii)\u2002When t = 0.5 s, \t 4.\tVelocity of projection (u): The initial velocity of vmax = 18 + 3 -1.5 = 19.5m\/s projectile is the velocity of projection. Example 2:\u2002 A particle moving along curved path has the \t 5.\tTime of flight (T): The total time taken by the projectile is termed as the time of flight. law of motion vx = 2t - 4, vy = 3t2 - 8t + 8 where vx and vy are the rectangular components of the total velocity in the x and \t 6.\t Horizontal range (R): It is the distance between the point of projection and the point where the trajectory y co-ordinates. The co-ordinates of a point on the path at an meets the horizontal plane. instant when t = 0, are (4, -8). The equation of the path is Equations of the Path of Projectile (A)\t x2 + 3x - 2\t (B)\t x3 + 4x + 2 Y 1 31 C (vertex) u (C)\t x 2 + 3x + 2\t(D)\tx 2 + 4x 2 + 2 P Solution: vx = 2t - 4 y vy = 3t2 - 8t + 8 Integrating both sides, we have \u222b vxdt = \u222b (2t - 4) dt a B AX x = 2\u00d7 t2 - 4t + C1 = t2 - 4t + C1 O 2 x \u222b vydt = \u222b (3t 2 - 8t + 8) dt P is the position occupied by the projectile after t sec and x and y are the two co-ordinates of P along the x-axis and y = 3\u00d7 t3 -8\u00d7 t2 + 8t + C2 = t3 - 4t 2 + 8t + C2 y-axis respectively. 3 2 Along the x-axis, ux = u cos a. Where C1 and C2 are constants Along the y-axis uy = u sin a Given x = 4, y = -8 when t = 0 The component ux remains constant all throughout uy retards due to the action of gravitational force. Substituting for x, y and t in equation 4 = 0 - 0 + C1 We know S = vt, for horizontal motion \u2234 C1 = 4 x = u cos a xt -8 = 0 - 0 + 0 + C2 t = u x a cos Now the equations \u2234 C2 = -8 are x = t2 - 4t + 4 and y s = ut + 1 at 2, for vertical motion of displacement 2 = t3 - 4t2 + 8t - 8 x = (t - 2)2 1 2 Therefore y = u sin at - gt 2 1 Substituting value of t we can write x2 = t -2 x 1 x2 cos 2 cos2 1 y = u sin a u a - g u2 a t = x 2 + 2 \b(1) y = t3 - 4t2 + 8t - 8\b (2) y = x tan a - gx2 2u2 cos2 Substituting the value of t from (1) in (2), we get a 31 This is the equation of the path of a projectile which repre- sents a parabola. y = x2 + 4x2 + 2x","3.78\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Horizontal range, Example 3:\u2002 Find the least initial velocity which a projectile may have so that it may clear a wall of 3.6 m high and 6 m R = 2u2 sin a cos a = u2 sin 2a distant and strike the horizontal plane through the foot of g g the wall at a distance of 3.6 m beyond the wall. The point of projection being at the same level as the foot of the wall. Time of Flight, T = \u23a1 2u sin a\u23a4 Take g = 9.81 m\/sec2 \u23a3\u23a2 g \u23a5\u23a6 (A)\t 10.2 m\/s \t(B)\t11 m\/s Maximum height when the vertical component of the veloc- (C)\t 12 m\/s \t(D)\t13.5 m\/s ity is zero. vy = 0 . ymax = uy2 Solution: 2g Let u be the least initial velocity of the projectile and a be the angle of projection with the horizontal plane. u2 sin2 a Horizontal range of projectile, R = 6 + 3.6 = 9.6 m 2g ymax = , (since uy = u sin a) R = 2u2 sin acos a g Co-ordinates of vertex C u2 sin2 a , u2 sin2 a \u2234 9.6 = 2u2 sin acos a 2g 2g g Now, Motion of a Projectile u2 = 9.6 g a on an Inclined Plane 2sin a cos Consider the motion of projectile with an initial velocity u Putting value, and making an angle a with the horizontal on an inclined plane of inclination q, taking the coordinate axes x, y the u2 = 4.8 g \u00d7 sec2 a \b(1) expressions for the distance r and height h can be derived. tan a Equation for the path of projectile u ymax = xtan a - gx2 a 2u2 cos2 r a q h = r sin q 3.6 = 6 tan a - 62 g a 2u2 cos2 Substituting for u2, we have, r cos q 62 tan a 9.6 x = u(cos\u2009a)\u2009t = r cos\u2009q 3.6 = 6 tan a - y = u(sin a)t - 1 gt 2 = h = r sin q 3.6 = tan a \u23a2\u23a16 - 62 \u23a4 2 \u23a3 9.6 \u23a5 \u23a6 By eliminating t, we get gr 2 cos2 q 3.6 = 2.25 tan a 2u2 cos2 a r sin q = r cosq tan a - 3.6 2.25 2u2 cos2 tan a = = 1.6 g cos q \u21d2 r = a (tan a- tan q). \b(1) a = 57.9\u00b0 \u2234 The distance r is given by equation (1) and thus the height From equation (1) h and the distance on the horizontal plane can be found. u2 = 4.8 g \u00d7 sec2 57.9 = 4.8 g \u00d7 3.54 = 104.57 tan 57.9 1.594 i.e., h = r sin\u2009q and x = r cos\u2009q. u = 10.2m\/s The maximum range possible on the inclined place is Example 4:\u2002 An aeroplane is moving horizontally at found out by differentiation equation 1 with respect to a and 108\u00a0km\/h at an altitude of 1000 m towards a target on the equating it to zero. ground which is intended to be bombed. \u2234 tan\u20092a = \u2013 cot\u2009q. \u2234 for maximum range the angle made by the velocity vec- tor a should be equal to (45\u00b0 + q\/2) with the horizontal.","108 mph Chapter 5\u2002 \u2022\u2002 Curvilinear Motion\u2002 |\u2002 3.79 under its own weight. After leaving the chute 1 at point D, the ball hits the wall as depicted in the figure. Wall 1000 m 1m A Ball V sin 60 1.5 m B B 60\u00b0 V cos 60 2.5 m (a)\t\u0007The distance from the target where the bomb must be 1.5 m D C released in order to hit the target is (A)\t 428.35 m\t (B)\t 450.54 m (C)\t 580.2 m\t (D)\t 800 m (b)\t The velocity, with which the bomb hits the target is (a)\u2002\u0007The time interval of the ball\u2019s travel from the point D to (A)\t 143 m\/s \t(B)\t148 m\/s the point of hit is (C)\t 150 m\/s \t(D)\t161.2 m\/s (A)\t 0.88 s\t (B)\t 0.92 s (C)\t 0.733 s\t (D)\t 0.898 s Solution: (b)\u2002\u0007The distance on the wall above the point D to the point (a)\u2002\u0007Let B be the point of target and A be the position of the of hit is aeroplane and the bomb is released from A to hit at B. The horizontal component of the bomb velocity, which (A)\t 0.21 m\t (B)\t 0.158 m is uniform, is (C)\t 0.32 m\t (D)\t 0.168 m v = 108 km\/h = 108 \u00d71000 = 30 m\/sec. Solution: 60 \u00d7 60 (a)\u2002\u0007The ball starts from point A. The vertical distance from Considering the vertical component of bomb velocity, A to C is equal to 3 m. Considering the motion of ball At A, u = 0, g = 9.81m\/sec2 from A to C, S = 1 gt 2 V 2 = 2as 2 Since initial velocity is zero, a = g = 9.81 m\/sec2 Let t be the time required to hit B, then or vC2 = 2 \u00d7 9.81\u00d7 3 1000 = 1 \u00d7 9.81 \u00d7 t2 vC = 7.67 m\/s, 2 This is the velocity of the ball at C. The motion of the ball from C to D 2000 t2 = 9.81 = 203.87 vD2 = vC2 - 2as 7.672 = 2 \u00d7 9.81\u00d71.5 \u2009 = 58.82 - 29.43 = 29.39 t = 14.278 sec vD = 5.42 m\/s Horizontal distance covered by the bomb S = Vt = 30 On reaching the point D, the horizontal component of the \u00d7 14.278 = 428.35 m i.e., the bomb is released from plane velocity of the ball when the horizontal distance is 428.35 from B = v cos 60\u00b0 = 5.42 \u00d7 1 = 2.71 m\/s (b)\u2002\u0007Vertical component velocity at B = u + gt = 0 + 9.81 \u00d7 2 14.278 = 140.06 m\/sec Resultant velocity at B = 302 +140.062 Let t be the time taken by the ball to hit the wall from point D. Then, = 20518.8 = 143 m\/sec \t\t Example 5:\u2002 A ball weighing 10 N starts from the position t = 2.5 = 0.922 sec . A as shown in figure and slides down a frictionless chute 2.71","3.80\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics (b)\u2002F\u0007inally considering the vertical motion of the ball Let D be the point reached by the bullet, 12 seconds after it is fired. Time taken by the bullet to reach point B from A beyond the point D (point at which it is fired from) = 10.6 sec. \u2234\u2002 Time taken by the bullet to travel from point B to point s = ut - 1 gt 2 D = 12 \u2013 10.6 = 1.4 sec. 2 5.42 m Horizontal velocity at B, vH = 120 cos 60\u00b0 Here u = vD = s = 120 \u00d7 0.5 = 60 m\/s = 5.42 \u00d7 0.922 - 1 \u00d7 9.81(0.922)2 2 The vertical velocity after 1.4 sec of travel from point B, = 4.327 - 4.169 = 0.158 m. vv = 0+ 1 \u00d7 9.81\u00d71.42 = 9.62 m\/s Hence the ball will hit the wall 0.158 m above the point 2 D after 0.922 sec Example 6:\u2002 From the top of a tower 60 m high, a bullet is Velocity at point D fired at an angle of 60\u00b0 with the horizontal, with an initial v = vH 2 + vv2 = 602 + 9.622 = 60.8 m\/s velocity of 120\u2009m\/s as shown in figure. Neglect air resistance. (a)\u2002\u0007The maximum height from the ground that would be attained by the bullet, is (A)\t 528 m\t (B)\t 611 m Kinematics of Rotation (C)\t 680 m\t (D)\t 720 m When a moving body follows a circular path it is known as circular motion. In circular motion the centre of rotation is (b)\u2002\u0007The velocity of bullet, 12 seconds after it is fired, is stationary. (A)\t 55 m\/s\t (B)\t 58 m\/s Angular Displacement and Angular Velocity (C)\t 61 m\/s\t (D)\t 80 m\/s Angular displacement is defined as the change in angular 120 m\/sec Bh D position (usually referred to as the angle \u03b8), with respect A 60 to time. q v Angular velocity is defined as the rate of change of angu- lar displacement with respect to time. Let a body, moving Tower 60 m along a circular path, be initially at P and after time t sec- onds be at Q. C Let \u2220POQ = \u03b8 Then angular displacement = \u2220POQ = \u03b8 Solution: Q (a)\u2002Height q Or \t\th = u2 sin2 a P 2g 120 \u00d7 120 \u00d7 (sin 60)2 120 \u00d7 120 \u00d7 3 \u00d7 3 Time taken = t 2\u00d7 9.81 2 2 = = Angular displacement q 2 \u00d7 9.81 Angular velocity = Time = t = 10800 = 551 m Mathematically, it is expressed as dq . 