PHYS Work Book

Chapter 8Integral Calculus8.1 SummaryThe eighth chapter of the text discusses integral calculus. We can use in-tegral calculus to help us do two kinds of problems. The first is to ‘addup’ a large number of small things, which could be vectors, or numbers, orpieces of a larger object whose mass, moment of inertia, or similar we wishto obtain. The second is to solve differential equations, convenient becausethe laws of physics are expressible in terms of differential equations. Some key points to remember are:• The fundamental theorem of calculus says that if the function g(x) is the derivative of f (x) (i.e. that d f (x) = g(x)) then dx b (8.1) g(x)dx = f (b) − f (a) a• An indefinite integral ‘undoes’ a differentiation. If d f (x) = g(x) then dx g(x)dx = f (x) + C (8.2) where C is an arbitrary constant.• The constant of integration C only appears in indefinite integrals.• There are a number of very useful integrals that often appear in 141

142 CHAPTER 8. INTEGRAL CALCULUSphysics: xndx = n 1 1 xn+1 + C for n = −1 + x−1dx = ln x + C cos axdx = 1 sin ax + C a sin axdx = − 1 cos ax + C a eaxdx = 1 eax + C (8.3) a8.2 Area in a region bounded by curvesExample Find the area of the region enclosed by the curves in the xyplane y = f (x) and y = g(x) for f (x) = x and g(x) = 2x2 − 4.Worked Solution The first job is to identify the shape we care about.When you draw y = f (x) you get a straight line on the xy plane, and whenyou draw y = g(x) you get a parabola opening upwards. The region we careabout is that enclosed by the two curves, with f (x) above and g(x) below.√These two curves meet at two different values of x, specifically x =1± 33 4 ≈ −1.186 or 1.686. If we imagine breaking the region up into anumber of small rectangles of width dx, and then adding their areas, we willneed to know the height of the rectangles. Since in this region we know thatf (x) ≥ g(x), the rectangle’s height is f (x) − g(x).The area enclosed is then upper x Area = [f (x) − g(x)] dx lower x 1.686 = x − 2x2 − 4 dx −1.186 1.686 = −2x2 + x + 4 dx −1.186 = − 2 x3 + 1 x2 + 4x 1.686 (8.4) 3 2 −1.186Substituting in the values of the upper and lower bound of integration givesus an area of 7.899.

8.3. ARC LENGTH 143You should have noticed that this is a straightforward integral. Themildly tricky parts were finding the bounds of integration and the heightof the rectangles we were adding together in the regular Riemann-sum way.The integral itself was a polynomial.Student Exercises Here are a few area examples: • Find the area bounded by the line y = sin(2x), y = 0, x = 0, and x = π . The area is 1, since the antiderivative of sin(2x) is − 1 cos(2x) + C 2 2 and cos(0) = 1 while cos(π) = −1. • Find the area bounded by the lines y = 1, y = 1 , x = 1, and x = 2. x The area is 1 − ln 2 ≈ 0.307. The antiderivative of the appropriate integrand is x − ln x + C, and ln 1 = 0. • Find the area bounded by the lines x = 1 + cos 2y and x = −1 + e−y, y = 2 and y = 4. In this case the natural way to proceed is to let y be the independent variable and x be the dependent variable, so the integral is over y rather than x. The integrand would be 2 + cos(2y) − e−y which has an antiderivative of 2y + 1 sin(2y) + e−y +C and between 2 and 4 2 the area is approximately 4.756.8.3 Arc Length8.3.1 A straight lineExample Suppose that a curve is defined as y = f (x) in the xy plane withf (x) = 2x − 3. What is the distance along the curve from x = 0 to x = 4?Worked Solution This is an example of a parametric curve. The waywe will approach it is to write both the x and y components of the positionvector as functions of another variable, perhaps s. We then write r(s) =x(s)ˆı + y(s)ˆ. After we have done this we find the displacement dr(s) =d r(s) ds. This tells us the displacement of the point as the parameterdschanges from s to s + ds. We find the magnitude of dr, and then add it upby integrating. In this case we have r(s) = sˆı + (2s − 3) ˆ (8.5)so the functions are x(s) = s and y(s) = 2s − 3. Note that y(s) = 2x(s) − 3by direct substitution. The range of s we care about is between s = 0 and

144 CHAPTER 8. INTEGRAL CALCULUSs = 4 because x was between 0 and 4. Knowing how to differentiate we getthat d ds r(s) = 1ˆı + 2ˆ (8.6)so we have that d ds dr(s) = r(s) ds = dsˆı + 2dsˆ (8.7)and this has a magnitude (8.8) √ |dr(s)| = |dsˆı + 2dsˆ| = 5ds Now that we know the value of |dr| we can calculate the arc lengthstraightforwardly as stop Length = |dr| start 4√ = 5ds √s=0 (8.9) = 4 5 ≈ 8.94You might be wondering why we set up this problem and did it in sucha convoluted way. You could have solved this using geometry and sayingthat we wanted the length of a straight line from 0ˆı − 3ˆ to 4ˆı + 5ˆ (the twoendpoints), which you can immediately check is 8.94. The reason we did itis so you can see that the technique we outline work well in a case whereyou know the answer from other considerations.Student Exercises• Consider the curve given by r(s) = 3 cos sˆı + 3 sin sˆ. What is the length of the arc swept out between s = π and s = 2π? Between s = 0 and s = 3π? In this, s is essentially an angle, in radians. The arc length in the first case is 3π ≈ 9.42, and the arc length in the second case is 9π ≈ 28.3. The second case is analogous to going around a circle one and a half times. This example is relevant to the case of circular motion. √• What is the length of the curve y = 9 − x2 for the interval between x = 0 and x = 3? Hint: the easiest way is to make a substitution so that y and x are both functions of the same parameter; note that y2 + x2 = 9. The arc length in this case is 3 π ≈ 4.71. 2

8.3. ARC LENGTH 1458.3.2 A curveExercise Determine the arc length of the curve y = f (x) with f (x) =2x2 − 3 between x = 1 and x = 3.Worked Solution We proceed in exactly the same way as in the previousexample. We make the parametrization r(s) = x(s)ˆı + y(s)ˆ (8.10)with the identification x(s) = s, and y(s) = 2s2 − 3. Then we calculate bothdr(s) and |dr(s)|: we find dr(s) = dsˆı + (4s) dsˆ (8.11) |dr(s)| = 1 + (4s)2ds (8.12)Note that the (4s)2 term in the above is d y(s) 2, so equation 8.12 can be dsreexpressed as |dr(s)| = 1+ d y(s) 2 (8.13) ds dswhich is likely to be the equation you are told when your math class teachesyou to calculate arc length. The key that makes that equation appropriate isthe parametrization: when you write y = f (x) you are implicitly identifyingthe parameter s with x exactly as we did above. The arc length is then 3 1 + (4s)2ds Length = (8.14) s=1This is not an integral which can be solved by the simple techniques outlinedin the text chapter. In subsequent Math classes you will learn techniquessuch as substitution which will allow you to evaluate integrals like this di-rectly. For now, notice that the derivative of  s2 + 1  s 2 16 + 1 ln s+ s2 + 1 (8.15)g(s) = 4  32 16 is exactly the integrand. Substituting the bounds of integration in we get alength of approximately 16.14.

