SYSTEM OF PARTICLES AND ROTATIONAL MOTION - Lecture Notes

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) CHAPTER - 00 SYSTEM OF PARTICLES AND ROTATIONAL MOTION Rigid Body A rigid body is that which has a definite size and definite shape. This is because distances between different pairs of particles of such a body do not change on applying any force on it. All forces exerted by various particles of the system on one another are called internal forces. The internal forces between any two particles of the system are mutual ie., internal forces between a pair of particles are equal and opposite. Hence such forces cancel out in pairs. Kinds of motion of a rigid body 1. Pure translational motion In such a motion, every particle of the body travels the same linear distance in given time interval. eg: sliding motion of a block on an inclined plane 2. Pure rotational motion In this motion a rigid body rotates about a fixed axis. Every particle of the body moves in a circle, which lies in a plane  to the axis and has its centre on the axis. eg : motion of a ceiling fan, gaint wheel in a circus  Pure rotation is also possible about a fixed point 3. Combined motion of translation and rotation In this kind of motion, instantaneous velocity of every particles are different throughout the motion eg : rolling motion 1

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Centre of mass The centre of mass of a s/m of particles is a fixed point at which the whole mass of the s/m may be supposed to be concentrated for describing its translatory motion. If a body possess atleast 2 axis of symmetry the COM will lie on the point of intersection of the s/m. If a system possess only axis of symmetry then COM will lie on the axis depending upon mass distribution of the system COM of ‘n’ particle s/m   Consider a s/m of ‘n’ particles having masses m1, m2 .... mn and position vectors r1, r2, ... rn relative to origin O Special Cases I. For a 2 particle s/m and m1 = m2 = m         Rcm mr1 mr 2 r1  r2 mm 2 Thus centre of mass of two equal masses lies exactly at the centre of the line joining the two masses III. For 3 particle s/m and m1 = m2 = m3 = m       Rcm r1 r2  r3 3 The CM of 3 equal masses lies at the centroid off the triangle formed by them Cartesian Coordinates of CM If (x1y1) and (x2y2) are the coordinates of the location of two particles, the coordinates of their CM are given by Xcm  m1x1  m2 x 2 Ycm  m1y1  m2y2 m1  m2 m1  m2 2

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Analysis of two particle s/m in the frame of CM  mixi  mi  Xcm  O   mixi cm  O m1x1 + m2x2 = O m1 (–r1) + m2 (r2) = ) m1r1  m2r2 The term mr = constant is called mass moment i.e., r  1 implies the centre of mass of the s/m lies closer to the heavier particles m also r1  m2d r2  m1d m1  m2 m1  m2 CM of rigid bodies The number of particles in a rigid body of finite size is so large that it is impossible to carry out summation over individual particles in these equations. So we can treat the body as continuous mass distribution. Let us subdivide the body into ‘n’ small elements of mass m1, m2, m3... mn  Xcm   mixi   mi xi  mi M Ycm   miyi M If the value of ‘n’ is bigger the mi will become smaller in that case we can replace summation by integration Xcm   xdm  dm Ycm   ydm  dm 3

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) 1. Centre of mass of a uniform rod Consider a uniform rod of mass M and length L it is held along x axis with its end at origin. M  L L  x2 L  x dx  X  xdm  0   2  L  dm  0 2 dx  x L 0 Xcm  L Ycm  0 2 2. Uniform Semicircular Ring X0 Y  2R  4

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) 3. Semicircular Disc X0 Y  4R 3 4. Solid Hemisphere X0 Y  3R 8 5

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) 5. Hollow Hemisphere X0 YR 2 6. Solid Cone X0 Y h 4 6

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) 7. Hollow Cone X0 Yh 3 8. Triangular Plate X0 Yh 3 7

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) CM of s/m of rigid bodies For a s/m of rigid bodies their common centre of mass is given by    mi  R ri  mi X   mixi  mi Y   miyi  mi Velocity and acceleration of CM   m1r1 m2r2  ...  mn rn We know that, Rcm   M differentiating w.r.t. time      dRcm 1 m1 dr1 dr 2 drn  M  dt  m2 dt  ...  mn  dt dt    1     ...   V cm M m1 v1 m2 v2 mn vn     mi  V Vi cm M differentiating w.r.t. time again 8