2 \u00d7 9.81 dt Maximum height above ground = 551 + 60 = 611 m. It is denoted by the symbol w (b)\u2002 Time of travel upto highest point B is given by, t = u sin a = 120 \u00d7 sin 60 = 10.6 sec . w = dq g 9.81 dt It is measured in radian\/sec or rad\/sec","Chapter 5\u2002 \u2022\u2002 Curvilinear Motion\u2002 |\u2002 3.81 Relation between Linear Velocity t = \u0007time (in seconds) during which angular velocity and Angular Velocity changes from w0 to w, Let v = linear velocity = Linear displacement v = linear speed in m\/s, Time The rotational speed is N revolutions per minute or N But linear displacement = Arc PQ = OP \u00d7 q = rq r.p.m. v = r \u00d7 q = r \u00d7 Angular velocity Example 7:\u2002 A wheel rotates for 5 seconds with a constant t acceleration and describes during the time 100 radians. It then rotates with a constant angular velocity and during the \u239b\u239c\u239d\u2235 q = angular velocity\u239f\u23a0\u239e next 5 seconds, it describes 70 radians. The initial angular t velocity and angular acceleration respectively are, (A)\t 15 rad\/s, 2.5 rad\/s2\t (B)\t 13 rad\/s, 2 rad\/s2 v = r\u00d7w (C)\t 15 rad\/s, -2 rad\/s2\t (D)\t 26 rad\/s, -2.4 rad\/s2 Where w = angular velocity Solution: Angular Acceleration Angular velocity It is defined as the rate of change of angular velocity. It is w = q = 70 = 14 rad\/s measured in radians per sec2 and written as rad\/sec2 and is t 5 denoted by the symbol a\u2009. a is constant angular acceleration and w0 be initial angular a = Rate of change of angular velocity velocity. a= dw = d \u239b dq\u239e \u239c\u239d\u239b\u2235 w = dq\u239e = d2q . q = w0t + 1 at 2 dt dt \u239c\u239d dt \u23a0\u239f dt \u239f\u23a0 dt 2 2 Also dw dw dq dw dw 100 = (w0 5) + 1 a \u00d7 52 dt dq dt dq dq 2 = \u00d7 = \u00d7 w = w \t\t\t\t 5w0 + 12.5a = 100\b (1) It has two components: \t\t\t\t w = w0 + a\u2009t V2 \t\t\t\t 14 = w0 + 5a\b(2) Normal component =r = w2r and tangential compo- aSo=lv-in2g.4erqauda\/tsieocn2s\u2002\u2002(\t1) and ((2R)ewta0r=da2ti6onra)d\/sec nent = dV = r dw = ra dt dt Example 8:\u2002 A wheel rotating about a fixed axis at 20 r.p.m. If a is the linear acceleration, then is uniformly accelerated for 80 seconds during which time it makes 60 revolutions. a = ra (a)\u2002 The angular velocity at the end of the time interval is Equations of Motion Along a Circular Path (A)\t 7.294 rad\/s\t (B)\u2002 8.384 rad\/s (C)\u2002 6.812 rad\/s\t (D)\u2002 7.829 rad\/s a = w - w0 (b)\u2002 The time required for the speed to reach 100 r.p.m. t (A)\t 3.65 min\t (B)\t 2.14 min (C)\t 1.85 min\t (D)\t 2.58 min q = w0t + 1 at 2 Solution: 2 (a) w2 - w02 = 2aq q = w0t + 1 at 2 2 If N is the r.p.m. w0 = initial angular velocity 2pN w = 60 radians\/sec w0 = 2p \u00d7 20 = 2.094 60 2pN p DN v = rw = 60 \u00d7r = 60 m\/s 2p \u00d7 60 = (2.094 \u00d7 80) + 1 a(80)2 2 Where, 2p \u00d7 60 = 167.52 + 3200a w0 = initial angular velocity in cycles\/sec, w = final angular velocity in cycles\/sec, a = 2p60 -167.52 = 0.065 rad\/sec2 3200","3.82\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Let w be the angular velocity at the end of 80 seconds Final angular momentum = Iw in rad\/sec. Then w = w0 + a\u2009t\u2002 w = 2.094 + (0.065 \u00d7 80) Change of angular momentum = I (w \u2013 w0) = 7.294 rad\/sec. Rate of change of angular momentum (b) = change of angular momentum Where Time 7.294 = 2p \u00d7 N 60 (w - w0 ) I a \u23a1\u23a3\u23a2\u2235 a = w - w0 = angular acceleration\u23a6\u23a5\u23a4 N = 69.65 r.p.m = I t = t w1 = w0 + a\u2009t1 From second law of motion of rotation, Torque a rate of change of angular momentum w1 = 2p \u00d7100 rad\/sec 60 T = Ia T = KIa, = 10.466 rad\/sec where K is a constant of proportionality. SI unit of torque 10.466 = 2.094 + 0.065 \u00d7 t1 is Nm. 8.372 t1 = 0.065 = 128.8 sec = 2.14 min . Angular momentum or moment of momentum: Moment Curvilinear and Rotary Motion of momentum of the body about O = Iw, Where the rigid body undergoes rotation about O. Kinetics of Curvilinear and Rotary Motion Angular momentum is the moment of linear momentum For a particle or a body moving in a curved path with particu- Rotational kinetic energy: Rotational kinetic energy lar emphasis to the circular path comes under this section. = 1 I w2 2 In order to maintain the circular motion, an inward radial Angular impulse or impulsive torque: Angular impulse or force called \u2018centripetal force\u2019 is acted upon the body, which impulsive torque = I dw is equal and opposite to the centrifugal force that is directed away from the centre of curvature. If r is the radius of the Work done in rotation: Work done in rotation = T \u00d7 q circular path, v is the linear velocity, w is the angular veloc- Kinetic energy in combined motion: Kinetic energy due ity and t is the time, then to translatory motion = 1 mv2 Angular acceleration dw 2 = dt Tangential acceleration = r dw , Kinetic energy due to rotation = 1 I w2 dt 2 Normal acceleration = v2 = w2r, Kinetic energy due to combined motion = 1 mv2 + 1 I w2 . r 2 2 Centripetal or centrifugal force = W \u00d7 v2 = W w2r. Conservation of Angular Momentum g r g The law of conservation of angular momentum states that Laws for Rotary Motion the angular momentum of a body or a system will remain First Law unaltered if the external torque acting on it is zero. It states that a body continues in its state of rest or of rotation D\u2019alemberts\u2019 Principle for Rotary Motion about an axis with constant or uniform angular velocity unless it is compelled by an external torque to change that state. D\u2019Alemberts\u2019 principle for rotary motion states that the sum of the external torques (also termed as active torques) acting Second Law on a system, due to external forces and the reversed active torques including the inertia torques (taken in the opposite It states that the rate of change of angular momentum of a direction of the angular momentum) is zero. rotating body is proportional to the external torque applied on the body and takes place in the direction of the torque. Suppose a disc of moment of inertia I rotates at an angu- lar acceleration a under the influence of a torque T, acting I = Mk2 in the clockwise direction. Inertia torque = Ia (acting in the where M = mass of the body and k = radius of gyration anti-clockwise direction) = moment of inertia \u00d7 initial angular velocity Initial angular momentum = Iw0 From D\u2019Alemberts\u2019 principle, T - Ia = 0, the dynamic equation of equilibrium for a rotating system.","Chapter 5\u2002 \u2022\u2002 Curvilinear Motion\u2002 |\u2002 3.83 Rotation caused by a weight W attached to one end of a T\u0007 orque due to inertia force on string passing over a pulley of weight W0 50 N = 50 \u00d7 a \u00d7 142 = 9.99 aNcm From D\u2019Alemberts\u2019 principle, it can be shown that, 981 a= gW , when the pulley is considered as a disc. Let T be the torque applied to the shaft for dynamic equilibrium \u03a3T = 0 \u239c\u239d\u239bW W0 \u239e + 2 \u239f\u23a0 T + 700 = 8400 + 32.62a + 359.63a + 9.99a T = 8400 + 312.33 = 9136 Ncm, Rotation caused due to two weights W1 and W2 attached Since a = 150 = 3.57 rad\/s2. to the two ends of a string which passes over a rough pulley 42 of weight W0 (b)\u2002\u0007Let F1 and F2 be the tensions in the strings. Applying g(W1 -W2 ) D\u2019Almberts\u2019 principle for linear motion, we get a = \u239d\u239b\u239cW1 W0 \u239e F1 - 200 - 200 \u00d7 1.5 = 0 2 \u23a0\u239f 9.8 + W2 + 50 F2 + 50 - F2 = 9.8 \u00d7 1.5 Example 9:\u2002 In a pulley system shown in figure the pulley F1 = 200 + 200 \u00d7 1.5 = 200 + 22.96 weighs 20 N and its radius of gyration is 40 cm. A 200 9.8 N weight is attached to the end of a string and a 50 N is attached to the end of the other string as shown in the = 180.6 N figure. F2 = 50 \u00d7 9.8 - 50 \u00d7 1.5 = 42.34N 9.8 42 cm Simple Harmonic Motion and Free Vibrations 14 cm Simple harmonic motion: It is defined as the type of 200 N 50 N motion in which the acceleration of the body in its path of motion, varies directly as its displacement from the equi- (a)\u2002\u0007The torque to be applied to the shaft to raise the 200 N librium position and is directed towards the equilibrium point. weight at an acceleration of 1.5 m\/s2 is Oscillation, Amplitude, (A)\t 6812 Ncm\t (B)\t 9136 Ncm Frequency and Period (C)\t 700 Ncm\t (D)\t 7832 Ncm. YP (b)\u2002 The tensions in the strings are respectively yw (A)\t 170.4 N, 35.6 N\t (B)\t 180 N, 40 N X1 O M X x (C)\t 190.2 N 35 N\t (D)\t 180.6 N, 42.34 N Y\u2032 Solution: In the above figure, when a particle P is describing a circu- (a)\u2002 Moment of inertia of the pulley I = W k2 lar path, M being the projection of P, it describes a simple g harmonic motion. I = 20 \u00d7 (40)2 Ncm2 = 32.62 Ncm2 The motion of M from X to X\u2032 and back to X is called an 981 oscillation or simple harmonic motion. T1 = Torque produced by 200 N \u2009= 200 \u00d7 42 = 8400 Ncm OX = OX\u2032 is the amplitude. This amplitude is the distance between the centre of sim- T2 = Torque developed by 50 N = 50 \u00d7 14 = 700 Ncm ple harmonic motion and the point where the velocity is zero. The period of one complete oscillation is the period of Inertia torque due to angular rotation of the pulley with simple harmonic motion. angular acceleration a = Ia = 32.62a Ncm. \u0007Torque due to inertia force on 200 N \u0007= (ma)r = 200 r ar = 200 \u00d7 a \u00d7 (42)2 981 981 = 359.63a Ncm","3.84\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics Thus the period of simple harmonic motion is the time in Oscillations of a Simple Pendulum which M describes 2p radians at w radians\/sec. Period of oscillation T = 1 = 2p l (for 2 beats) T = 2p , where T is the time period in seconds. f g w l = length of pendulum. Half of an oscillation is called a Velocity and Acceleration beat or swing. A pendulum executing one half oscillation The simple harmonic displacement per second is called seconds pendulum. Time of one beat or X = r sin w\u2009t swing = p l = T . For n number of beats, time = np l. v = w r2 - x2 g 2 g Acceleration = d2x = -w2r sin wt For a compound pendulum T = 2p KG2 + h2 dt 2 gh a = -w2x Where h is the distance between the point of suspension Frequency = 1 a and centre of gravity. Where kG = radius of gyration about 2p x O, the centre of suspension. A compound pendulum is a rigid body free to oscillate about a smooth horizontal axis passing through it. Frequency of Vibration of a Spring A simple pendulum whose period of oscillation is the Mass System same as that of a compound pendulum is called as a simple Consider a helical spring subjected to a load W. The static equilibrium position is 0-0. Let S be the stiffness of the equivalent pendulum 1= kG 2 +h. spring which is defined as force required to cause one unit h extension. If the weight is displaced and stretched to posi- tion 1-1\u2032 by an amount \u2018y\u2019, as shown in the below figure, Example 10:\u2002 A body performing simple harmonic motion then the acceleration with which the load springs back, has a velocity 12 m\/s when the displacement is 50 mm and 3 m\/s when the displacement is 200 mm, the displacement being measured from the mean position. (a)\u2002 Calculate the frequency of the motion. w a = - sy (A)\t 35 cycles\/sec\t (B)\t 40.5 cycles\/sec g (C)\t 31.8 cycles\/sec\t (D)\t 35.5 cycles\/sec s\u00d7g (b)\u2002 What is the acceleration when the displacement is 75 mm. -W \u2234 a= \u22c5 y (A)\t 15 m\/s2 \t (B) 16.5 m\/s2 (C)\t 13.8 m\/s2\t (D) 15.6 m\/s2 This is of the form a = -wn2 y Solution: sg g, (a)\u2002 In simple harmonic motion w d Where wn2 = = V2 = w2(r2 - x2) V = velocity, r = amplitude d being w x = distance from mid positions s x1 = 50 mm, x2 200 mm V1 = 12 m\/s V2 = 3 m\/s 1 g. 2 Frequency f = wn = 2p d \u23a1 \u239b 50 \u239e \u23a4 2p 122 = w2 \u23a2 r 2 - \u239c\u239d 1000 \u239f\u23a0 \u23a5 \u23a6\u23a5 \u23a2\u23a3 \u23a1 \u239b 200 \u239e 2 \u23a4 \u23a2r \u239c\u239d 1000 \u23a0\u239f \u23a5 32 = w2 2 - \u23a6\u23a5 \b(1) \u23a3\u23a2 2 2 Dividing we get 144 = r2 1 w 9 r2 - 400 0 0 y - 4 w 100 1 1w 16 = r2 - 1 r2 - 400 4 100","Chapter 5\u2002 \u2022\u2002 Curvilinear Motion\u2002 |\u2002 3.85 16r 2 - 16 \u00d7 4 = r2 - 1 a = w2 x = \u239b 9 \u00d7 75 \u239e = 15 m\/s2 . 100 400 \u239d\u239c 0.045 1000 \u23a0\u239f 15r 2 = 16 \u00d7 2 - 1 Example 11:\u2002 The amount of seconds a clock would loose 50 400 per day, if the length were increased in the ratio 800 : 801 is 2 \u00d7 64 \u00d7 4 1 511 15r 2 = 400 - 400 = 400 (A)\t48 s\t (B)\t54 s (C)\t50 s\t (D)\t60 s r2 = 511 = 0.085 Solution: 400 \u00d715 Given I = 800 units r = 0.29, m = 290 mm. I + dI = 801 units Putting the value of r2 in equation (1), we get \u2002 dI = 1 unit 9 = w2[0.085 - 0.04] We get, dl = I I 800 Or, w2 = 9 dn = -dI = I 0.045 n 2I 1600 We get, w = 200 rad\/s n 86400 1600 1600 So, f = w = 200 = 31.83 cycles\/sec. dn = - = - = -54 2p 2p (b)\u2002 If a be the acceleration when displacement x = 75 mm Where n = 86400, as a seconds pendulum will beat 86400 times\/day. The clock will loose 54 seconds a day.) Exercises Practice Problems 1 surface is subjected to a sudden application of a force of 300 N at a point of its periphery. Direction for questions 1 to 10:\u2002 Select the correct alterna- tive from the given choices. \t 5.\t The angular acceleration is Direction for questions 1 to 3: A force of 2t Newton, where \t (A)\t 0.75 rad\/s2\t (B)\t 1.5 rad\/s2 t in seconds, acts on a mass of 100 kg initially at rest, for a period of 20 seconds. \t (C)\t 2 rad\/s2\t (D)\t 2.5 rad\/s2 \t 1.\t The impulse on the mass is \t 6.\t The acceleration of mass centre is \t (A)\t 400 Ns\t (B) \t300 Ns \t (A)\t 1 m\/s2\t (B)\t 1.5 m\/s2 \t (C)\t 350 Ns\t (D)\t 500 Ns \t (C)\t 2 m\/s2\t (D)\t 3 m\/s2 \t 2.\t The velocity of the mass is \t7.\t A particle traveling in a curved path of radius of curva- \t (A)\t 1 m\/s\t (B)\t 2 m\/s ture 500 m with a speed of 108 km\/h and a tangential acceleration of 4 m\/s2. The total acceleration of the par- \t (C)\t 3 m\/s\t (D)\t 2.5 m\/s ticle is \t 3.\t The average force, which would have resulted in the \t (A)\t 4.38 m\/s2\t (B)\t 5 m\/s2 same velocity, is \t (C)\t 3.5 m\/s2\t (D)\t 8 m\/s2 \t (A)\t 15 N\t (B) \t30 N Direction for questions 8 and 9: A solid cylinder 80 cm in diameter is released from the top of an inclined plane 2.0\u00a0m \t (C)\t 20 N\t (D)\t 10 N high surface and rolls down the inclined surface without any loss of energy due to friction. \t 4.\t A car of mass 1500 kg descends a hill of 1 in 5 incline. The average braking force required to bring the car to rest from a speed of 80 km per hour in a \t8.\t The energy equation for the system is distance of 50 m is (take the frictional resistance as 300 N) 1 mv2 \t(B)\tmgh = 13 mv2 2 \t (A)\t 10 N\t (B)\t 15 N \t(A)\tmgh = \t (C)\t 8 N\t (D)\t 12 N Direction for questions 5 and 6: A thin circular ring \t(C)\tmgh = 3 mv2 \t(D)\tmgh = 23 mv 2 of mass 200 kg and radius 2 m resting flat on a smooth 4","3.86\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics \t 9.\t The linear and angular speeds, at the bottom respec- pulley and supports a 4 kg mass. The angular accelera- tively are tion of the pulley (g = 10 m\/s2 ) is \t (A)\t 6.1 m\/s and 12.75 rad\/s \t (A)\t 1 rad\/ s2 \t(B)\t12 rad\/ s2 \t (B)\t 5.5 m\/s and 34 rad\/s 4 \t (C)\t 5.1 m\/s and 12.75 rad\/s \t (D)\t 6.1 m\/s and 34 rad\/se 1\t 0.\t A disc shaped frictionless pulley I = 1 MR2 has a mass \t (C)\t 1 rad\/ s2 \t(D)\t3 rad\/ s2 2 4 of 80 kg and radius of 2 m. A rope is wound round the Practice Problems 2 \t (C)\t Ratio of the velocity vectors before and after collision \t (D)\t\u0007Negative of the ratio of the energies of the bodies Direction for questions 1 to 10:\u2002 Select the correct alterna- tive from the given choices. before and after the impact \t1.