146 CHAPTER 8. INTEGRAL CALCULUSThe main point of this example is not the actual integral but rather thecalculation of the general formula for the arc length of a curve.Student exercise• A particle moves with a position as a function of time given by r(t) = 5 m tˆı + 3m cos 2 1 t ˆ+ 3m sin 2 1 t kˆ (8.16) s s s How far does it move between t = 2s and t = 3s? Calculate the arc length just like we did for two dimensions, but instead for three. You will find that it has gone along a pat 7.81m long.8.4 Velocity and displacementProblem A particle has a velocity which is given as a function of time as v(t) = 5 mˆı + 2 m e−0.2 1 t ˆ (8.17) s ssWhat is the particle’s displacement between t = 2s and t = 3s?Worked solution The differential equation that relates position and ve-locity is d r(t) = v(t). Using the fundamental theorem of calculus this then dttells us that t2 v(t)dt = r(t2) − r(t1) (8.18) t1This is exactly the quantity (the displacement) that we are asked for.We need to integrate. The integrand is a vector but that is not a problembecause we can use the fact that vectors add together linearly and unitvectors are essentially just constants. So, we have 3sDisplacement = v(t)dt = = 2s = 3s 5 m ˆı + 2 m e−0.2 1 t ˆ dt 2s s s s 5 ms ˆı 3s m 3s e−0.2 1 s s dt + 2 ˆ tdt 2s 2s 5 ms ˆı (3s − 2s) + 2 m ˆ 1 e−0.6 − e−0.4 s −0.2s−1 (8.19) = 5mˆı + 1.21mˆ

8.4. VELOCITY AND DISPLACEMENT 147Notice how we substituted in the values for t at the ends of the integrationregion.A key point in all this is that the integral of a vector is just a number ofindividual integrals. One for each component. This would not be true if theintegral is that of a scalar such as you get in arc length or when you take adot product.Student exercises• Suppose that a mass experiences a constant acceleration of a= 2 m ˆı + s2 3 m ˆ. s2 – By how much does the velocity change between t = 2s and t = 4s? It changes by 4 ms ˆı + 6 m ˆ. s – If the velocity were 4 m ˆı at t = 0s what is it at t = 3s? It is s ms ˆı m 10 + 9 s ˆ. – How much does the position change between t = 2s and t = 4s? Can you solve this if you don’t assume the velocity at a particular instant? What if you assume that the velocity were 4 ms ˆı at t = 0s? You can’t determine this without knowing the velocity at an instant. The velocity at a particular instant is essentially the integration constant that appears. If you knew that f (x) = x + C 2 you couldn’t work out 1 f (x)dx without knowing C. Once the initial condition is given the displacement can be worked out as ∆r = 20mˆı + 18mˆ. – Where is the mass at t = 4s? We don’t know without specifiying another constant of integration. This is why the expression for position as a function of time for constant acceleration is r(t) = r0 + v0t + 1 at2. r0 and v0 at these constants. 2• A mass has a velocity of v(t) = 4 m cos 2 1 t ˆı + 3 m sin 3 1 t ˆ (8.20) s s s s What is the mass’s displacement between t = 0s and t = 2s? The displacement is about −1.51mˆı + 0.04mˆ.

148 CHAPTER 8. INTEGRAL CALCULUS

Chapter 9Momentum9.1 SummaryRead the ninth chapter of the text, which introduces momentum. The reasonwe discuss momentum is as an alternative formulation of Newton’s secondlaw: a net force results in a change in momentum. The particular usefulnessof this new idea is that in the absence of external forces, an interactingsystem will have a constant net momentum. Momentum is an example of a‘conserved’ quantity.• The momentum of a point particle is p = mv. Momentum is a vector quantity, and all calculuations using momentum must treat it that way.• Newton’s second law can be formulated in terms of momentum. A netforce on a point particle results in a change its momentum: Fnet = d p. dt• A system of particles, in the absence of external forces, will have a constant net momentum. This is an expression of Newton’s third law: A pair of interacting particles will have their momentum change in opposite ways because the forces they exert on each other are equal in magnitude and opposite in direction.• The forces that the colliding objects exert on each other are typically much larger than the other forces they are subjected to, so the ob- jects can be treated as approximately isolated. We analyze collisions based on the assumption that momentum is conserved (constant) in the interaction. The momentum that is conserved is the sum of all the momenta of the individual particles. 149

150 CHAPTER 9. MOMENTUM • For a system of particles, the motion of their center of mass can be explained by considering only the external forces. • The position of the center of mass is the mass-weighted average posi- tion. If there are n particles then rCM = m1r1 + m2r2 + . . . + mnrn (9.1) m1 + m2 + . . . mn If the mass is continuously distributed with a density ρ(r) then the position of the center of mass is rρ(r)d3r (9.2) rCM = ρ(r)d3r The quantity d3r is the little element of volume. • The change in a particle’s momentum (often called the impulse, and sometimes denoted J) on a particle between times t1 and t2 is t2 (9.3) ∆p = J = F dt = Favg∆t t1 This relationship defines the average force exerted between the two times; ∆t = t2 − t1. • The idea of ‘impulse’ is often useful in situations when we do not know the exact details of the force being exerted.9.2 A ball hitting a batProblem A 0.25kg ball, initially travelling horizontally at 20 m is struck sby a bat. After the collision it is travelling back in the opposite directionthat it was initially travelling at an angle of 30◦ above the horizontal, and ata speed of 35 m . If the collision lasted for 1.50×10−3s, what is the magnitude sof the average force on the ball during the collision?Worked Solution If you have ever seen a slow-motion picture of a base-ball colliding with a bat (or a golf ball being hit by a club, or a tennis ballbeing hit by a racket, or a football being kicked, or anything like that) youwill have noticed that the ball itself is deformed during the collision – itchanges shape. The fact that it chages shape (and then returns to the orig-inal shape) suggests that it will act something like a spring: there is a force

9.2. A BALL HITTING A BAT 151which is related to how much it is deformed. Our assumption with springswas that the force was proportional to the deformation, but we do not knowenough about how balls are constructed to be sure that is a reasonable as-sumption. We are going to ignore this subtlety and simply calculate theaverage force, since we do not know the details of the actual forces exerted. The fact we will use is that the impulse, or change in momentum isrelated to the average force. In fact: Favg∆t = ∆p = pf − pi = mvf − mvi Favg = mvf − mvi ∆t m Favg = ∆t |vf − vi| (9.4) In this question, ∆t is the duration of the collision, m is the mass, andvi and vf are the initial and final velocities respectively. From the statement of the question we have: vi = 20 ms ˆı m vf = 35 s − cos 30◦ˆı + sin 30◦kˆ m = 0.25kg ∆t = 1.5 × 10−3s (9.5)This means that vf − vi = −50.3 m + 17.5 m kˆ and hence that |vf − vi| = s s53.3 m , which gives that Favg = 8.88 × 103N . Note that the direction of sthis force was 19◦ above the horizontal and ‘back’ in the direction the ballwas originally going.We talked about this because needed an excuse to calculate a changein momentum. Those calculations are very similar to changes in any vectorquantity, so emphasizing that it is a vector is worthwhile. When you are looking at the average force exerted, the relation we de-rived (equation 9.4) tells us that (all things being equal) a bigger changein velocity requires a bigger force, a shorter contact requires a bigger force,and if the mass is larger the force required for a given change in velocity willbe larger.

152 CHAPTER 9. MOMENTUMStudent Exercises • Why are airbags so effective in cars? Consider an 80kg man travelling in a car horizontally in the x-direction at 30 m . (This is about 110 km .) s h Suppose the car collides with something, comes to a stop, and exerts a force on the man to stop him. What is the magnitude of the average force exerted on him if: – The thing exerting the force is the steering wheel column, and it stops him in 0.006s. In this case the average force is 4.0 × 105N . – The thing exerting the force is an airbag which stops him in 0.12s. In this case the magnitude of the average force is 2.0 × 104N .9.3 An inelastic collisionExample A particle of mass m1 travels at velocity v1 and collides with asecond particle of mass m2 travelling at initial velocity v2. The two massesstick together. Immediately after the collision, what is the velocity (directionand speed) of the wreckage? Obtain a numerical answer is m1 = 2kg, m2 = 1kg, v1 = 3 m ˆı, and s mv2 = 8 s ˆ, which is illustrated in figure 9.1. What is the change of momentum of each of the individual masses in thecollision?Worked Solution This is a key type of problem because it illustratesthe idea of conservation of momentum. When we say that momentum isconserved in a collision we usually mean that for the duration of the collision,the forces that the two particles exert on each other are so much larger inmagnitude than the other forces they experience that for all intents andpurposes the only forces are ones between the two colliding objects. In thissituation, the momentum immediately before the collision is the same as themomentum immediately after. We can write this in mathematical language, knowing that the twomasses have combined after the collision: pbef ore = paf ter p1,bef ore + p2,bef ore = paf ter m1v1 + m2v2 = mcombinedvcombined vcombined = m1v1 + m2v2 (9.6) mcombined

9.3. AN INELASTIC COLLISION 153 Before Collisiony x After Collisiony xFigure 9.1: The figure shows the momentum vectors before a collison oftwo particles. The star indicates the collision point. After the collision, theparticles have merged and there is one final momentum vector.

