PY105 (C1) Rolling Rolling simulation • Assignment 9 is due tonight at 10pm. • We can view rolling motion as a superposition of pure • Assignment 10 has been posted on WebAssign. rotation and pure translation. It is due the Weds. after the Thanksgiving break. Pure rotation Pure translation Rolling • Have a nice holiday. +rω +rω v ω ω v -rω +=v -rω v • For rolling without slipping, the rotational speed of the rim of the wheel (i.e., rω) equals the translational speed (i.e., v). • The net instantaneous velocity at the bottom of the wheel is zero (= v – rω = 0), while at the top it is twice the translational velocity of the wheel (= v + rω = 2v).

1 • A point on the rim of the wheel traces out a cycloid. 2 Modified from Duffy’s.

Condition for Rolling Without Big yo-yo Slipping If the yo-yo's axle is half the radius of the yo-yo, and the When a disc is rolling without slipping, ω = v/r and α = a/r yo-yo moves a distance L to the right when the rope where v and a is the horizontal velocity and acceleration of is pulled from above the axle, how far does the end the center of mass of the disc, respectively. of the rope move? ω rω 1. it doesn't move at all ω 2. L/3 6. 4L/3 v 3. L/2 7. 3L/2 4. 2L/3 8. 5L/3 -rω 5. L 9. 2L If there is no slipping, the net acceleration of the point of contact between the wheel and the surface is zero. Hence, v - rω = 0 or ω = v/r. Similarly, we can show that α = a/r. 3 4

The distance moved by the rope Another example of rolling without slipping: If the axle is half the yo-yo's radius, a point on the outer edge An accelerating car of the axle has a rotational speed equal to half the yo-yo's You are driving your front-wheel drive car on Comm. translational speed. Let’s call the translational speed v. Above Ave. You are stopped at a red light, and when the light the axle, where the rope is unwinding, the net velocity is 1.5 turns green you accelerate smoothly so that there is no slipping between your car tires and the road. During the v, because the rotational and translational velocities have the acceleration period, in what direction is the force of same direction. friction from the road acting on your front tires? Is it If the yo-yo moves a distance L, the end of the rope would static friction or kinetic friction? move a distance 1.5 L. This also illustrates that rolling equals 1. The frictional force is kinetic friction acting in the translation (L) plus rotation (0.5L). direction you are traveling. ω 2. The frictional force is kinetic friction acting opposite to the direction you are traveling. 3. The frictional force is static friction acting in the direction you are traveling. 4. The frictional force is static friction acting opposite to the direction you are traveling. 5 6

1 An accelerating car An accelerating car During the acceleration period, in what direction is the Car simulation force of friction from the road acting on your rear tires? Let’s first turn friction off. With no friction at all, pushing down Is it static friction or kinetic friction? on the accelerator makes the front wheels spin clockwise. They spin on the frictionless surface, the rear wheels do nothing, and the car goes nowhere. 1. The frictional force is kinetic friction acting in the Friction on the front wheels opposes the spinning, so it must direction you are traveling. point in the direction the car wants to go. For the front wheels 2. The frictional force is kinetic friction acting opposite to to roll without slipping, the friction must be static. the direction you are traveling. If we turn on friction to the front wheels only, the car 3. The frictional force is static friction acting in the accelerates forward with the back wheels dragging along the direction you are traveling. road without spinning. Friction opposes this motion, so it must 4. The frictional force is static friction acting opposite to point opposite to the way the car is going. Again, it must be the direction you are traveling. static friction. The static friction force acting on the front wheels is the force that accelerates the car forward. It is much larger than the friction force on the rear wheels, which just has to give the 7 rear wheels the correct angular acceleration. 8

Example 1: an accelerating cylinder Example 1: An accelerating cylinder (cont’d) A cylinder of mass M and radius R has a string wrapped around it, with the string coming off the cylinder We would expect the acceleration to be bigger than F/M only above the cylinder. If the string is pulled to the right if there is another horizontal force acting in the direction of F. with a force F, what is the acceleration of the cylinder What could such a force be? if the cylinder rolls without slipping? (Hint: Use the It is a force of static friction. Let’s draw it in as pointing to example of an accelerating car as a guide.) the right and solve for the acceleration.