\t A bullet is projected so as to graze the top of two walls \t 5.\t A cylinder of radius of r and mass m rest on a rough each of height 20 m located at distances of 30 m and horizontal rug. If the rug is pulled from under it with an 180 m in the same line from the point of projection as acceleration, a perpendicular to the axis of the cylinder, shown in figure. The angle and speed of projection of the angular acceleration of the centre of mass of the the bullet, respectively, are cylinder, assuming that it does not slip, is \t(A)\t23 Ar \t(B)\t13 Ar P1 P2 \t(C)\t3 A \t(D)\t23 A V0 4r 20 m O a 20 m 108 m Direction for questions 6 to 8: A soldier positioned on a hill 30 m fires a bullet at an angle of 30\u00b0 upwards from the horizontal as shown in the figure. The target lies 60 m below him and the bullet is fired with a velocity of 200 m\/s. \t (A)\t34.1\u00b0 and 44 m\/s 200 m\/s \t (B)\t38.2\u00b0 and 48 m\/s 30\u00b0 \t (C)\t35.29\u00b0 and 49.5 m\/s 60 m \t (D)\t37.87\u00b0 and 46.1 m\/s \t2.\t For a given value of initial velocity for a projectile, the maximum range, on an inclined plane inclined to the \t 6.\t The maximum height, to which the bullet will rise horizontal at an angle of b (in degrees), can be obtained above the position of the soldier, is if the angle of projection is \t (A)\t 615 m\t (B)\t 490 m \t (A)\t45\u00b0\t (B)\t90\u00b0 \u2013 0.5b \t (C)\t 509.7 m\t (D)\t 710.6 m \t (C)\t45\u00b0 + 0.5b\t (D)\t45\u00b0 \u2013 0.5b \t7.\t The velocity with which the bullet will hit the target is \t3.\t A shell bursts on contact with the ground and the pieces \t (A)\t 202.9 m\/s\t (B)\t 245.3 m\/s of it fly off in all directions with speeds up to 40 m\/s. A \t (C)\t 312.7 m\/s\t (D)\t 343.6 m\/s person, standing 40 m away from the point of burst, can be hit by a piece in a time duration of \t8.\t The time required to hit the target is \t (A)\t 1.5 sec\t (B)\t 1 sec \t (A)\t 21.7 sec\t (B)\t 20.97 sec \t (C)\t 2 sec\t (D)\t 3 sec \t (C)\t 15.6 sec\t (D)\t 23 sec \t4.\t The coefficient of restitution is defined as the \t9.\t A carpet of mass m made of an inextensible material is \t (A)\t\u0007Negative of the ratio of the velocity of separation rolled along its length in the form of a cylinder of radius R and is kept on a rough horizontal floor. When a small to the velocity of approach push, of negligible force, is given to the carpet, it starts \t (B)\t Ratio of the velocity components in the line of impact","Chapter 5\u2002 \u2022\u2002 Curvilinear Motion\u2002 |\u2002 3.87 unrolling without sliding on the floor. The horizontal ground level and the top of the track is 8.3 m above velocity of the axis of the cylindrical part of the carpet the ground. The distance on the ground, with respect to the point B (which is vertically below the end of the is 63 gR when the radius of the carpet reduces to track), where the sphere would land is 3 A \t(A)\t3R \t(B)\tR 44 \t 8.3 m C \t \t(C)\tR2 \t(D)\tR5 D 2m \t10.\t A small sphere rolls down without slipping from the (A)\t 6 m\t B top most point of a track, with an elevated section and a (C)\t 3 m\t horizontal part, as shown in the following figure, in (B)\t 10 m a vertical plane. The horizontal part is 2 m above the (D)\t 2 m Previous Years\u2019 Questions \t1.\t A circular disk of radius R rolls without slipping at a velocity v. The magnitude of the velocity at point P M (see figure) is \b [2008] R P H R 30\u00b0 V (A)\t gH \t (B)\t 2gR \t(A)\t3v \t(B)\t3 v (C)\t 2gH \t (D)\t0 2 \t5.\tA mass of 2000 kg is currently being lowered at a \t(C)\tv \t(D)\t2v velocity of 2 m\/s from the drum as shown in the fig- 23 ure. The mass moment of inertia of the drum is 150 kg-m2. On applying the brake, the mass is brought to \t2.\tAn annular disc has a mass m, inner radius R and rest in a distance of 0.5 m. The energy absorbed by the outer radius 2R. The disc rolls on a flat surface with- brake (in kJ) is _______.\b [2016] out slipping. If the velocity of the centre of mass is v, the kinetic energy of the disc is\b [2014] \t(A)\t196 mv2 \t(B)\t1161 mv2 2m \t(C)\t1163 mv2 \t(D)\t1165 mv2 \t3.\tConsider a steel (Young\u2019s modulus E = 200 GPa) 2 m\/s column hinged on both sides. Its height is 1.0 m and 2000 kg cross-section is 10 mm \u00d7 20 mm. The lowest Euler \t6.\tA circular disc of radius 100 mm and mass 1 kg, initially at rest at position A, rolls without slipping critical buckling load (in N) is ______.\b [2015] down a curved path as shown in figure. The speed v of the disc when it reaches position B is _______ m\/s. \t4.\tA point mass M is released from rest and slides down \b [2016] a spherical bowl (of radius R) from a height H as Acceleration due to gravity g = 10 m\/s2. shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of the mass at the bottom of the bowl is:\b [2016]","3.88\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics A V2 B 30 meters V1 y, \u2227j \u03b8 = 45\u00ba Bv A \t7.\tA rigid rod (AB) of length L = 2 m is undergoing x, i\u2227 translational as well as rotational motion in the x-y plane (see the figure). The point A has the velocity The magnitude of the velocity V2 (in m\/s) at the end B is V1 = i\u02c6 + 2 \u02c6j m\/s. The end B is constrained to move ______.\b [2016] only along the x direction. Answer Keys Exercises Practice Problems 1 \t1.\u2002A\t2.\u2002B\t3.\u2002C\t4.\u2002A\t5.\u2002A\t6.\u2002B\t7.\u2002A\t8.\u2002C\t9.\u2002C\t10.\u2002A Practice Problems 2 \t1.\u2002D\t2.\u2002C\t3.\u2002B\t4.\u2002A\t5.\u2002A\t6.\u2002C\t7.\u2002A\t8.\u2002B\t9.\u2002B\t10.\u2002A Previous Years\u2019 Questions \t1.\u2002A\t2.\u2002C\t 3.\u2002 3285 to 3295\t4.\u2002C\t5.\u2002 14.1 to 14.3\t6.\u200220\t7.\u20023","Test Engineering Mechanics\b Time: 60 Minutes Direction for questions 1 to 30:\u2002 Select the correct alterna- \t 9.\t Moment of Inertia of a square of side \u2018a\u2019 about an axis tive from the given choices. passing through its C. G is equal to \t1.\tTwo equal and opposite co-planar couples \t (A)\t a3 \t(B)\t1a24 \t (A)\t Balance each other. 12 \t (B)\t Produce a couple and unbalanced force. \t (C)\t Cannot balance each other. \t (C)\t a3 \t(D)\ta4 \t (D)\t Give rise to a couple of double the magnitude. 36 36 \t2.\t In a perfect frame, the number of members are 1\t 0.\tAccording to the law of the machine, the relation \t (A)\t2j \u2013 3\t (B)\t 2j + 3 between effort \u2018P\u2019 and load W is given by \t (C)\t2j \u2013 2 \t (D)\t 2j \u2013 1 \t (A)\t W = mP + C\t(B)\tW = mP - C \t(C)\tP = mW + C\t(D)\tP = mW - C \tWhere j = number of joints. 1\t 1.\t Weight of 150 kN is being supported by a tripod whose \t3.\tThe state of equilibrium of a body implies that the body must (with respect to some inertial frame) be: leg is of the length of 13 m. If the vertical height of the \t (A)\t At rest or with uniform acceleration. point of attachment of the load is 12, the force on the \t (B)\t Uniform velocity or uniform acceleration. \t (C)\t At rest or with uniform velocity. tripod leg would be \t (D)\t\u0007At rest or with uniform velocity or uniform accel- \t (A)\t 48.24 N\t (B)\t 54.16 N eration. \t (C)\t 50.8 N\t (D)\t 45.3 N \t12.\t The resultant of two forces 4P and 3P is R. If the first \t 4.\t The distance of the centroid of a semicircle of radius \u2018r\u2019 force is doubled the resultant is also doubled. The angle from its base is between the two forces is 4r \t(B)\t34pr \t (A)\t48.25\u00b0\t (B)\t95.73\u00b0 3p \t (A)\t \t (C)\t32.5\u00b0\t (D)\t45.53\u00b0 \t(C)\t43pr \t(D)\t2p \t13.\t In the truss shown the force in the member BC is 3r PP \t 5.\t A machine requires an effort of 10 Kg to lift a load of BC 200 Kg an effort of 12 Kg for a load of 300 Kg. The A 60\u00b0 60\u00b0 D effort required to lift a load of 500 Kg will be \t (A)\t 16 Kg\t (B)\t 15 Kg \t (C)\t 14 Kg\t (D)\t 17 Kg \t (A)\t 0 \t (B)\t 0.577 P(T) \t (C)\t0.577 P (comp)\t (D)\t 0.866 P (comp) \t 6.\t The moment of a force \t (A)\t Ocures about a point Direction for questions 14 and 15: A body is weighing \t (B)\t Measures the capacity to do useful work. 500 N is just moved along a horizontal plane by a pull of \t (C)\t Occurs only when bodies are in motion 100 2 N making 45\u00b0 with horizontal. \t (D)\t\u0007Measures the abilities to turning or twisting about \t14.\t Find the value of normal reaction R axes. \t 7.