154 CHAPTER 9. MOMENTUM For the values we have: p1,before = m1v1 = 6kg ms ˆı m p2,before = m2v2 = 8kg s ˆso we have that vcombined = 2 ms ˆı + 2.7 m ˆ. This velocity has a magnitude of s m 53◦3.3 s and it makes an angle of counterclockwise from the positive x-axis. To find the change in momentum for the 2kg object (which after thecollision travels locked together with the 1kg object) we have to take thedifference between pfinal and pinitial. The initial momentum is m1v1, andthe final momentum of this mass is m1vcombined = 2kg 2 ms ˆı + 2.7 m ˆ . Then s 2kg Mass : ∆p1 = pfinal − pinitial = 4kg ms ˆı + 5.3kg m ˆ − 6kg ms ˆı s m m = −2kg s + 5.3kg s ˆ (9.7)Similarly, for the second mass the change in momentum is the differencebetween its pfinal and pinitial. These are m2v2 and m2vcombined respectively.This means 1kg Mass : ∆p2 = pfinal − pinitial = 2kg ms ˆı + 2.7kg m ˆ − 8kg m ˆ s s ms ˆı m = 2kg − 5.3kg s ˆ (9.8) Note that for the numbers provided ∆p2 = −∆p1 (9.9)This effect is generic: when two object which are otherwise isolated interactthe change in one’s momentum is compensated by the change in the other’smomentum. The total momentum is unchanged. We have a numerical answer for the change in momentum of the 2kg pieceand the 1kg piece. They are equal in magnitude and opposite in direction.Some things to notice: The thing that should be jumping out at you ishow relatively straightforward calculations involving momentum are. Thatis why physicists like the concept - it is straightforward and easy to apply.

9.4. AN EXPLOSION 155In this case, by ‘straightforward’ we really mean ‘linear’. You just have toadd and subtract vectors, which we have done a lot.The second thing that is important to notice (but not surprising) is thatthe two changes in momentum were opposite. This means that the totalmomentum was the same before and after the collision. It is a reflection ofNewton’s third law, together with the fact that Fnet = d p. dtStudent Exercises: Suppose that an object of mass m1 travels with speed|v1| and collides with an object of mass m2 travelling with an initial speed|v2|. The two masses stick together after the collision. The angle betweenv1 and v2 is θ.• Find the final speed of the combined mass if m1 = 4kg, m2 = 2kg, |v1| = 6 m , |v2| = 8 m and the angle between the masses’ velocities is s s 53◦. m θ = The combined speed is 5.99 s .• Find the angle the combined mass’s velocity makes with the vector defined by the velocity of mass m1 for the numbers provided above. The angle is 20.9◦.• Symbollically, find the final speed of the combined mass in terms of m1, m2, |v1|, |v2|, and θ. Find the angle φ that the final velocity of the combined mass makes with initial velocity v1 in terms of the same variables. It is probably going to make your life easier if you assume that v1 travels along the x-axis; make sure your answer matches the result from equation 9.6 if θ = 90◦, and make sure you reproduce the results above.9.4 An explosionExample A particle initially at rest explodes into three pieces: A 3kgpiece which travels along the x-axis at 5 m , a 2kg piece which travels along a s 135◦ mline making an angle of with the positive x-axis in the x-y plane at 8 s ,and a 1kg piece. What is the velocity of the 1kg piece after the explosion?This is illustrated in figure 9.2.Worked Solution This is another application of conservation of momen-tum. The explosion produces forces act between the three fragments. After

156 CHAPTER 9. MOMENTUM Beforey x Aftery xFigure 9.2: The figure shows a single stationary particle in the before part.The particle explodes into three pieices.

9.4. AN EXPLOSION 157the explosion, the momenta are 1kg : p1 = m1v1 = 1kgv1 We want v1 ms ˆı m 2kg : p2 = m2v2 = 2kg −5.66 + 5.66 s ˆ 3kg : p3 = m3v3 = 3kg 5 ms ˆı (9.10) Using conservation of momentum, together with the fact that the particleis initially at rest, we have pinitial = pf inal 0 = p1 + p2 + p3 p1 = −p2 − p3 v1 = −p2 − p3 (9.11) m1 (9.12)and with the values we have for p2 and p3 we have v1 = −3.7 ms ˆı − 11.3 m ˆ sWhat you should have noticed: Once again, we had to apply conser-vation of momentum. That is really the theme of this chapter: Momentumis a vector, and it is conserved in isolated interactions.Student Exercise This question is intended to get you to understandhow rocket engines work.A rocket has total mass 3000kg, of which 2000kg is fuel. When therocket is turned on, the gas goes out at a rate of 1.0 kg at a speed of 1500 m s srelative to the rocket. If the rocket is initially at rest, what is the rocket’sspeed after 30 minutes?To solve this follow the following procedure:• Suppose a spaceship of total mass Mtot is travelling at velocity vˆı as shown in the top half of figure 9.3. Now suppose there is an explo- sion and it ejects a small amount of mass ∆m so that it is travelling at velocity (v−vex)ˆı. What is the change in speed of the rest of the space- ship? (in this set-up, vex is the speed at which ‘gas’ is being ejected from the spaceship.) Note, ignore terms where two ∆ quantities are multiplied together.

158 CHAPTER 9. MOMENTUM Rocket Coastingy M tot vˆı x After Firing Enginey ∆ m M tot − ∆ m ( v − vex )ˆı ( v + ∆ v)ˆı xFigure 9.3: The upper half of the figure shows a rocket of mass Mtot movingwith velocity vˆı. The lower half of the figures shows the situation after amass ∆m has been expelled from the rocket motor with a velocity vex.

9.5. CENTER OF MASS AND PROJECTILE MOTION 159• If the mass ∆m was ejected in a short amount of time ∆t, what is the average magnitude of acceleration of the remainder of the spaceship during that time? Express this in terms of vex, M , ∆m, and ∆t.• Now, imagine that Mtot is the mass at some instant of time, t, then we can see that Mtot is M (t) the (changing) mass of the spaceship. Explain why ∆m ≈ − d M (t) using the definition of the derivative, ∆t dt dM (t) = lim M (t + ∆t) − M (t) = M (t) − ∆m − M (t) (9.13) dt ∆t ∆t ∆t→0 Note that the mass of the spaceship is changing because it is exhausting gas out the back - it is the same principle as how a bottle full of water and compressed air works: some water is shot out the back, reducing the mass but pushing the rest in the opposite direction.• If ∆t is very short, explain why the expression for acceleration mag- nitude can be expressed as |a| = −vex 1 dM M (t) dt• If the rocket starts from rest at mass M1, exhausts all its fuel at a constant speed vex and when its fuel is exhausted has a mass M2 show that its final speed is vex ln M1 by integrating. M2After all that work, you should get a speed of 1374 m . s9.5 Center of Mass and Projectile MotionProblem Suppose a ball is thrown, and part-way through its flight, itexplodes into two pieces. Argue convincingly that the center of mass ofthese two pieces will follow the same path the ball would have if it had notmet with an unfortunate demise.Worked Solution We have analyzed constant acceleration motion (suchas by gravity) in chapter 5. At the instant before the ball explodes it isat a location ri, and travelling at a velocity vi. It experiences a constantacceleration g, so it follows the path: r(t) = ri + vit + 1 gt2 (9.14) 2assuming that we call the time at which it explodes t = 0s.