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) d     d  1 dt Vcm   mi dt Vcm M    mi  Acm ai M Linear momentum for a s/m of particle For a s/m of n particles, the linear momentum of the s/m          Ps/m P1  P2  ...Pn  m1v1  m2 v2  ...  mn vn   mi vi MVcm   Ps/m M Vcm Motion of Centre of Mass   mi ai We know that Acm  M      MAcm  m1a1  m1a2  ...  mn an  F1  F2  ...  Fn  Fext   MAcm  Fext i.e., CM of a s/m of particles moves as if the entire mass of the s/m were concentrated at the centre of mass and all the external forces were applied at that point Conservation of linear momentum   d  [According to newtons law] dt Ps Fext m   MVcm  M dVcm   d dt dt in an isolated s/m, the vector sum of external forces acting on the s/m is zero ie   0   0 Fext  M dVcm dt  Vcm  const Velocity of centre of mass remains constant when total external force on the s/m is zero. 9

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Example 1. When a bomb at rest explodes, the fragments move in opposite directions such that the CM will remain at rest [explosion is due to internal forces only]. 2. When a fire cracker explodes in mid air the individual fragments flies off in different directions in such a way that their centre of mass will continue to move along the initial parabolic path [external force is gravity alone] 3. When a bullet is fired from a gun, the gun recoils backwards so that the CM will remain at the same position 4) Decay of radioactive nucleus If the CM of s/m is initially at rest and Fext=0  mixi  0 then  miyi  0 Moment of inertia (I) It is the rotational analogy of mass. A quantity that measures the inertia of rotational motion of a body is called rotational inertia or moment of inertia of the body. Moment of inertia of a body about a given axis of rotation as the sum of the products of masses of all the particles of the body and square of their respective  distances from the axis of rotation I   mir12 unit  kgm2  Tensor quantity 10

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Moment of inertia of rigid bodies I   dmr2 Parallel axis theorem The moment of inertia of a body about any axis is equal to its moment of inertia about 2 parallel axis through its centre of mass plus the product of the mass of the body and the square of its perpendicular distance b/w the two parallel axes. IAB  Icm  md2 Perpendicular axis theorem Moment of inertia of a plane lamina about any axis perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually  axes in the plane of lamina, meeting at a point where the given axis passes through the lamina Iz  Ix  Iy 11

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Moment of inertia of uniform rod Consider a uniform rod of mass M and length L rotating about an axis shown in the figure dI  dm x2  Mdx xp L L I  M 2 x2dx  ML2 L L 12 2 M.I of a circular disc Area of the disc = R2 M Mass per unit area of the disc  R2 Area of ring = 2xdx dm  M  2xdx  2Mxdx R2 R2 dmx2 R 2Mxd  x2 MR2  I   0 R2  2 12

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) About diameter Ix  Iy  MR2 2 Id  Id  MR2 2 Id  Mr 2 4 * about tangent in its plane = 5MR2 4 * about tangent  to its plane = 3MR2 2 I  ML2  M  L 2 12  2   I  ML2 13 3

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Moment of inertia of uniform ring Length of ring = 2R M Mass per unit length of ring  2R dm  M dx 2R 2R M dx  R2  MR2 dmR2   I  0 2R About diameter Ix  Iy  MR2 Id  Id  MR2 Id  MR2 2 14

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Hollow cylinder I1  MR2 I2  m  L2  R2   12 2    15

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Rectangular Lamina I2  m2 I3  mb2 12 12 I1  m 2  b2  12 I4  m2 I5  mb2 3 3 m l 2b2 6 2  b2  I6 16

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Solid sphere I  2 MR2 5 Hollow sphere I  2 MR2 Solid cone 3 I  3MR2 10 17

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Hollow cone I  MR2 2 Solid cylinder I1  MR2 2 I2  M  L2  R2  12 4   18

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Triangular lamina Mh2 I Square lamina 6 I1  I2  I3  I4  ml 2 I5  ml 2 12 6 Moment of inertia for a s/m of rigid bodies ml 2 2 mr2 2 3 5  IAB l  I1  I2   m r 19

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Radius of gyration Radius of gyration of a body about a given axis is the  distance of a point from the axis, where if whole mass of the body were concentrated, the body shall have the same moment of inertia as it has with the actual distribution of mass I  mr12  mr22  ...  mrn2  m r12  r22  ...  rn2   n n   nm   r12  r22  ...  rn2   n    I  M  r12  r22  ...  rn2     (1)  n    If total mass of the body M is concentrated at P I  MK2    (2) Comparing (1) and (2) K  r12  r22  ...rn2 n Torque The torque or moment of force is the turning effect of the force about the axis of rotation. It ismeasured as the product of magnitude of the force and the  distance b/w the line of action of the force and the axis of rotation.   F ON  F d Unit  Nm  it is a vector quantity  torque is the rotational analogy of force. • Clockwise torque is taken as negative and anticlockwise torque is taken as positive. 20