\t The required condition of equilibrium of a body is that \t (A)\t 300 N\t (B)\t 400 N \t (A)\t\u0007The algebraic sum of horizontal components of all \t (C)\t 200 N\t (D)\t 500 N the forces must be zero. \t (B)\t\u0007The algebraic sum of the vertical components of \t15.\t Find the coefficient of friction all the forces must be zero. \t (A)\t0.32\t (B)\t0.33 \t (C)\t\u0007The algebraic sum of moments about a point must \t (C)\t 0.25\t (D)\t0.28 be zero. \t (D)\t All the above. Direction for question 16, 17, 18: For the mass-pulley \t 8.\t The unit of the moment of Inertia of an area is system shown, the mass m2 = 5 Kg is placed on a smooth inclined plane of inclination \u03b8 where as mass m1 = 5 Kg is (B)\tKg-m2 a hanging force. If acceleration of the system is 1.5 m\/s2. \t (A)\tKg-m\t \t (C)\tKg-m4\t (D)\tm4","3.90\u2002 |\u2002 Part III\u2002 \u2022\u2002 Unit 1\u2002 \u2022\u2002 Engineering Mechanics T a = 1.5 m\/s2 \t23.\t The reaction at the hinge when a rigid rod of mass \u2018m\u2019 and length \u2018L\u2019 is subjected to a force \u2018P\u2019 as shown M2-5 Kg q M1 5 Kg O \t16.\t The inclination of the plane will be 2L L 3 \t (A)\t41.52\u00b0\t (B)\t35.50\u00b0 \t (C)\t52.15\u00b0\t P (D)\t43.96\u00b0 \t17.\t The tension in the string will be \t (A)\t 41.55 N\t (B)\t 35.15 N \t (A)\t\u2013P\t (B)\t0 \t (C)\t 21.5 N\t (D)\t 25.28 N \t18.\t How the acceleration of the system would be affected \t(C)\tP \t(D)\t2P 33 of each mass is doubted \t24.\t In the figure shown tension in the member QR is \t (A)\t 3 m\/s2\t (B)\t 2 m\/s2 \t (C)\t 1.5 m2\t (D)\t 2.5 m\/s2 \t19.\t A block g4is.sTlihdeinngthdeowkinneatnicinccoleinffeicoiefn3t0o\u00b0fwfriitchtiaonnaicscel- F eration P 105\u00b0 \t (A)\t 3 \t(B)\t1 Q 45\u00b0 30\u00b0 R 2 3 \t(C)\t1 \t(D)\t1 2 23 \t (A)\t 0.732 F\t (B)\t 0.63 F \t (C)\t 0.433 F\t (D)\t 0.75 F Direction for question for 20 and 21: A 600 N weight is 2\t 5.\t Force in member QR (B)\t 0.75 F suspended by flexible cables as shown in figure \t (A)\t 0.633 F\t (D)\t 0.433 F \t (C)\t 0.732 F\t A 30\u00b0 60\u00b0 B 90\u00b0 2\t 6.\t A force of 600 N is applied to the brake drum of 0.6 m C diameter in a band brake. System as shown in below W = 600 N figure, where the wrapping angle is 180\u00b0c. If the coef- \t20.\t The tension in the wire BC will be ficient of friction between the drum and band is 0.25, \t (A)\t519.6\t (B)\t613.4 the breaking lorque applied, in Nm is \t (C)\t 318\t (D)\t435.5 600 N 2\t 1.\t The tension in the wire AC will be \t (A)\t256\t (B)\t300 \t (C)\t 311\t (D)\t288 \t22.\t The smallest angle \u03b8 for equilibrium of the homoge- \t (A)\t 97.8 N\t (B)\t 16 N nous ladder of length l is, when coefficient of friction \t (C)\t 22.1 N\t (D)\t 15.7 N for all surfaces is assumed as m: (A)\t tan -1 \u239b 1 -m 2 \u239e \u239c\u239d 2m \u239f\u23a0 \t \t(B)\ttan-1 m2 \t27.\t A circular roller of weight 200 N and radius of 0.8 m 2 hangs by a tie rod of length 2 m and rests on a smooth vertical wall as shown in figure. The tension \u2018T\u2019 in the \t(C)\ttan -1 \u239b 2m \u239e tie rod will be \u239d\u239c 1- m \u23a0\u239f \t (A)\t219.78 \t(D)\ttan -1 \u239b m 2- 1\u239e \u239d\u239c 2 \u23a0\u239f \t (B)\t239.2 \t (C)\t310.30 \t (D)\t250.5","Test\u2002 |\u2002 3.91 2\t 8.\t A mass of 50 kg is suspended from a weight less bar \t(C)\tRx = 1080 N; Ry = 0 \u2018AB\u2019 which is supported by a cable BC and pinned at \t(D)\tRx = 755. N, Ry = 0 \u2018A\u2019 as shown in figure. The Pin reactions at \u2018A\u2019 on the bar AB are Direction for questions 28 and 29: All the forces acting on a particle are situated at the origin of the two dimensional refer- C ence frame. One force has a magnitude of 10 N acting in the positive \u2018X\u2019 direction, whereas the other has a magnitude of 125 mm T 5 N acting at an angle of 120\u00b0 directed away from the origin 2\t 9.\t The value of the resultant force will be. A B \t (A)\t 5.88 N\t (B)\t 7.2 N 75 mm 50 kg \t (C)\t 7.98 N\t (D)\t 8.66 N \t(A)\tRx = 343.4 N, Ry = 755.4 N \t(B)\tRx = 343.4 N, Ry = 0 \t30.\t The value of a made by resultant with the horizontal force will be \t (A)\t43\u00b0\t (B)\t30\u00b0 \t (C)\t78\u00b0\t (D)\t80\u00b0 Answer Keys \t1.\u2002D\t2.\u2002A\t3.\u2002C\t4.\u2002A\t5.\u2002A\t6.\u2002A\t7.\u2002D\t8.\u2002D\t9.\u2002B\t10.\u2002C \t11.\u2002B\t12.\u2002B\t13.\u2002C\t14.\u2002B\t15.\u2002C\t16.\u2002D\t17.\u2002A\t18.\u2002C\t19.\u2002D\t20.\u2002A \t21.\u2002B\t22.\u2002A\t23.\u2002C\t24.\u2002A\t25.\u2002A\t26.\u2002A\t27.\u2002A\t28.\u2002C\t29.\u2002D\t30.\u2002B","Engineering Mechanics Test 1 Number of Questions 35\bTime:60 min. Directions for questions 1 to 35: Select the correct alterna- \t8.\t Two particles with masses in the ratio 1 : 9 are moving tive from the given choices with equal kinetic energies. The magnitude of their lin- ear momentums will conform to the ratio. \t 1.\t A body is moving in a curved path with speed of 10 (A)\t 1 : 3\t (B)\t 1 : 9 m\/s and tangential acceleration of 3 m\/s2. If radius of curvature be 25 m, the total acceleration of body in (C)\t 3 : 1\t (D)\t 3 : 1 m\/s2 is (B)\t 4 \t9.\t Match List \u2013 I with List \u2013 II (A)\t3\t (C)\t5\t (D)\t6 LIst \u2013 I List \u2013 II \t2.\t A stone of mass m at the end of a string of length \u2018\u2193\u2019 P. Collision of particles 1. Euler\u2019s equation of motion is whirled in a vertical circle at a constant speed. The tension in the string will be maximum when the stone is Q. Stability 2. Minimum kinetic energy (A)\t at the top of the circle (B)\t half way down from the top R. Satellite motion 3. Minimum potential energy (C)\t quarter \u2013 way down from the top (D)\t at the bottom of the circle S. Spinning top 4. Impulse momentum principle \t3.\t For maximum range of a projectile, the angle of projec- \t P\tQ\t R\t S tion should be (A)\t1,\t 2,\t 3,\t 4 (A)\t30\u00b0\t (B)\t 45\u00b0 (B)\t4,\t 2,\t 1,\t 3 (C)\t60\u00b0\t (D)\t90\u00b0 (C)\t3,\t 1,\t 4,\t 2 (D)\t4,\t 3,\t 2,\t 1 \t4.\t The resultant of two forces P and Q inclined at angle q \t10.\t A spring scale indicates a tension to 10 N in the right will be inclined at following angle with respect to P. hand cable of the pulley system shown in the figure. Neglecting the mass of the pulleys, ignoring friction (A)\t tan\u22121 \uf8eb P Q sin q \uf8f6 \t(B)\ttan\u22121 \uf8eb Q Cos q \uf8f6 between the cable and pulley the mass m is (Take g = \uf8ec\uf8ed + Q Cos q \uf8f8\uf8f7 \uf8ec\uf8ed Q + P Sin q \uf8f8\uf8f7 10 m\/s2) (C)\t tan\u22121 \uf8eb Q P Sin q \uf8f6 \t(D)\tTan\u22121 \uf8eb P Cos q \uf8f6 \uf8ec\uf8ed + P Cos q \uf8f7\uf8f8 \uf8ec\uf8ed Q + P Sin q \uf8f8\uf8f7 \t5.\t If n = number of members and j = number of joints, then for a perfect frame n = (A)\t j \u2013 2\t (B)\t 2j \u2013 1 (C)\t2j \u2013 3\t (D)\t 3j \u2013 2 \t6.\t A body of mass 15 kg moving with velocity of 2 m\/s is acted upon by a force of 75 N for two seconds. The final velocity will be T (A)\t 10 m\/s\t (B)\t 11 m\/s Spring scale (C)\t 12 m\/s\t (D)\t 15 m\/s \t7.\t Two blocks with masses 10 kg and 5 kg are in contact m with each other and are resting on a horizontal friction- less floor as shown in figure. When horizontal force (A)\t 10 kg\t (B)\t 40 kg 600 N is applied to the heavier, the blocks accelerate to (C)\t 1 kg\t (D)\t 4 kg the right. The force between the two blocks is 1\t 1.\t Two bodies htoh1feamnradatsihso2mroe1fsaptniemdcetmiv2teaalkyree.nNdrteoogpldeprceotdpinfgtrhotomhuegdehiffffteehcre-t ent heights 600 N of friction, 10 kg given heights would be 5 kg \t(B)\t\uf8ec\uf8ed\uf8eb hh12 \uf8f6\uf8f7\uf8f81\/2 (A)\t m1 m2 (A)\t 300 N\t (B)\t 200 N \uf8eb h1 \uf8f6 2 \t(D)\t\uf8ec\uf8ed\uf8eb hh12 \uf8f6\uf8f8\uf8f71\/2 (C)\t 100 N\t (D)\t 50 N \uf8ed\uf8ec h2 \uf8f8\uf8f7 (C)","3.6\u2002|\u2002Engineering Mechanics Test 1 1\t 2.\t A sphere \u2018M\u2019 impinges directly on to another identical O sphere \u2018N\u2019 at rest. If the co\u2013efficient of restitution is 0.5, 200 60\u00b0 the ratio of velocities VN after the impact would be VM (A)\t 1 : 3\t (B)\t 3 : 1 (C)\t 1 : 2\t (D)\t 2 : 1 1\t 3.\t A block of steel is loaded by a tangential force on its (A)\t 150 mm\t (B)\t 140 mm top surface while the bottom surface is held rigidly. The (C)\t 128 mm\t (D)\t 108 mm deformation of the block is due to (A)\t Shear only\t (B)\t Torsion only 1\t 9.\t A uniform rod PQ remains in equilibrium position rest- (C)\t Bending only\t (D)\t Shear and bending ing on a smooth inclined plances PO and QO which are at an angle of 90\u00b0 as shown in figure. \t14.