160 CHAPTER 9. MOMENTUM The explosion happens fast, and breaks the ball into two pieces. At thisinstant, we apply conservation of momentum: pinitial = p1 + p2 (9.15) (m1 + m2)vi = m1v1 + m2v2 The two masses will travel immediately after the explosion with velocitiesv1 and v2 respectively. This means that the trajectories they will follow are: r1(t) = ri + v1t + 1 gt2 2 r2(t) = ri + v2t + 1 gt2 (9.16) 2Both have their initial position at ri because the two fragements came fromthe same ball. The general expression for the center of mass of two objectsis m1r1 + m2r2 m1 + m2 rcm = (9.17)Applying this to our case: rcm(t) = m1r1(t) + m2r2(t) m1 + m2 = m1 ri + v1t + 1 gt2 + 2 m1 + m2 m2 ri + v2t + 1 gt2 2 m1 + m2 = ri + m1v1 + m2v2 t + 1 gt2 m1 + m2 2 = ri + vit + 1 gt2 (9.18) 2for the last equality, we are using the result from equation 9.15, and wenotice that this expression is the same as equation 9.14, so the center ofmass travels in the same path that the ball was going to (until some of ithits something, which would exert a force on at least one fragment.)What you should notice: we did this example primarily to highlighthow you calculate the center of mass for a set of objects that are movingin time. We also wanted to show that you can use momentum ideas evenif there is a net force (as there is for projectile motion in the presence ofgravity.)

9.6. QUESTIONS 161Student Exercise Three particles are located at the following positions:mass one is 3 kg and located at r1 = 4mˆı+2mˆ, mass two is 1 kg and locatedat r2 = −4mˆı − 2mˆ, and mass three is 3kg and located at r3 = −1mˆı + 3mˆ.Find the location of the center of mass of this set of three particles. Thecenter of mass can be shown to be at 0.71mˆı + 1.86mˆ.9.6 Questions• What average force is required to change the velocity of a 1.5kg ball from 20 ms ˆı + 10 m ˆ to −10 ms ˆı + 20 m ˆ in 3.0s? s s• A ball of mass m1 travelling with velocity v1,i hits a ball of mass m2 which is initially at rest. The two balls have masses that do not change during collision. The first ball travels with velocity v1,f af- ter the collision. This situation is illustrated in figure 9.4. Find the velocity (speed and direction) of the second ball after the collision if m1 = 5kg, m2 = 15kg, v1,i = 20 ms ˆı, and v1,f = 5 ms ˆı + 15 m ˆ. s9.7 Answers• If this case the average force required is −15Nˆı + 5N ˆ.• The final speed of the ball is |v| = 7.07 m , along the unit vector 0.71ˆı − s 0.71ˆ.

162 CHAPTER 9. MOMENTUM Before Collisiony x After Collisiony xFigure 9.4: The figure shows the momentum vectors before a collison of amoving particle with a stationary particle. The particles retain their originalmasses after the collision.

Chapter 10Angular Momentum10.1 SummaryRead the tenth chapter of the text. In previous chapters, we have learnedabout how objects in static equilibrium experience no net force as well as nonet torque. Here we will learn how objects change their rotational state inresponse to non-zero torques. We will find that a quantity called ‘angularmomentum’ changes in response to non-zero net torques the same way thatmomentum changes in response to net forces. Some of the key points are:• For point at distance r from a pivot, when the angle from the pivot to that point changes by θ (measured in radians from a fixed axis) then the point has moved an arc length rθ.• The tangential speed of a point which is at a fixed distance from aaxis of rotation is v = d rθ =r d θ . The quantity d θ ≡ ω is often dt dt dtcalled the angular speed. Note that v = rω.• The tangential acceleration of a point which it at a fixed distance rfrom an axis of rotation is atan = d v = r d |ω|. If this quantity is dt dtpositive, the rotation is speeding up, and if it is negative the rotationis slowing down.• The angular momentum of a particle around a point is L = r × p. The vector r is the vector from the point around which you are measuring L to the particle.• The angular momentum of an object rotating along an axis can be found by imagining breaking the object up into little pieces, calculating 163

164 CHAPTER 10. ANGULAR MOMENTUM v m v m post post ad ad dFigure 10.1: A ball of mass m is moving in a horizontal circle of radius duntil the rope to which it is attached hits a post at a distance ad from theoriginal center of the circle. the angular momentum of each, and adding them together. Since the speed of each piece is related to the angular speed, it turns out that the magnitude of the angular momentum is L = Iω.• The direction of the angular momentum of a rotating rigid object is – in the cases we study in this course – along the axis of rotation. The magnitude is given by Iω and the direction is given by the right-hand rule.• I is a measure of the distribution of the mass of an object. You calculate I by adding, for each ‘piece’ of an object, the mass of that piece multiplied by the square of how far it is from the rotation axis. I = r2dm = |r|2 ρ(r)d3r (10.1) all the object10.2 A ball swinging on a ropeProblem You are swinging a ball of mass m in a horizontal circle of radiusd at speed v. The ball is attached by a rope which will break if it is subjectedto a tension bigger than T . There is a post a distance ad from you whichwill get in the way of the rope, as shown in figure 10.1• If m = 1kg, v = 3 m , d = 2m, and the maximum tension before the s rope breaks is T = 10N , what is the biggest possible value of a such that the rope does not immediately break?

10.2. A BALL SWINGING ON A ROPE 165 • If the post had circumference c = 2cm and the post was placed at a = 1 , how many times would the rope wind around the post before 2 it broke?Worked Solution Before we start to work through the problem, we haveto have a vision for what the issue will be. We know that the ball is movingin a circle, and then something is in the way of the rope. This means thatthe rope will wrap around the post it hits, and this in turn means that therope will then be holding the ball in a smaller circle. In fact, a smaller circleof radius (1 − a)d. A smaller radius means, all other things being equal, alarger tension on the rope because the circle is smaller with the same speed,so the centrepetal acceleration is larger. When the acceleration gets toolarge, so the required tension is too large, the rope will break. The first part of this question does not actually require the concept ofangular momentum, but we will use it to set up the second part. The ball travels, immediately after the rope hits the post, in a circle ofradius (1 − a)d. This means that the tension in the rope must be enough tosupply the net force of magnitude (recall the material on circular motion)m v2 . This means that r T = Fnet < Tmax mv2 < Tmax r mv2 < Tmax (1 − a)d mv2 < (1 − a) (10.2) dTmaxFor the given values of m, v, d, and Tmax, 1 − a < 0.45, so the largestpossible a = 0.55, and that means that if the post is more than 1.1m awayfrom the place you are swinging the ball, the rope will break immediatelyupon contact with the post. v2 r The interesting part here was the critical condition that m < Tmax.Now that we have seen this, we will look at what happens when the ropestarts to wrap around the pole. Obviously, the ‘effective’ length of the ropewill get shorter, and in the condition we have, that means r gets smaller, andwe get closer to the rope breaking. We haven’t worked out what happensto v (the speed) as the radius of the circle r gets smaller. In the first part,we treated v as though it did not change, because we were trying to figureout what would happen immediately upon the rope hitting the post.