F 1. a = F m F F 2. a < α α m Simulation F F 3. a > s m 9 10

Example 1: An accelerating cylinder (cont’d) Example 1: An accelerating cylinder (cont’d)

Take positive to the right, and clockwise positive for torque. Forces Torques 1 +FF−=+ Ma FF+=− Ma The normal force cancels Mg vertically. Apply Newton's S S 2 Second Law for horizontal forces, and for torques: 3 Forces Torques Adding these two equations gives 2 FMa = , 2 v v v ∑FMax = ∑τ = Iα 4F which leads to the surprising result, a = (> F/m). 3M +−FF+ = Ma 1 2 ⎛⎞a S ++RF− RFS == Iα MR ⎜⎟ 2 ⎝⎠R We can make sense of this by solving for the force of static friction. 11 1 FMaF=− =− FF−+= Ma S ++ S 2 43 The positive sign means that our initial guess for the direction a * Only for rolling without slipping can we use α = of the friction force being in the same direction as F is correct. R 11 12

2 Rotational Kinetic Energy Racing Shapes

Energy associated with rotation is given by an equation We have three objects, a solid disk, a ring, and a solid analogous to that for straight-line motion. sphere, all with the same mass, M and radius, R. If we release them from rest at the top of an incline, 1 For an object that is moving but not rotating: Kmv= 2 which object will win the race? Assume the objects 2 roll down the ramp without slipping. 1 For an object that is rotating only: KI= ω2 2 1. The sphere 2. The ring For an object that is translating and rotating simultaneously, the total kinetic energy of such as a rolling object is: 3. The disk 4. It’s a three-way tie 11 5. Can't tell - it depends on mass and/or radius. KmvI=+22ω 22

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Example 2: Racing shapes Example 2: Racing shapes (cont’d) Question: Which shape will have the biggest velocity after rolling down the slope? Find the frictional force. 11 Mgh=+ Mv22 Iω (1) 22 Solution: Let’s use conservation of energy to analyze the race v between two objects that roll without slipping down the ramp. Because the object rolls without slipping, we can use ω = R Let’s analyze a generic object with a mass M, radius R, and a Next, substitute IcMR = 2 , where c = ½ for the disc, 2/5 for rotational inertia of: the sphere and 1 for the ring. IcMR= 2 11 v 2 Start with the usual five-term energy conservation equation. Mgh=+ Mv22() cMR 22 R2 UKWii++ ncff =+ UK Both the mass and the radius cancel out! Eliminate the terms that are zero: U = K (Why is W = 0?) i f nc 11 gh=+ v22 cv Insert the expressions for the various terms. 22 2gh 1122 Solving for the speed at the bottom: Mgh =+Mv Iω (1) v = 22 15 1+ c 16

Example 2: Racing shapes (cont’d) What does this tell us? Let’s now find the frictional force:

Mgh = ½Mv2 + ½cMv2 Forces Torques α 2 Translational Rotational Mgsinθ – fs = Ma fsR = cMR (a/R) fs mgsinθ KE KE α (2): gsinθ – f /M = a (3): f /cM = a s s a 2gh h v = mg 1+ c (2) and (3) ⇒ gsinθ – fs/M = fs/cM θ ⇒ Mgsinθ = f (1/c+1) This result shows that the larger the value of c, the slower the s object is, because a larger fraction of the potential energy is ⇒ f = Mgsinθ/(1/c+1) directed toward the rotational kinetic energy, with less s available for the translational kinetic energy. Note that fs, which produce the torque that causes rotation, Simulation reduces the linear acceleration of the object. From here, we see 17 that fs is smaller for objects with smaller c. This is a different 18 way of seeing why objects accelerates faster down the incline.