\t The co-efficient of friction depends on (A)\t Nature of the surface\t (B)\t Area of contact P (C)\t Strength of surface\t (D)\t All of the above \u03b8 \t15.\t The force induced in member PQ due to load W in fig- Q ure will be 90\u00b0 RQ \u03b1 O If the plane QO makes angles of a with the horizontal, then what is the inclination q of the rod PQ with the \u03b8 plane PO P W (A)\t equal to a\t (B)\t less than a (A)\t W secq\t(B)\tW cosq (C)\t greater than a\t (D)\t equal to 90\u00b0 (C)\t W tanq\t(D)\tW cosecq 1\t 6.\t A 10 m long ladder is placed against a smooth vertical 2\t 0.\t A uniform wheel of 500 mm diameter, weighing 5 kN wall with its lower end 3 m from the wall. For ladder to rests against a rigid rectangular block of 100 mm height remain in equilibrium as shown in figure, what should as shown in figure. The least pull, through the centre be the co-efficient of friction between ladder and floor? of the wheel, required just to turn the wheel over the corner A of the block is A 10 m O 500 mm A 100 mm \u00b5W W (A)\t 2 kN\t (B)\t 3 kN 3m (C)\t 4 kN\t (D)\t 5 kN \u03b8 B (A)\t0.16\t (B)\t 0.25 2\t 1.\t A hollow semicircular section has it outer and inner (C)\t0.36\t (D)\t0.45 diameter of 200 mm and 150 mm respectively shown in figure. The moment of inertia about the base AB in \t17.\t Three forces acting on a particle in equilibrium are 2P mm4 is and 3 P. Angle between 2P and P is 120\u00b0. What will be angle between P and 3 P. (A)\t45\u00b0\t (B)\t 60\u00b0 150 mm (C)\t90\u00b0\t (D)\t135\u00b0 200 mm \t18.\t A plane lamina of 200 mm radius is shown in figure (A)\t5 \u00d7 106\t (B)\t10 \u00d7 106 given below. The centre of gravity of lamina from the (C)\t17 \u00d7 106\t (D)\t27 \u00d7 106 point O","2\t 2.\t What is the maximum load (W) which a force P equal Engineering Mechanics Test 1\u2002|\u20023.7 to 6 kN will hold up, if the co\u2013efficient of friction at C is 0.2 in the arrangement shown in figure, neglect other 12 kg friction and weight of the member? A P 1m 0.5m 40mm 60mm B 6 kg W (A)\t 39.2 N\t (B)\t 25.0 N (C)\t 12.5 N\t (D)\t 6 N (A)\t 2.7 kN\t (B)\t 3.5 kN (C)\t 4.0 kN\t (D)\t 5.0 kN \t26.\t A simple pendulum consists of a 500 mm long chord and a bob of mass 2 kg is suspended inside a train, accelerating smoothly on a level track at the rate of 3.2 \t23.\t A load of 3 kN is to be raised by a screw jack with m\/s2. Find the angle which the chord will make with the mean diameter of 60 mm and pitch of 10 mm. The co\u2013 efficient of friction between the screw and nut is 0.075. vertical. (A)\t12\u00b0\t (B)\t 14\u00b0 The efficiency of screw jack is (C)\t16\u00b0\t (D)\t18\u00b0 (A)\t38.54%\t (B)\t 41.24% (C)\t42.25%\t (D)\t44.15% \t27.\t A body of mass 0.6 kg oscillates about an axis at a dis- tance 300 mm from the centre of gravity. If the mass \t24.\t A system of masses connected by string, passing over moments of inertia about the centroidal axis, parallel to pulley A and B is shown in figure. the axis of rotation, be 0.125 kg-m2, the length of the equivalent simple pendulum is. (A)\t 0.6 m\t (B)\t 0.8 m (C)\t 0.9 m\t (D)\t 1.0 m A 2\t 8.\t A conical pendulum 2 m long is revolving at 35 revolu- tions per minute. Find the angle which the string will make with the vertical, if the bob describes a circle of 500 mm radius. (A)\t15.2\u00b0\t (B)\t 29.8\u00b0 (C)\t32.5\u00b0\t (D)\t35\u00b0 20 kg B \t29.\t A rod of length 2 m is sliding in a corner, as shown in figure. At an instant when the rod makes an angle of 55 5 kg degrees with the horizontal plane, the angular velocity of the rod is 5 rad\/s. The velocity of the rod at point B is 7 kg The acceleration of mass 20 kg is A (A)\t 2.45 m\/s2\t (B)\t 2.01 m\/s2 2m (C)\t 1.89 m\/s2\t (D)\t 1.255 m\/s2 55\u00b0 \t25.\t A solid body A of mass 12 kg, when it is being pulled B by another body B of mass 6 kg along a smooth hori- zontal plane as shown in figure. The tension in the string is (Take g = 9.8 m\/s2)","3.8\u2002|\u2002Engineering Mechanics Test 1 (A)\t 1.50 m\/s\t (B)\t 2.50 m\/s \t32.\t The minimum value of P for which the equilibrium can (C)\t 5.01 m\/s\t (D)\t 8.19 m\/s exist. (A)\t 45.25 kN\t (B)\t 55.65 kN 3\t 0.\t A mass 40 kg is suspended from a weightless bar AB (C)\t 85.55 kN\t (D)\t 105.25 kN which is supported by a cable CB and a point at A, as shown in figure. The tension in the cable is \t33.\t The maximum value of P for which the equilibrium can exist. C (A)\t 115.5 kN\t (B)\t 250.5 kN (C)\t 350.5 kN\t (D)\t 451.5 kN 100 mm Linked Answer for Questions 34 and 35: A The figure below shown a pair of pin jointed gripper tongs B holding an object weighting 1500 N. The co-efficient of friction (m) at the gripping surface is 0.1. X \u2013 X is the line of 200 mm action of the input force and Y \u2013 Y is the line of application of gripping force. m Assuming pin joint is friction less. (A)\t 876.5 N\t (B)\t 755.5 N XF F (C)\t 654.5 N\t (D)\t 500 N X \t31.\t An elevator weighting 1000 kg attains an upward veloc- 200 mm ity of 4 m\/sec in two sec with uniform acceleration. The tension in the supporting cables will be (A)\t 2000 kg\t (B)\t 1200 kg Pin (C)\t 1000 kg\t (D)\t 800 kg Y 100 mm Y Common Data for Questions 32 and 33: A body of weight 600 N is lying on a rough plane inclined 150 N at an angle of 25\u00b0 with the horizontal. It is supported by an effort (P) parallel to the plane as shown in figure. The angle of friction is 20\u00b0 P \t34.\t The reaction force at the gripping surface is (A)\t 10,000 N\t (B)\t 7500 N (C)\t 5000 N\t (D)\t 3750 N \t35.\t The magnitude of force F required to hold the weight is (A)\t 7500 N\t (B)\t 5000 N (C)\t 3750 N\t (D)\t 2000 N F 600 N 25\u00b0 Answer Keys \t 1.\u2002C\t 2.\u2002D\t 3.\u2002B\t 4.\u2002A\t 5.\u2002C\t 6.\u2002C\t 7.\u2002B\t 8.\u2002A\t 9.\u2002D\t 10.\u2002D 11.\u2002B\t 12.\u2002B\t 13.\u2002D\t 14.\u2002A\t 15.\u2002A\t 16.\u2002A\t 17.\u2002C\t 18.\u2002C\t 19.\u2002A\t 20.\u2002C 21.\u2002D\t 22.\u2002A\t 23.\u2002B\t 24.\u2002A\t 25.\u2002A\t 26.\u2002D\t 27.\u2002C\t 28.\u2002B\t 29.\u2002D\t 30.\u2002A 31.\u2002B\t 32.\u2002B\t 33.\u2002D\t 34.\u2002B\t 35.\u2002C","Hints and Explanations \t1.\t Radial acceleration ar = V2\/r = 102 = 4 m\/s2 \t11.\t s = ut + gt 2 25 2 Tangential acceleration at = 3 m\/s2 u = 0, H1 = gt 2 \u2234\t Total acceleration a = ar2 + at2 2 \t= 42 + 32 = 5 m\/s2\b Choice (C) H2 = gt 2 2 Q Sin q \uf8eb t1 \uf8f62 \uf8eb h1 \uf8f6 t1 \uf8eb h1 \uf8f6 1\/ 2 + Q Cos q \uf8ed\uf8ec t2 \uf8f7\uf8f8 \uf8ed\uf8ec h2 \uf8f7\uf8f8 t2 \uf8ec\uf8ed h2 \uf8f8\uf8f7 \t4.\t Tan q = P \b Choice (A) = ; = \b Choice (B) \t6.\t Force = mass \u00d7 acceleration Choice (C) \t12.\t let sphere M = body 1 75 = (15) \u00d7 a ; a = 5 m\/s2 sphere N = body 2 Velocity after 2 seconds vm11,avn2darme2thareevmelaoscsitoiefsspbheeforereMimapnadcstphere N v = u + at v11, v21 are the velocities after impact = 2 + 5 \u00d7 2 = 2 + 10 = 12 m\/s\b \t7.\t Let N force b\/w the block. Form free body diagram. (i) \u2234\t m1v1 = m1 . v11 + m2 v21 600 \u2013 N = (10)a \b N = (5)a \b (ii) \t As the balls are identical m1 = m2 From (i) & (ii) \t v1 = v11 + v21 600 \u2013 N = 10 \uf8eb N \uf8f6 \t Given that 0.5 = v21 \u2212 v11 = v21 \u2212 v11 \uf8ed\uf8ec 5 \uf8f7\uf8f8 v1 \u2212 v2 V1 3N = 600 Choice (B) \t(\\\\ V2 = 0) N = 200 N\b v12 \u2212 v11 \u2234\t 0.5 = v11 + v12 \t8.\t KE1 = m1 v12 \t v12 + v11 = v12 \u2212 v11 2 2 2 KE2 = m2 v22 \t 3 v11 = v12 2 2 2 Given that KE1 = KE2 v21 m1 1 v11 m2 = 9 \t = 3 : 1\b Choice (B) m1v12 = m2v22 1\t 3.\t Choice (D) \uf8eb v1 \uf8f62 m2 v1 1\t 4.\t Choice (A) \uf8ec\uf8ed v2 \uf8f7\uf8f8 m1 v2 = = 9; = 9 = 3. 1\t 5.\t Cosq = W PQ Momentum ratio = m1v1 = m1 \u00d7 v1 PQ = W = W Secq\b Choice (A) m2v2 m2 v2 Cos q = 1 \u00d7 3 = 1 \b Choice (A) \t16.\t Taking moments about A 9 1 3 10 W \u00d7 3\u2212W \u00d7 2 Cos q = mW \u00d710Sin q 1\t 0.\t referring figure 4T = Mg 3 91 10 10 M= 4T 3 \u2212 5 \u00d7 = m \u00d710 \u00d7 g 91 = 4 \u00d710 From the figure Cosq = 3\/10 , Sinq = 10 10 3 \u2013 3\/2 = m \u00d7 9.53 M = 4 kg\b Choice (D) 1.5 m= 9.53 = 0.157 \u2248 0.16\b Choice (A)","3.10\u2002|\u2002Engineering Mechanics Test 1 1\t 7.\t 3P = 2P Sinq = 150 = 0.6 Sin120 Sin q 250 2P q = 36.86\u00b0 AB = (250)2 \u2212 (150)2 = 200 mm P O 120\u00b0 \u03b8 250 mm P \u221a3 P \u03b8 150 mm RA 3 = 2P B sin120 Sin q Taking moments about A 5 kN P \u00d7 250 = 5 \u00d7 200 3 2 = 2P \u21d2 Sinq = 1 3\/ Sin q q = 90\u00b0\b Choice (C) 0 \t18.