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Torque acting on a particle    Consider a particle P in XY plane suppose its position rector is OP  r . Let F be the force acting on the particle. Then the torque action or the particle about orgin. o  F OP  F r sin     o  rF sin  o  r F • When   0o , the line of action passes through orgin sin   0  o  0 • When   90o   rFsin 90  rF [torque is max imum] Fradial  F cos  Ftan gential  F sin    Fsin  r ie torque is due to tangential component of force. 21

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Force couple A pair of equal and opposite forces acting along two different lines of action constitute a force couple Fnet  F  F  0  net   0 Fd  Fd A Fnet  0 net  0 • Moment of force = Force × perpendicular distance b/w two forces • Torque ofa couple is independent of choice of point of rotation. If the net force acting on a system is zero then net torque about any point remains the same. Conditions for equilibrium of rigid body 1) Arigid body is said to be in translational equilibrium, if it remains at rest or moving a constant velocity in a particular direction. For this, the net external force or the vector sum of all the external forcesacting on the body must be zero.  ie;  F  0 it implies  Fx  0  Fy  0  Fz  0 2) Arigid body is said to be in rotational equilibrium, if the body does not rotate or rotates with constant angular velocity. For this, the net external torque or the vector sum of all the torques acting onthe body is zero. • A body is in rotational equilibrium, when algebraic sum of moments of all the forces acting on the body about a fixed point is zero.  ext  0 Examples for force couple • When we open the lid of a bottle by turning it, our fingers apply a couple on the lid. • When a compass needle is held arbitrarily in any direction in earth’s magnetic field, a couple actson the needle and aligns it along north- south direction. 22

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Centre of gravity The centre of gravity of a body is a point where the weight of the body acts and total gravitationaltorque on the body is zero. G  0 • If the body is extended such that value of g varies from part to part of the body the centre of gravity shall not coincide with centre of mass of the body . Rotational Kinematic equations • w  wo  t v  u  at •   w o t  1 t2 s  ut  1 at 2  2 2  • w2  w 2  2 v2  u2  2as 0 • w  d  v  ds  dt  dt  dw d2 wdw •    dt dt2 d  dv d2s vdv  a    dt dt 2 ds   Angular momentum It is the rotational analogy of linear momentum. The angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis. It is measured as the product of linear momentum and  distance of its line of action from the axis of rotation.   pr  unit  unit  kgm2 / s or Js it is a vector quantity. 23

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Angular momentum in vector form Consider a particle P of mass m rotating about an axis through O as shown. sin   r  ; r  r sin  ; o  p  r  p r sin  r    o  r  p   The direction of angular momentum  is perpendicular to plane of r and p in the sense given by right hand thumb rule. Relation between torque and moment of inertia Suppose a rigid body consists of n particles of masses m , m , .........m situated at distances r , r , ...... r 12 n 12 n from the axis of rotation AB. Linear acceleration of first particle a1  r1 Force acting on first particle F1  m1a1  m1r1 Torque acting on first particle 1  F1r1  m1r12 Total torque acting on the rigid body is   1  2  3......  n  m1r12  m2r22  ....  mnrn2     m1r12  m2r22  ....  mnrn2   m1r12       I which is analogues to F  ma 24

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Relation between angular momentum and moment of inertia Consider a rigid body rotating about a fixed axis with uniform angular velocity w. The body consists of n particles of masses m , m ......m 12 n. Linear momentum of first particle, P1  m1v1  m1r1w Angular momentum of first particle   p1r1  m1r12 w 1 Angular momentum of rigid body L    2  ....  n  m1r12 w  m 2 r22 w  ....  m n rn2 w 1  L  m1r12  m2r22  ...  mn rn2 w    L  Iw which is analogues to p  mv Newtons law in rotation Angular momentum of a system        L  1  2  ....  n   1   ri  pi  differentiating wrt time           dL   Pi  ri  dPi =  Vi  mi vi  ri  Fi     dri  dt dt dt       i  total  ext  int but int  0    which is ana log us    dL ext to Fext dP dt dt Law of conservation of angular momentum Suppose the external torque acting on a rigid body due to external forces is zero. Then   dL  0 dt L  cons tan t Applications of law of conservation of angular momentum 1) A man carrying heavy weights in his hands and standing on a rotating turntable can change the angular speed of the turn table by stretching his arms. 2) A diver jumping from a spring board exhibits somersaults in air before touching the water surface. 3) An ice- skater or a ballet dancer can increase her angular velocity by folding her arms and bringing the stretched leg close to the other leg. 4) The speed of the inner layers of the whirlwind in a tornado is alarmigly high. 25