3 A race Angular momentum

If we take the winner of the rolling race (the sphere) and The angular momentum of a spinning object is represented by L. race it against a frictionless block, which object wins the race? Assume the sphere rolls without slipping. v 1. LI= ωv 1. The sphere 2. The block 2. Angular momentum is a vector, pointing in the direction of 3. It’s a tie the angular velocity. 4. Can't tell 3. If there is no net torque acting on a system, the system's angular momentum is conserved.

4. A net torque produces a change in angular momentum that is equal to the torque multiplied by the time interval during 19 which the torque was applied. 20

Example 1: A figure skater Example 1: A figure skater (cont’d)

A spinning figure skater is an excellent example of angular momentum conservation. The skater starts spinning with her Question: When the figure skater moves her arms in arms outstretched, and has a rotational inertia of Ii and an closer to her body while she is spinning, what initial angular velocity of ωi. When she moves her arms close happens to the skater’s rotational kinetic energy? to her body, she spins faster. Her moment of inertia decreases, so her angular velocity must increase to keep the angular momentum constant. 1. It increases 2. It decreases Conserving angular momentum: 3. It must stay the same, because of conservation of vv energy LLif= vv IIiiω = fω f

Question: In this process, what happens to the skater's kinetic energy? 21 22

Example 2: A bicycle wheel Example 1: A figure skater (cont’d) Question: A person standing on a turntable while holding a 11 bicycle wheel is an excellent place to observe angular KI==ω 2 () Iωω × iiiiii22 momentum conservation in action. Initially, the bicycle wheel is rotating about a horizontal axis, and the person is at rest. Can 112 you predict what happens when the person flips the wheel to KIf ==ffω () I ffωω × f 22 bring rotational axis vertical?

The terms in brackets are the same, so the final kinetic Solution: The initial angular momentum about a vertical axis is energy is larger than the initial kinetic energy, because zero. If the person re-positions the bicycle wheel so its rotation

ωif< ω . axis is vertical, the wheel exerts a torque on the person during the re-positioning that makes the person spin in the opposite direction. The angular momenta cancel, so L = 0 at all times Where does the extra kinetic energy come from? about a vertical axis. The skater does work on her arms in bringing them closer to her body, and that work shows up as an increase in kinetic Flipping the bike wheel over makes the person spin in the energy. 23 opposite direction. 24

4 Example 3: Jumping on a Merry-go-around Example 3: Jumping on a Merry-go-around (cont’d)

The system clearly has angular momentum after the This is an example involving a rotational collision: completely inelastic collision, but where is the angular Question: Sarah, with mass m and velocity v, runs toward a momentum beforehand? playground merry-go-round, which is initially at rest, and jumps on at its edge. Sarah and the merry-go-round (mass M, radius R, and I = cMR2) then spin together with a constant It’s with Sarah. Sarah’s linear momentum can be converted to an angular momentum relative to an axis through the center angular velocity ωf. If Sarah's initial velocity is tangent to the of the merry-go-around. circular merry-go-round, what is ωf? Simulation L = Iω = (mR2)(v/R) Sarah Solution: Alternatively, we may also think of the linear momentum like a force gives rise to a torque: Lrp= sinθ What concept should we use to attack this problem? LRmv= sin(90°= ) Rmv Conservation of angular momentum. In this case, i . 25 The angular momentum is directed counter-clockwise. 26

Example 3: Jumping on a Merry-go-around (cont’d)

vv Conserving angular momentum: LLif=

Let’s define counterclockwise to be positive.

+=+Rmv Itotalω f

22 +=++Rmv() cMR mR ωf

We can treat Sarah as a point, a distance R from the center.

mv Solving for the final angular speed: ωf = cMR+ mR

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