\t As the lamina is symmetrical about y \u2013 y axis, bisecting 150 mm the lamina, its centre of gravity lies on the axis. 100 mm A p a = 30\u00b0 = 6 Centre of gravity of the lamina y = 2r Sin a 1000 3a 250 P = = 4 kN\b Choice (C) = 2 \u00d7 200 \u00d7 Sin 30 = 400 \u00d7 0.5 2\t 1.\t D = 200 mm, R = 100 mm 3 p 3 d = 150 mm, r = 75 mm \uf8eb\uf8ec\uf8ed p \uf8f8\uf8f7\uf8f6 I=AB0=.3903.3[9130(0R4 4\u2013\u20137r544)] = 26.86 \u00d7 106 mm4\b 6 6 = 128 mm\b Choice (C) Choice (D) 1\t 9.\t \t22.\t Let R = normal reaction of the pulley on the beam at C R \u00d7 1 = 6 \u00d7 1.5 R R = 9 kN Maximum force of friction at C = mR P = 0.2 \u00d7 9 = 1.8 kN RP \u03b8 Taking moments about the centre of pulley G W \u00d7 40 = 1.8 \u00d7 60 Q W = 1.8 \u00d7 60 = 2.7 kN\b Choice (A) RQ 40 \u03b1 2\t 3.\t Load (W) = 3 kN O Mean diameter of the screw (d) = 60 mm Pitch (p) = 10 mm PR ^ PO and QR ^ QO therefore PR || QO m = tanf = 0.075 QR || PO \u2220PRQ = 90\u00b0 tan(a) = P = 10 = 0.053 pd p \u00d7 60 PG = QG \u2220GPO = \u2220GOP tan a q = a\b Choice (A) Efficiency h = tan (a + j) 2\t 0.\t Diameter of the wheel = 500 mm = tan a = 0.053 Weight of wheel = 5 kN tan a + tan j 0.053 + 0.075 Height of the block = 100 mm Let P = Least pull required just to turn the wheel in kN 1\u2212 tan a tan j 1\u2212 (0.053 \u00d7 0.075) 0.4124 = 41.24%\b Choice (B)","Engineering Mechanics Test 1\u2002|\u20023.11 2\t 4.\t LFreotmm1th=e2s0ysktgem, mo2f=p7u0llekygs, amn3d=m5akssges, we find that at 0.125 =00.01..6265KG=2 0.208 pulley A, the 20 kg mass will come down with some KG2 = acceleration as the total mass on the other side of the Length of equivalent simple pendulum string is less than 20 kg At pulley B, the 7 kg mass will come down with some KG2 0.208 acceleration. L= h+ h = 0.3 + 0.3 Acceleration of 20 kg mass is = 0.993 m\b Choice (C) = g (m1 \u2212 m2 ) = 9.81(20 \u2212 (7 + 5)) 2\t 8.\tL = 2 m N = 35 rpm m1 + m2 20 + (7 + 5) r = 500 m = 0.5 m = 2.45 m\/s2\b Choice (A) angular velocity of the bob w = 2pN = 2p \u00d7 30 60 60 2\t 5.\t gmLe2=t=9m.681 k=mg1\/s22 kg w = 3.35rad\/s Tension in the string, T = m1 m2 g = 12 \u00d7 6 \u00d7 9.8 tanq = w2r = (3.35)2 \u00d7 0.5 = 0.572 m1 + m2 12 + 6 g 9.8 = 39.2 N\b Choice (A) q = tan\u20131(0.572) = 29.8\u00b0\b Choice (B) 2\t 9.\t \t26.\t Let q = Angle, which the chord will make with the vertical A I 3m\/s2 \u03b8T 2m T 55\u00b0 ma OB mg Weight of the hob = mg = 2 \u00d7 9.8 = 19.6 N w = 5 rad\/s Choice (D) Inertia force acting on the hob (Opposite to the accel- VVI\t\u2234IABAB\t====OOVV=VAeBeB5llo==o=\u00d7ccLwiLi2ttsyyc\u00d7i\u00d7noaaIqlsSlBooqi=nnn=g5g252tt\u00d7h\u00b0h\u00d7ee=sivchn8eoo5.rs1r5t5ii9z\u00b0c5oam\u00b0nl \/tsa\bl Choice (A) eration of the train) Choice (B) = ma = 2 \u00d7 3.2 = 6.4 N \t30.\t T Cos(90 \u2013 q) = mg \u03b8 mg T Sinq = mg 100 ma Tanq = 200 \\\\\t Tanq = 6.4 q = 26.56\u00b0 40 \u00d7 9.8 19.6 mg T = Sin q = Sin (26.56) \t q = 18\u00b0\b Choice (D) 2\t 7.\t m = 0.5 kg, h = 300 m = 0.3 m = 876.5 N\b O 3\t 1.\t Velocity = acceleration \u00d7 time h 4=a\u00d72 G a = 2 m\/sec2 W P g Tension in Cable = (g + a) IG = 0.125 kg-m2 = 1000 (9.81 + 2) = 11, 810 IG = m KG2 9.81 9.81 \u2248 1200 kg","3.12\u2002|\u2002Engineering Mechanics Test 1 3\t 2.\t For the minimum value of P, the body is at the point of 2mR = 1500 N sliding downwards. 1500 R= 2 \u00d7 0.1 = 7500 N\b Choice (B) Sin (a \u2212 j) Choice (C) \u2234\t P = W \u00d7 3\t 5.\t Cos j \t = 600 \u00d7 Sin (25 \u2212 20) F 200 Cos 20 100 \t = 55.65 kN\b Choice (B) R \t33.\t For the maximum value of P, the body is at the point of sliding upwards. \u2234\t P = W \u00d7 Sin (a + j) P Cos j \t\u2002 = 600 \u00d7 Sin (25 + 20) Cos 20 \t\u2002 = 451.5 kN\b Choice (D) \t34.\t \u00b5R \u00b5R Taking moments about Pin (R) \u00d7 100 = F \u00d7 200 7500 \u00d7100 RR F= 200 F = 3750 N\b 1500N","Engineering Mechanics Test 2 Number of Questions 35\bTime:60 min. Directions for questions 1 to 35: Select the correct alterna- (A)\t 5 m\/s2\t (B)\t 2.5 m\/s2 tive from the given choices. (C)\t 10 m\/s2\t (D)\tZero \t1.\t Two forces of 500 N and 600 N are acting simultane- \t7.\t If two bodies one light and other heavy have equal ously at a point. If the angle between them is 60o then kinetic energies and equal mass then which one has a the resultant of these two forces is greater momentum? (A)\t 781 N\t (B)\t 954 N (A)\t Heavy body (C)\t 1063 N\t (D)\t 881 N (B)\t Light body (C)\t Both have equal momentum \t2.\t A flywheel 400 mm in diameter is brought uniformly (D)\t None of these from rest to a speed of 240 rpm in 16 seconds. The tangential acceleration of a point on the rim (in m\/s2) is \t8.\t A ball of mass 9.81 kg is thrown with an angle a to (A)\t1.57\t (B)\t 0.628 the horizontal with a velocity of 9.905 m\/s. What is the (C)\t0.314\t (D)\t0.419 maximum range the ball reaches. (A)\t 96.2361 m\t (B)\t 9.81 m \t3.\t A stone of mass 5 kg is tied to a spring of length 2 m (C)\t 1 m\t (D)\t 10 m and whirled in a horizontal circle at a constant angular speed of 10 rad\/sec. The tension in the spring will be \t9.\t A rescue airplane flying at a height of 500 m from (A)\t 1000 N\t (B)\t 750 N ground for a flood affected area drops a rescue kit trav- (C)\t 500 N\t (D)\t 250 N eling at 200 m\/s. How much distance does the airplane travel from the point of releasing the kit to the point of \t4.\t Ratio of moment of inertia of a sphere and that of a the kit hitting the ground. (Neglect air resistance) cylinder having same radius and mass about their cen- troidal axis is (A)\t 20.387 km\t (B)\t 20.387 m (C)\t 2.0192 m\t (D)\t 2.019 km (A)\t 1 \t(B)\t52 5 \t10.\t A thin solid circular disc of 10 kg is applied by a torque through a shaft connecting at the center of the disc. If 2 4 the angular velocity reached is 5 rad\/sec what is the (C)\t 5 \t(D)\t5 amount of angular impulse acted upon the circular disc. \t5.\t A pulley and rope arrangement is shown below (take r = 4 m and initially the disc is at rest). (A)\t 200 Nms\t (B)\t 400 Nms (C)\t 500 Nms\t (D)\t 1000 Nms 1\t 1.\t The angular speed of the seconds hand in a clock in rad\/min is (A)\t p \t (B)\t120 p 30 (C)\t2 p\t (D)\t60 p 40 kg P 1\t 2.\t Determine the point of action of the resultant of forces (Hold by a person) acting on the inclined plane as shown in figure. If coefficient of friction between pulley and rope is 0.25 50 N then the holding load by the person will be (A)\t 178.91 N\t (B)\t 860.64 N 30 N (C)\t 294.21 N\t (D)\t 741.23 N B \t6.\t The velocity-time graph of a body is shown in the fig- A AB = 40 mm ure. The acceleration at point A will be 30\u00b0 V A (A)\t 20 mm from A\t (B)\t 20 mm from B (m\/s) (C)\t 25 mm from A\t (D)\t 25 mm from B 10 1\t 3.\t A car is moving along a straight road according to the equation x = 4t3 + t + 7, where x is in meters and t is in seconds. What is the average acceleration during the 2 t (seconds) fifth second?","3.14\u2002|\u2002Engineering Mechanics Test 2 (A)\t 108 m\/s2\t (B)\t 109 m\/s2 6N (C)\t 110 m\/s2\t (D)\t 112 m\/s2 10\u00b0 5 m\/s \t14.\t The magnitude of the force of friction between two P 20\u00b0 bodies, one lying above the other depends upon the roughness of the 30\u00b0 (A)\t upper body (B)\t lower body (C)\t both the bodies (D)\t the body having more roughness 1\t 5.\t If the sum of all the forces acting on a body is zero, then the body may be in equilibrium provided the forces are (A)\tParallel\t (B)\t Concurrent (A)\t 612 N\t (B)\t 574 N (C)\t Coplanar\t (D)\t Unlike parallel (C)\t 438 N\t (D)\t 451 N 1\t 6.\t Two sphere of same radius of 100 mm and same mass 1\t 9.\t Particle A of mass \u2018m\u2019 is tied with 2 m cord at the instant of 0.5 kg are in equilibrium within a smooth cup of shown in figure. At this instant angular velocity is 2.83 radius 300 mm as shown in the figure. Reaction force rad\/sec. What will be the angular velocity (in rad\/sec) between the cup and one sphere will be when the angle turned by cord is 45o? O O 30\u00b0 Cord (2m) Sphere A B cup A C D (A)\t2.54\t (B)\t 2.83 R R (C)\t2.62\t (D)\t2.93 (A)\t 5.14 N\t (B)\t 4.91 N 2\t 0.\t A 200 mm diameter pulley on a generator is being (C)\t 6.41 N\t (D)\t 5.66 N turned by a belt moving with 25 m\/s and accelerating with 8 m\/s2. A fan with an outside diameter of 300 mm is attached to the pulley shaft. Linear acceleration of 1\t 7.