166 CHAPTER 10. ANGULAR MOMENTUM vf post m xfFigure 10.2: The geometry of the ball from figure 10.1 as it wraps around. We want to figure out the angular momentum of the ball as it goesaround the post: As we see in figure 10.2, initially the angular momentumof the ball around the post isL = r×p (10.3) = [(1 − a)dˆı] × (mvˆ) = (1 − a)dmvkˆSince the radius vector (from the post to the ball) and the velocity vectorare perpendicular to each other, we could have gotten this from the wayof calculating cross products by obtaining the magnitude from A × B = A B sin θ, and gotten the direction using the right hand rule. Now, we look at figure 10.2, and determine out the angular momentumafter the rope has wrapped a bit. In this case, the mass is xf from thepole, and it is travelling at speed vf . The two vectors of which these arethe magnitudes are still perpendicular. Calculating the angular momentumusing the right-hand rule (and assuming that the paper is the x-y plane, aswe did before) we get L = xf mvf kˆ. Now for the key part: we equate the two values we got for angularmomentum. Why? Because there is no torque around the pole exerted bythe rope on the mass because the force is in towards the pole, and the radiusvector which we care about because τ = r × F is in the opposite directionto the force, so the angle between them is 180◦, and so the torque has 0

10.2. A BALL SWINGING ON A ROPE 167magnitude. This tells us that Laf ter = Loriginal (1 − a) dmvkˆ = xf mvf kˆ (1 − a)dv = xf vf vf = (1 − a)d v (10.4) xfIn other words, as the ball wraps (i.e. as xf gets smaller) the speed of theball gets larger. Note that (1 − a), d, and v are quantities that are known.Now, we can find out how much wrapping will happen: At the instantthe rope breaks m v2 = Tmax (10.5) rWe have called the speed of the ball once it is wrapping vf , and the distancefrom the pole to the ball is xf , so the point we are looking for is when therequired force is the same as the maximum force m vf2 = Tmax (10.6) xfbut finally we have a relation between vf and xf : m vf2 = Tmax xf (1−a)dv 2 m xf = Tmax xf m (1 − a)2d2v2 = Tmax x3f xf = m(1 − a)2d2v2 1/3 (10.7) TmaxPutting in the known values: m = 1kg, a = 0.5, d = 2m, v = 3 m , and sTmax = 10N , we get xf = 0.965m. Since the circumference of the rod was0.02m, this means it will wind about 1 and two thirds time around beforeit breaks.

168 CHAPTER 10. ANGULAR MOMENTUMSome things to notice The really big points of this example were: • In the case of a central force (one directed inwards towards the center of rotation) the angular momentum is unchanged. • If you have a point particle, and it has constant angular momentum, as it moves closer to the axis of rotation, it will speed up. • If you thought about this instead as being a rigid body with all the mass at one end and rotating about the other, the change (reduction) of the moment of inertia as the rope winds (so the ball gets closer) would mean a corresponding increase in the angular speed.Student Exercises Suppose a dumbbell can be modelled as a pair ofballs, each with mass m = 0.5kg, and the two masses are joined by a rigidrod of length L = 0.4m. The two balls rotate around their center of massalong an axis perpendicular to the rod at an angular speed ω = 10 1 . s• What is the speed of each mass? It is 2.0 m . s• What is the magnitude of the angular momentum of one of the balls around the center of mass? It is 0.2 kg m2 . s• What is the moment of inertia of the dumbbell around its axis of rotation? The moment of inertia is 0.04kg m2.• Does the magnitude of the angular momentum calculated from the moment of inertia and the angular speed match the total angular mo- mentum by adding up the angular momentum of the two balls? It should be clear that it does.10.3 Constant Acceleration Merry-Go-RoundExample A merry-go-round (essentially, a uniform disk in the horizontalplane that is free to rotate around a vertical axis) has a radius of r0. Asmall block of mass m is placed a distance ar0 from the center of the merrygo round. It has a coefficient of static friction µs with the surface on whichit rests.The merry-go-round starts rotating from rest at a constant angular ac-celeration α. If r0 = 2m, a = 0.333, m = 5kg, µs = 0.5, and α = 0.1 1 how s2long after the start of rotation does the mass start to slide?

10.3. CONSTANT ACCELERATION MERRY-GO-ROUND 169Worked Solution The first thing we have to do is think through why themass is going to slide. We know that there is a constant angular accelerationα, which means that the angular speed ω is increasing. If ω is increasing,that means that the speed v is also increasing. Since the speed is increasing,this means that the force required to give the centripetal acceleration willalso increase. At some point, the required force will be bigger than the forcethat friction can provide; then it will start to slide. We can start by looking at a free-body diagram for the mass. There arethree forces being exerted on it: the downwards force of gravity, the upwardsnormal force, and the force of static friction. The force of static friction actshorizontally, and this force is what must provide the acceleration of themass: Fnet = ma = Fs + Fn + Fg. As usual for a horizontal surface, themagnitude of Fn is mg, so the maximum possible magnitude of the force offriction is µsmg. Now, what does this force have to do? It must supply the change (in-crease) in speed, and it must also supply the centripetal acceleration. Theincrease in speed is supplied by the tangential acceleration; we know that themagnitude of this is rα. Similarly, we know that the centripetal acceleration v2has magnitude r , and the speed v = rω, so the centripetal accelerationcould be expressed as (rω)2 = rω2. This can be seen in figure 10.3. The rmagnitude of the acceleration is |a| = a2tan + ac2en (10.8) = (ar0α)2 + (ar0ω2)2 = ar0 α2 + ω4but since the rotation started from rest, we have that ω = αt. Also, themaximum force magnitude is µsmg, so the maximum acceleration it providesis µsg, and so we have to find the largest value of t that ar0 α2 + (αt)4 ≤ µsg (10.9)and this gives us α2 + (αt)4 ≤ µsg 2 ar0 1 + t4 ≤ µsg 2 α2 α2ar0 µsg 2 1 α2ar0 α2 t4 ≤ − (10.10)

170 CHAPTER 10. ANGULAR MOMENTUM ac = − mv2 rˆ rv r ac rˆ Fn Ff FgFigure 10.3: A mass is a distance |r| from the center of a merry-go-round.The force of friction provides the centripetal acceleration as well as anychange in speed.

10.3. CONSTANT ACCELERATION MERRY-GO-ROUND 171 L mωFigure 10.4: A ball of mass m and velocity v is heading horizontally towardsthe bottom end of a ballistic pendulum of length L which is hanging ver-tically. The ball will stick to the bottom of the pendulum, which will thenswing with angular speed ω as shown.so the largest t that satisfies this inequality (t4 ≤ 63580s4 for the numbersgiven) is 15.9s.More things to note: The things that are especially critical here are therelation between the angular and linear kinematics. Knowing the angularacceleration told us about the tangential component of acceleration; knowingthe angular speed told us the speed of the rotating object, which in turn toldus the centripetal component of acceleration. It was very important thatthese two components of acceleration had to be treated together to giveus the magnitude of the overall acceleration. Since we knew the relationbetween angular speed and angular acceleration, we were able to find themagnitude of the overall acceleration in terms of given quantities and t, theunknown quantity.Student Exercise• A particular ballistic pendulum consists of a rigid rod of length L suspended vertically from one end. This rod has moment of inertia I around its suspension point. A ball of mass m is shot horizontally at speed v, as shown in figure 10.4. The ball collides and sticks to the rod.Suppose that v = 15 m , m = 0.1kg, and the moment of inertia (of just sthe rod ) I = 0.75kg m2, and L = 0.3m. What is the angular speed of

172 CHAPTER 10. ANGULAR MOMENTUM r0 mFigure 10.5: A cylinder of radius r has moment of inertia I0 around its axisof symmetry, about which it is free to rotate. A massless, inextensible ropeis attached to a ball of mass m and wound around the cylinder.this ball and rod combination around the pin at the top immediatelyafter the collision? The angular speed after the collision is ω = 0.593 1 . s• A ball of mass 4.0kg is held up by a rope of length L = 0.5m andis at rest in equilibrium. A ball of mass 1.0kg is travelling at 10 m ˆı sand hits and bounces off the first ball. After the collision the 1.0kgball is travelling with velocity −2.0 ms ˆı. Find the tension in the ropeimmediately after the collision. It is 111N .10.4 An Atwood MachineExample A mass m is attached to a rope which is wound around a cylinderof radius r0 and which has a moment of inertia I0. This is shown in figure10.5.