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Analogy between translation and rotation S  V 1) Work done, W  F.S a F  mI  W  . pL Kt  Kr   2) Power, P  F.V   P  . 3) Work energy theorem FS  1 mv2  1 mu2 22   1 I2  1 I20 2 2 Rotational kinetic energy Kinetic energy of first particle k1  1 m1v12  1 m1r12w 2 2 2 Kinetic energy of rigid body, Kr  k1  k2  ...  kn  1 m1r12w 2  1 m2r222  ....  1 mnrn2w 2 2 2 2  1 w2 m1r12  m2r22  ...  m n rn2  Kr  1 I2 2 2 Analogy between translation and rotation s  v 1) Work done, w  F.S a F  mI W  . PL Kt  Kr   2) Power, P  F.V   P.w 3) Work energy theorem FS  1 mv2  1 mu2 22   1 Iw 2  1 Iw 2 2 2 0 26

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Angular impulse   Linear impulse, J  F t  mV  P Angular impulse, A.I  t  I F r  t  I ; P  r  I A.I  J  r  Iw Rolling motion (pure rolling motion) Rolling motion can be regarded as the combination of pure rotation and pure translation. In pure rolling without slipping at any instant of time the contact point of the body is at rest. ie; there is no relative motion between contact point and surface. Pure rolling = Pure translation + pure rotation. For pure rolling the contact point must be at rest V B= 0 V – Rw = 0 cm Vcm  Rw  VA  Vcm  Rw  2Rw VB  Vcm  Rw  0 if Vcm  Rw, the body undergoes forward slipping if Vcm  Rw, the body undergoes backward slipping Acceleration of a point in rolling motion Consider a rolling wheel, each point or the periphery of the wheel is rotating in a circle of radius R, due to which centripetal acceleration of each point on the wheel is w 2R . 27

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Velocity of a point in rolling wheel Consider a point P on a rolling body, the velocity of point P is the vector sum of velocity due to translation and due to rotation.    Vp  Vt  Vr Vp  Vcpm  Vc2m  2Vc2m cos   2 Vcm 1 cos     2 Vcm  2 cos2 2  Vp  2Vcm cos  2 Role of friction in pure rolling When a uniformly rotating wheel is kept on a sufficientlyrough horizontal plane, the wheel undergoes backward slipping. As a result kinetic friction acts in the forward direction provind linear acceleration aswell as angular decceleration. After some time V = Rw, the contact point will be at rest and force of static friction will be cm zero. The wheel executes pure rolling motion. Kinetic energy in rolling motion  Krolling  Kt Kr cm  1 MVc2m  1 Icmw 2  1 MVc2m  1 MK 2 Vc2m 2 2 2 2 R2 28

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) K rolling  1 MVc2m 1  K2  2  R2   Let   1 K2 R2 K rolling  Kt   Rt  1 Kr  1 1 K rolling  K rolling  Angular momentum in combined motion     Lo  Lt  Lr cm Lt  mVcmr Lr  Icm w Rolling motion on an inclined plane Consider a body of mass m and radius R rolling down an inclined plane of inclination  . Due to action of mg sin  component the contact point has a tendency to slip downwards as a result the force of static friction acts in the upward direction which provides angular acceleration. So if the inclined plane is sufficiently rough at any instant of the motion Vcm= Rw. The body executes pure rolling motion. Acceleration down an inclined plane mg sin   fs  ma  1 fs  R  I  2 a  r  3 29

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) from (2) fs  R  mk2  a fs  ma K2 4 R R2 1  mg sin   ma K2  ma R2 a  g sin   g sin  1  K2    R2   Minimum friction required for pure rolling fs  ma K 2  m g sin  K2 R 2 1 K2   R2    fs min  mg sin  1 R2   K2   Minimum coefficient of friction required for pure rolling s   tan  min R2 1 K2 Velocity on reaching bottom Loss of potential energy = gain in kinetic energy mgh  1  2mv2 V  2gh  30

BBrilliant STUDY CENTRE LT -2023M MED (ONLINE CLASS NOTES) Time taken to reach bottom Vcm  ucm  acmt  0  acmt tV 2gh   a  g sin  t  1 2h sin  g Concept of reduced mass Two particle s/m connected by central forces can always be reduced into one body problem using concept of reduced mass. The reduced mass of a two particle system is given by  m1m 2 m1  m2 31