\t A A-Frame is shown in the given figure. Floor reaction the tip of the fan (in m\/s2) is at A and vertical pin reaction at D are respectively. (A)\t12\t (B)\t 3475 C (C)\t5671\t (D)\t9375 B D \t21.\t A homogenous cylinder of radius \u2018R\u2019 and mass \u2018m\u2019 is 4 cm 5 cm acted upon by a horizontal force \u2018P\u2019 applied at various positions along a vertical centre line as shown in the 60\u00b0 45\u00b0 E figure. Assume movement upon a horizontal plane. At A 12 cm 9 cm what radius above the centre (h) should the force \u2018P\u2019 is applied so that the frictional force \u2018F\u2019 is zero? 180 y (A)\t 757 N and 1142 N P (B)\t 612 N and 1013 N h (C)\t 757 N and 1241 N (D)\t 612 N and 1142 N R 1\t 8.\t A block of mass 50 kg is placed on an inclined surface, mg as shown in the figure. Coefficient of friction between block and surface is 0.3. Find the value of force \u2018P\u2019 F required to be applied on the block to maintain uniform R velocity of 5 m\/s?","Engineering Mechanics Test 2\u2002|\u20023.15 (A)\t R\t(B)\tR3 Bar sphere O G (C)\t R \t(D)\tR4 2 \t22.\t Figure shows the line diagram of connecting rod AB of 0.5 m 0.2 m a slider crank mechanism. I is the instantaneous center of rotation of the rod. (A)\tZero\t (B)\t 11.13 (C)\t16.17\t (D)\t19.67 VA = 12 m\/s I \t27.\t ABCD is a square which forms a plane truss with load 60\u00b0 AI = 1.2 m P at point A. What is the axial force in the bar 1. A BI = 1.6 m AB = 2 m B1 B 5 2 30\u00b0 A 4C VB P6 3 Relative velocity of A and B is D (A)\t 20 m\/s\t (B)\t 18 m\/s (A)\t2 P\t (B)\tP (C)\t 16 m\/s\t (D)\t 14 m\/s (C)\t 0.707 P\t (D)\t 1.414 P Statement for linked answer questions 23 and 24: 2\t 8.\t A stool rests on a smooth horizontal floor and is loaded with a load P. What is the value of a to have maximum 730 N\/m 2 cm (radius) shear force at the point E. O 40 kg \u03b1\u00d7a P AB A 50 kg cylinder of radius 0.4 m rolls without slipping under a the action of an 40 kg force. A spring is attached to a cord E that is wound around the cylinder. The spring is streched when the 40 kg force is applied. CD 2\t 3.\t When the cylinder is moved by 0.15 m then the total a\/2 a\/2 work done will be (B)\t 26 N-m (A)\t0\t (B)\t 1 (A)\t 33 N-m\t (C)\t 0 or 1\t (D)\t None of these (C)\t 92 N-m\t (D)\t 59 N-m \t29.\t A body of mass m is suspended by a string of length L. 2\t 4.\t What is the speed of the center of the cylinder after it The body traces a horizontal circle of radius 2 m when has moved 0.15 m? (A)\t 0.613 m\/s\t (B)\t 0.921 m\/s the semicone angle at the top is 30o. If the centrifugal force of the non-suspended body of same mass is 23 N (C)\t 0.833 m\/s\t (D)\t 0.731 m\/s while rotating in a circle of radius 4 m with the same \t25.\t A uniform chain of mass 10 kg and length 1 m lies on a angular velocity, what is the tension in the string. smooth table such that one-fourth of its length is hang- ing vertically down over the edge of the table. Work (A)\t23 N\t (B)\t 11.328 N (C)\t 19.91 N\t (D)\t 11 N done to pull the hanging part of the chain on the table is (A)\t 3.065 J\t (B)\t 12.625 J 3\t 0.\t During the replacement of machines in a workshop (C)\t 24.525 J\t (D)\t 6.131 J floor a nail was protruding 5 mm from the floor level. A hammer of 5 kg mass of the head is used to strike \t26.\t A homogenous sphere of mass 1 kg is attached to the the nail to make it level with the floor. Consider the bar of negligible mass. In the horizontal position shown hammer as a free fall from a height of 100 mm and in the figure, the angular acceleration of the system completes the job in single strike. What is the mass of (in rad\/s2) is","3.16\u2002|\u2002Engineering Mechanics Test 2 the nail if the resistance offered by the floor is 1.032 \t34.\t The wheel of a trolley bag which is being pulled by a N\/mm force of 10 N (horizontal force) is of 50 mm radius. If (A)\t 45.87 gms\t (B)\t 98.77 gms the weight of the bag is 100 N what is the coefficient of (C)\t 198.77 gms\t (D)\t 99.385 gms rolling resistance in meters Common data for questions 31 and 32: P p \u03c6 = 30\u00b0 \u03b1 W = 30 N \u03c6R W W A block of 30 N weight is being pulled by a force P making (A)\t0.1\t (B)\t 0.005 an angle a with the horizontal. The reactive force R makes (C)\t0.5\t (D)\t0.001 an angle of 30o with the vertical (angle of friction, f). 3\t 5.\t A plane truss is loaded as shown in the figure. What is 3\t 1.\t What is the minimum force Pmin required to impede the magnitude of force in the member CD and is it in com- block to move pression or tension (A)\t15 3 N\t (B)\t 30 N B 100 N 100 N 20 D (C)\t 45 N\t (D)\t 15 N \t32.\t What is the value of a interms of the angle of friction w(Ah)e\tn2Pfm\tin is acting on the blo(cBk).\tf\/2 (C)\t f\t (D)\t 90 \u2013 f 20\u221a3 E \t33.\t Two balls of weights 6 N and 2 N are made to collide A 20 20 with each other. The velocities of the balls before col- lision are 4 m\/s and 8 m\/s respectively and the 2 N ball 500 N C is moving in opposite direction to 6 N ball. What is the ratio of velocities of the 6 N ball, after the collision, when the impact is considered to have a coefficient of (A)\t 692.82 N, compression restitution of 0.5 to when the impact is perfectly elastic. (B)\t 692.82 N, Tension (A)\t1\t (B)\t 0.25 (C)\t 1385.64 N, Compression (C)\t0.5\t (D)\t0.75 (D)\t 1385.64 N, Tension Answer Keys \t 1.\u2002B\t 2.\u2002C\t 3.\u2002D\t 4.\u2002D\t 5.\u2002B\t 6.\u2002D\t 7.\u2002C\t 8.\u2002D\t 9.\u2002D\t 10.\u2002A \t11.\u2002C\t 12.\u2002C\t 13.\u2002A\t 14.\u2002C\t 15.\u2002B\t 16.\u2002D\t 17.\u2002A\t 18.\u2002C\t 19.\u2002A\t 20.\u2002D \t21.\u2002C\t 22.\u2002A\t 23.\u2002B\t 24.\u2002C\t 25.\u2002A\t 26.\u2002C\t 27.\u2002B\t 28.\u2002C\t 29.\u2002A\t 30.\u2002C \t31.\u2002D\t 32.\u2002C\t 33.\u2002B\t 34.\u2002B\t 35.\u2002B Hints and Explanations \t1.\t Resultant force, R = F12 + F22 + 2F1F2 Cos q now Tangential acceleration, at = r \u00d7 a Choice (C) = 0.2 \u00d7 1.57 = 0.314 m\/s2\b Choice (D) or R = 5002 + 6002 + 2 \u00d7 500 \u00d7 600 \u00d7 Cos60o \t3.\t F = mw2 = 5 \u00d7102 R = 953.94 N ~ 954 N\b Choice (B) r2 \t2.\t Angular acceleration, a = w \u2212 wo t \u21d2\t F = 250 N\b \t4.\t For sphere, IGS = 2 mR2 5 \uf8ec\uf8ed\uf8eb 2p \u00d7 240 \uf8f7\uf8f8\uf8f6 \u22120 mR2 or a = 60 sec2 For cylinder, IGC = 2 =1.57 rad 16","Engineering Mechanics Test 2\u2002|\u20023.17 2 mR2 1\t 2.\t 5 \u2234 IGS = = 4 \b Choice (D) 50 IGC 5 mR2 30 AB = 40 mm 2 PB P = e0.25\u00d7 p Ax 40 \u00d7 9.81 ( )\t 5.\t R 50 \u221a3 2 \u2234\t P = 860.64 N\b Choice (B) 30\u00b0 Choice (D) 30 \t6.\t acceleration = dV 2 \u221a3 dt Let the point P be the point of action of the resultant \u2234\t Velocity is constant Taking moments around P. \u2234\t acceleration = 0\b \u03a3MP = 0 \t 7.\t \uf8ee1 mV 2 \uf8f9 = \uf8ee1 mV 2 \uf8f9 ( )30 50 \uf8f0\uf8ef 2 \uf8fb\uf8fa H \uf8ef\uf8f0 2 \uf8fb\uf8fa L 2 2 3\u00d7x \u2212 3 \u00d7(40 \u2212 x) = 0 NVHow= mVLomentum = m \u00d7 V \u21d2 30 3x = 50 3 (40 \u2212 x)\u21d280x = 200 \u21d2 x = 25 mm \u2234\t Both has same mass and velocities 2 2 \u2234\t Momentum is same for both. \b \b Choice (C) Choice (C) \t13.\t V = dx =12t 2 +1 dt \t8.\t Range of a projectile is given as V02 Sin2a g a dV = 24t = dt It is maximum when a = 45o \u2234\t change in acceleration during 5th second is V02 9.9052 98.109025 \u2234Rmax = g = 9.81 = 9.81 \u2206V = V5 \u2212V4 \uf8f0\uf8ee(12 \u00d7 25) +1\uf8fb\uf8f9 \u2212 \uf8ee\uf8f0(12 \u00d716) +1\uf8fb\uf8f9 \u2206t 1 Rmax = 10.00091 \u2243 10 m\b Choice (D) \t a= = 1 \t9.\t \t a = 108 m\/s2\b Choice (A) 200 m\/s 1\t 4.\t Choice (C) \t15.\t Choice (B) 500 m 1\t 6.\t Free body diagram of sphere A R R 2 \u00d7 500 RB 120\u00b0 60\u00b0 9.81 Time taken for the kit to reach ground = = 90\u00b0 10.096 sec Distance travelled by air plane in the time = 200 \u00d7 10.096 = 2019.2 m = 2.0192 km\b Choice (D) (0.5 \u00d7 9.81) N \t10.\t Angular impulse = I(w2 \u2013 w1) From geometry: OC = 300 mm, OA = 200 mm, OB = 200 mm, I= mr 2 10 \u00d7 42 = 40 kg m2 AB = 200 mm 2 =2 \u2234\t \u2206OAB is equilateral triangle. \u2234\t Applying Lami\u2019s theorem Angular Impulse = 40 \u00d7 (5 \u2013 0) = 200 Nms\bChoice (A) 1\t 1.\t A seconds hand rotates 2p radians in 60 seconds i.e., in ( )( )\t 1 minute. R = 0.5 \u00d7 9.81 Sin 90o Sin 120o \u2234\t wsec = 2p rad sec = 2p rad min \b Choice (C) 60 \u21d2\t R = 5.6638 N\b Choice (D)"]
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