10.4. AN ATWOOD MACHINE 173 r0 FT m FT FgFigure 10.6: Sketches of the free-body diagrams for both the cylinder andthe suspended mass from 10.5. If m = 0.5kg, r0 = 0.4m, and I0 = 50kg m2, what is the acceleration ofthe mass?Worked Solution The purpose of this question is to illustrate why it wasso important to have a massless frictionless pulley when we talked aboutAtwood Machines before. The big picture of the question is the same as what we had for Atwoodmachines. We will apply Newton’s 2nd law to the suspended mass to getits acceleration; the acceleration will be related to the tension in the rope(unknown) and the mass’s weight (known in terms of m). We will applythe rule that τ = d L (that torque changes angular momentum) to find dtthe rate of angular acceleration of the cylinder; this will relate the angularacceleration to the tension in the rope (since the tension is what providesthe torque). Finally, we will relate the angular acceleration to the mass’sacceleration. There will be three relationships between three things we don’tknow: |a|, α, and the tension T . Three relationships between three unknownthings will give us the chance to solve for each. We will give this a shot! First, we look at the free-body diagrams, shownin figure 10.6.

174 CHAPTER 10. ANGULAR MOMENTUM The case of the mass, applying Newton’s laws are easy: Fnet,m = FT + Fg ma = T kˆ + −mgkˆ m −akˆ = (T − mg) kˆ a = g − T (10.11) mIn this, we have set it so that a is the vertical component of the acceleration,and we chose a particular sign convention (see the − sign in the 3rd line)that if a > 0 the mass is falling down. This was not essential, but it willmake our algebra easier. Now, we need to find out what is happening with the cylinder. Thereis going to be an upwards force on it from the axle. This will keep it intranslational equilibrium, but the cylinder is subject to forces exerted atdifferent points, so it will change its rotational state. We calculate thetorque around the axle (axis of rotation of the cylinder) and we will use thefact that the torque and the angular momentum both have only componentsalong the line perpendicular to the paper. (If you don’t believe this, checkit yourself.) τnet = d L dt |τ | = d L dt r0T = I d ω dt r0T = Iα (10.12)In this, we have gone ahead and calculated the magnitude of the torqueusing the method of simply obtaining the magnitude. The direction is intothe paper (in the positive y-direction if ‘up’ the page is the z-direction). The final relationship is between α and a. The rope does not stretch,so as the distance the mass goes down corresponds to the amount the rope‘unwinds’ from the cylinder. This means that the amount the mass goesdown is d = r0θ when the cylinder unwinds by an angle θ. Differentiating d2the distance with respect to time twice, we find that a = r0 dt2 θ → a = r0α.Note that something is implied here: the mass’s displacement is −dkˆ, so the d2 dkˆ.second time derivative of that is − dt2 If you compare to how we wrote

10.4. AN ATWOOD MACHINE 175the acceleration this says that a = d2 d, so we have taken care of the signs dt2appropriately. There are three relations: a = g − T m r0T = Iα a = r0α (10.13)substituting the third into the second we then have a = g − T m a r0T = I r0 (10.14)Solving the second equation for T in terms of a and substituting gives a = g − 1 Ia m r02 a 1 + I =g mr02 g a = (10.15) 1 + I mr02When we plug in the numbers given in the statement of the question, weget a = 1.6 × 10−2 m . s2You should have noticed how similar this was to the previous Atwoodmachine problems: we used considerations that let us find the acceleration orangular acceleration, and we also used a relation between the motion of thetwo objects to find a relation between the angular and linear accelerations.This set of linear equations allowed us to solve for the unknown quantities.Notice that we used, almost without second thought, the relations betweenlinear and angular quantities.Student Exercises • For the problem done above, using the given values for m, r0, and I, find the tension T in the rope. The tension works out to 4.892N .

176 CHAPTER 10. ANGULAR MOMENTUM• Use the tension you calculated in the previous question to calculatethe time rate of change of angular momentum (about the axis of thecylinder) for both the disk and the falling mass. We find that for thedisk dL = 1.9578 kg m2 and that for the mass dL = 0.0032 kg m2 . dt s2 dt s2Note that the sum of these two components is the same as the nettorque on this system. (calculated around the cylinder’s axis)10.5 Questions• Two masses m = 0.5kg are attached to a rod of variable length. Themasses rotate attached to the rod and initially spin around their centerof mass at a rate ω = 10 1 . The length of the rod is initally 0.4m, but sit changes to 0.14m. – What is the new moment of inertia Inew of the dumbbell? – What is the new angular speed of the rotating dumbbell?• Consider the Atwood machine illustrated in figure 10.7. Assume that m1 = 1kg, m2 = 2kg, I = 10kgm2, R = 0.3m, and then – Find the magnitude of the mass’s acceleration. – Find the magnitude of the angular acceleration. – Find the two tensions T1 and T2 in the rope holding the masses up. Are they the same?• Consider the situation illustrated in figure 10.8. If the stick is released from rest, and the ball is not attached to the stick, what is the ratio of the accelerations of the ball and the end of the stick? (Does the ball stay at the end of the stick as they both fall?)10.6 Answers• For the change in configuration of the dumbbell we find – The new moment of inertia is Inew = 4.9 × 10−3kg m2. – The new angular speed is ωnew = 81.6 1 . s• For the Atwood machine illustrated we find: – The magnitude of the mass’s acceleration is 8.59 × 10−2 m . s2

10.6. ANSWERS 177 r0 T1 T2 M1 M2Figure 10.7: A mass m1 and another mass m2 are attached to each othervia a rope which passes over a disk of moment of inertia I and radius R.

178 CHAPTER 10. ANGULAR MOMENTUM m Lpin aLFigure 10.8: A ball of mass m is placed on the end of a stick of length L.The other end of the stick is a pivot point. The moment of inertia of thestick is I = 1 M L2, and the mass is M. 3 – The magnitude of the disk’s angular acceleration is 0.286 1 . s2 – The two tensions are T1 = 9.88N and T2 = 19.4N . Note that they cannot be the same, otherwise there would be no torque on the disk.• The angular acceleration of the end of the stick (assuming it is hori- zontal) is 3 g . This means that the end of the stick will fall faster than 2 L the ball.

Chapter 11Work and Kinetic Energy11.1 SummaryRead the eleventh chapter of the text. What we have focussed on so far isthe application of laws relating forces: what the requirements for rotationaland translational equilibrium are; what accelerations (angular or linear) areproduced by net torques and net forces; the mechanics of the forces producedby various mechanisms; and how motion responds to different forces. Whatwe are going to do now is try and develop a way of evaluating motion wherethe details of exactly what force is exerted when does not matter, or is notwhat we are interested in. In some ways, this is similar to momentum, inthat the details of the force are not what is important; rather the ‘net’ effect,as determined by an integral. Some of the key points are:• The work done by a constant force on an object that undergoes a displacement ∆r is defined by W = F · ∆r (11.1)W can be positive, negative, or zero. The units of W are the Joule [J];1[J] = 1 kg m2 . s2• The work done by a non-constant force on an object that undergoesa displacement can be calculated by breaking up the path the objectfollows into a large number of little displacements (which we call dr)in such a way that we can approximate the force as constant over thattiny displacement. We add up the word done in each displacement to endget the total work; this is the same as saying that W = starting F · dr. 179

180 CHAPTER 11. WORK AND KINETIC ENERGY• Newton’s second law, combined with what we know about the rela- tionship between acceleration and velocity, tells us that if work is done the speed of an object will change.• The relation between net work and speed is given by Wnet = KEfinal − K Einitial .• The quantity denoted KE is called ‘Kinetic Energy’ and defined, for a point particle, as KE = 1 m |v|2. 2• For an extended object, we can imagine dividing it into a large num- ber of small pieces and finding the Kinetic Energy for each piece and then adding the energies together. For an extended object that is not rotating, this will just give 1 M |vcm|2 where vcm is the center of mass 2 velocity.• For a rotating extended object the kinetic energy is KE = 1 I ω2 , where 2 ω is the rate of rotation and I is the moment of inertia. This is strictly true only for the rotation around one of the ‘principle axes’, however in this course we will always be considering only that case.• Forces for which the work done being does not depend on the path taken are called ‘conservative’ forces, while forces for which the work done does depend on the path taken between the end-points are called ‘non-conservative’ forces.• The reason to use work and energy considerations is to simplify the calculations; sometimes the details of how the force is applied, or what the path taken are not relevant, or would lead to things that are cal- culationally harder.• In ‘elastic’ collisions the final kinetic energy is the same as the initial kinetic energy.11.2 Work done along different pathsExample A mass m is moved over a horizontal surface with which it hasa coefficient of kinetic friction µk. Assuming that the mass is m = 10kg andµk = 0.5, what is the work done by friction in the following cases: • The mass is moved from 1mˆı to 4mˆı along a straight line.

11.2. WORK DONE ALONG DIFFERENT PATHS 181Figure 11.1: Three different paths from 1mˆı to 4mˆı.• The mass is moved from 1mˆı to 2mˆı + 1mˆ and then to 3mˆı + 1mˆ and from there to 4mˆı.• The mass is moved from 1mˆı to 4mˆı along the path given byr(t) = 2.5m − 1.5m cos(πs−1t ˆı + 1.5m sin(πs−1t)ˆ (11.2)in the time between t = 0s and t = 1s.These three paths are shown in figure 11.1. Note that the paths all have thesame starting and ending points.Worked Solution The critical part of this is to think through what thework done by the friction force means. The force of kinetic friction is alwaysin the opposite direction to the motion. We will use that here. The first thing to figure out is the magnitude of the force of friction.We could draw a free-body diagram, but we have done this particular caseenough before. A box is on a horizontal surface, and is not accelerating inthe vertical direction. This means that the downward force of gravity isexactly counteracted by the upwards normal force, so the magnitude of thenormal force is Fn = mg, and hence the magnitude of the force of frictionis µk Fn = µkmg. Now, we are in a position to calculate the work done along the straight-line path from 1mˆı to 4mˆı. Since the force is constant we know that the

182 CHAPTER 11. WORK AND KINETIC ENERGYwork done is Wfric = Ff · ∆r (11.3) = Ff · (rf − ri) = Ff · (4mˆı − 1mˆı)We can find the direction of the force of friction by finding the unit vectorin the opposite direction to that piece of displacement. In this case, thevector describing the displacement is ∆r = 3mˆı, and the unit vector in thisdirection is ∆r = 3mˆı = 3mˆı = ˆı. To get the unit vector in the opposite |∆r| |3mˆı| 3mdirection, you multiply by −1, since negative signs tell us about direction.For this displacement, Ff = (µkmg) (−ˆı), so we can write the work as Wfrice = Ff · ∆r = Ff − ∆r · ∆r |∆r| = −µkmg |∆r| = −(0.5)(10kg) 9.8 N (3m) = −147J (11.4) kg The fundamental result that we got here: that kinetic friction will dowork equal to −µkmg |∆r| for straight-line displacements over a horizontalrough surface. This will be useful in subsequent calculations for other partsof this problem. Now, we calculate the word done on the second path. The key thing toremember here is that the definition we have of work as W = F · ∆r onlyworks for a constant force. Constant means both direction and magnitude,so there are three constant forces at work in the second path; one for eachof the three segments. We find √the displacements. For 1mˆı to 2mˆı + 1mˆ, ∆r = 1mˆı + 1mˆ,and |∆r| = 2m; for the 2mˆı + 1mˆ to 3mˆı + 1mˆ segment, ∆r = 1mˆı, so|∆r| = √1m; and for the 3mˆı + 1mˆ to 4mˆı segment, ∆r = 1mˆı − 1mˆ, and|∆r| = 2m.

11.2. WORK DONE ALONG DIFFERENT PATHS 183 We can now calculate the work done by friction in the three steps as Wf ric = Wstep 1 + Wstep 2 + Wstep 3 = −µkmg |∆r|step 1 − µkmg |∆r|step 2 − µkmg |∆r|step 3 = −µkmg |∆r|step 1 + |∆r|step 2 + |∆r|step 3 = −(0.5)(10kg) 9.8 N (1.41m + 1m + 1.41m) = −188J kg (11.5) Finally, we will calculate the work done following third (curved) path.This is a bit harder, since the path does not obviously split up into discretechunks. What we do is take the path we know, break it up into a bunchof tiny segments, calculate the work done by friction in each of them, andadd them up. The key idea is that if the displacement is small, the force offriction is approximately constant in direction. We know that the positionis given by r(t), and that in the time between t and t + dt the object willhave a displacement of vdt (remember that this is the same as saying thatd r = v). Then, using the rule that the work done is −µkmg |∆r| in a smalldtstep, this means that dW (the small amount of work done in a small timeinterval) is going to be −µkmg |vdt| = −µkmg |v| dt. Now, we just have tocalculate what |v| is: v = d r(t) dt = d 2.5m − 1.5m cos(πs−1t) ˆı + 1.5m sin(πs−1t)ˆ dt m m = 1.5π s sin(πs−1t)ˆı − 1.5π s cos(πs−1t)ˆ |v| = 1.5π m sin(πs−1t)ˆı − 1.5π m cos(πs−1t)ˆ s s = 1.5π m sin(πs−1t) 2 −1.5π m cos(πs−1t) 2 s s + = 1.5π m (11.6) s This tells us that dW = −µkmg 1.5π m dt. This is the amount of work sdone in time interval dt. We have to add up all the appropriate dW s (theones that make up part of our time interval for going between the two end-

184 CHAPTER 11. WORK AND KINETIC ENERGYpoints). stop W= dW start = 1s 1.5π m dt s −µkmg 0s = −µk mg1.5π m 1s s dt 0s = −(0.5)(10kg) 9.8 N 1.5π m (1s) = −231J (11.7) m s The different paths resulted in different amounts of work done. This tellsus that friction is not a conservative force.Things you should notice here • The work done by friction depended on the path taken. Since the work done depended on the path taken, this tells us that kinetic friction is a non-conservative force. • We calculated work done for a non-straight-line path by adding up the work done along each little straight bit that made up the path. • In the case of a curve which was ‘parametrized’ we ended up calculating its length by finding the length of each ‘tiny bit’ that occurred between some time t and another time t + dt.Student Exercises • Find the work done by gravity moving a box of mass m = 3kg from 1mˆı to 4mˆı + 3mkˆ along the paths specified – A straight line from 1mˆı to 4mˆı+3mkˆ. The work done is −88.2J. – A horizontal line from 1mˆı to 4mˆı, and then a vertical line from 4mˆı to 4mˆı + 3mkˆ. The work done in the first step is 0J, and in the second step it is −88.2J, so the total is −88.2J. – Along the path xˆı + x − 1m + 2m sin(xπm−1) kˆ from x = 1m to x = 4m. There are three terms you have to integrate. The xˆı term the integrand F · dr vanishes. The (x − 1m)kˆ term will give an integral of −88.2J, and the sin term has a non-zero indefinite integral, but when you substitute in the bounds of integration the definite integral turns out to be 0J. The sum is −88.2J.

11.3. BLOCK SLIDING ON A ROUGH SLOPE 185 Are the results the same or different? What does this tell you about whether the force of gravity is a conservative force? The results are all the same. Gravity is conservative.• Find the work done by the electric force on a charge q which is moved in a region with a constant electric field E: – If q = 1.6 × 10−19C, E = 5000 N ˆı, and the charge is moved in a C straight line from the origin to 1mˆı + 2mˆ. – If q = 1.6 × 10−19C, E = 5000 N ˆı, and the charge is moved from C the origin to 2mˆ, and then in a straight line to 1mˆı + 2mˆ. – If q = −1.6 × 10−19C, E = 5000 N ˆı, and the charge is moved from C the origin to 2mˆ, and then in a straight line to 1mˆı + 2mˆ. Can you tell if the electric force is conservative or not from this?11.3 Block sliding on a rough slopeExample A block of mass m is sliding with initial speed vi along a roughsurface with which it has a coefficient of kinetic friction of µk. The surfacemakes an angle of θ with respect to the horizontal. This is shown in figure11.2. If m = 3kg, µk = 0.4, θ = 20◦, vi = 5 m , how far along the surface will s mthe mass have gone when it has a speed of vf = 2 s ? In other words, forwhat d does vf = 2 m ? sWorked Solution We are going to attack this problem with the work-energy theorem. This tells us that ∆KE = Wnet 1 m |vf |2 − 1 m |vi|2 = Fnet · ∆r (11.8) 2 2Now, what we have to do is figure out the things that make up the net force,and express the displacement as an appropriate vector. In this step, we haveused a piece of intution - that all forces on the block are constant. We wouldhave to calculate the work as an integral if the forces were varying. The free-body diagram for this mass shown in figure 11.3. There area total of three forces: gravity, the normal force, and the friction force.Using the pˆ, nˆ coordinate system illustrated in figure 11.3 we can express

186 CHAPTER 11. WORK AND KINETIC ENERGY dFigure 11.2: A block of mass m is sliding along a rough surface which makesan angle of θ with the horizontal. The initial state, and the final state wherethe block is travelling at speed vf while being a distance d farther along theslope are illustrated. Fn θ θθ (b) FfF(a) gFigure 11.3: Part (a) shows the free-body diagram for the mass moving asdesribed in figure 11.2. Part (b) shows the pˆ, nˆ coordinate system we areusing for simplicity here.

11.3. BLOCK SLIDING ON A ROUGH SLOPE 187the individual forces as Fn = Fn nˆ Ff = − Ff pˆ pˆ (11.9) Fg = −mg cos θnˆ − mg sin θpˆ Fnet = Fn + Ff + Fg = Fn − mg cos θ nˆ + −mg sin θ − FfAs usual, the condition that the mass is not accelerating in the nˆ direc-tion tells us that Fn = mg cos θ. The net force simplifies as Fnet =− (mg sin θ + µkmg cos θ) pˆ. We have used the usual relationship betweenthe normal force and the force of kinetic friction here. The mass’s displacement up the slope is dpˆ. This means that the total work is 1 m |vf |2 − 1 m |vi|2 = Fnet · ∆r 2 2 = (− (mg sin θ + µkmg cos θ) pˆ) · (dpˆ) = −mgd (sin θ + µk cos θ) vf2 − vi2 = −2gd (sin θ + µk cos θ) (11.10)Here we have used the notation |v| = v (the magnitude of the velocity is thespeed).Using the known values of vf = 2 m , vi = 5 m , µk = 20◦, and θ = 20◦, s swe find that d = 1.5m.Things to notice in this problem: Here, we could have gotten the workdone by each individual force (note that the work would have been 0 for thenormal force, and would have been negative for gravity and for friction).Even had we done this, we still would have needed to use the second lawanalysis to get the magnitude of the force of friction. Also note that thisresult tells us what would happen for something going horizontally on arough surface (so θ = 0) or straight up (so that θ = 90◦)Student Exercises The problem of blocks sliding on a plane is a rich one.Here are some similar questions:• Consider the system described in figure 11.2. If µk = 0, θ = 30◦, andvi = 4 m , what is the maximum distance along the slope the block will s

188 CHAPTER 11. WORK AND KINETIC ENERGYFigure 11.4: A block of mass m slides along a rough surface with which ithas a coefficient of kinetic friction of µk. The surface makes an angle of θbelow the horizontal.travel? It will travel along the slope until it stops, when its velocitywill vanish. The maximum distance is 1.63m.• Consider the system shown in figure 11.4. If m = 3kg, vi = 4 m , sµk = 0.2, and θ = 30◦, what is the speed of the block when it hasreduced its height above the ground by 2m? It will be travelling at6.45 m . s• In the system shown in figure 11.4, for m = 3kg, and θ = 30◦, what value of µk is such that the speed of the block remains constant as it slides down? The required coefficient of kinetic friction if µk = 0.577.11.4 A mass in a loop-the-loopExample A ball of mass m slides down a frictionless path and enters aloop of radius R. If the initial height of the ball was h, and the ball startsfrom rest, what is the normal force on the ball by the track when the anglefrom the center of the circle to the ball is θ? This is illustrated in figure11.5. If m = 0.1kg, R = 0.2m, h = 0.6m, and θ = 30◦, find the numericalvalue of the normal force.Worked Solution Our strategy for this example is to use what we knowabout the work-energy theorem to find the speed of the ball at the pointspecified. From this, we can find the needed magnitude of the perpendicularcomponent of the total force. This total force is supplied by the normal forceand by the force of gravity; we can find the centripetal component of the

11.4. A MASS IN A LOOP-THE-LOOP 189 v θ h RFigure 11.5: A ball slides down from initial height h and enters a circle ofradius R along a frictionless track.force of gravity, and the rest is going to have to be supplied by the normalforce. We need to do a little geometry first. By looking at figure 11.5 we see thatthe z-component of the initial position of the ball is h, and the z-componentof the final position of the ball is R + R cos θ. The calculation that you didin a previous example should have convinced you that the work done by theforce of gravity is easy to calculate. We find that ∆KE = Wnet1 mvf2 − 1 mvi2 = Wnormal + Wgrav2 2 = 0 + Fg · ∆r = −mgkˆ · doesn t matter ˆı + (R + R cos θ − h) kˆ = −mg (R(1 + cos θ) − h) (11.11)Assuming that the initial speed is zero, we get that vf2 = 2mgh − 2mgR (1 + cos θ) (11.12) It’s worth talking about how we got the two works that went into thenet work: The work done by gravity is the simple one since gravity was aconstant force. For a constant force we know that W = F ·∆r; we didn’t careabout the x-component of ∆r because we knew (or can easily verify) thatchanging the x-component doesn’t change the value of the work done. Thework done by the normal force must, in principle, be calculated by an integralbecause the direction of the normal force changes: W = F · dr howeverthis can be simplified a lot: remember that this integral really means ‘break

190 CHAPTER 11. WORK AND KINETIC ENERGY Fn (b) (a) FgFigure 11.6: In part (a) there is a free body diagram for the ball on thefrictionless track from 11.5. In part (b) we see a convenient coordinatesystem: nˆ points to the center of the circle, perpendicular to the track, andpˆ point along the direction of motion of the ball.the path into a whole bunch of tiny steps, each with a constant force, andadd up the tiny amount of work for each tiny step.’ This saves us, becausethe normal force is for a small step perpendicular to that small step, soF · dr = 0 for each dr; and so W = 0 since we are adding up a whole lot of0s. From the relationship in equation 11.11 we can figure out the speed thatthe ball is moving at this point. We will draw a free-body diagram of theball at this point, displaced by θ from the top in figure 11.6. We can analyzethe free-body diagram and find that Fnet = Fg + Fn m v2 nˆ + something pˆ = [mg cos(θ)nˆ − mg sin(θ)pˆ] + Fn nˆ(11.13) RThe nˆ component of the net force must be m v2 since the ball is moving in Ra circle. The pˆ component of the net force tells us something about whatis happening to the speed; we do not care for the purposes of this questionwhat it is, since we only want to obtain the magnitude of the normal force,but notice that it’s negative. The ball is slowing down because the ball isgoing up which should seem reasonable.The nˆ component of equation 11.13 is m v2 = mg cos θ + Fn (11.14) Rand into this we can put the result of equation 11.12 combined the identifi-