1. Bowling Balls - Qualitative a)The ball begins by sliding; kinetic friction dissipates energy and reduces v. A small amount of rotation begins immediately. v0v0 f  f v '' v'v' b) Torque due to the frictional force results in an ang. accel. .  increases and v decreases. c) When v =  R, the ball starts to rotate without sliding at all. There is no friction force and the angular speed remains constant.

2. Bowling Balls - Quantitative Assumptions: m = 5 kg; 2R = 25 cm; v 0 = 9 m/s;  k = 0.1 Newton’s 2 nd Law:Since we can solve to find a cm = -1.0 m/s 2 Thus, the speed decreases due to friction, At the same time, the angular speed is increasing due to the torque: which leads to when we use The angular speed is given by We can solve the two equations to find the time at which the non-slip condition v =  R will take effect. The result is t = 1.5 s. After that time, which corresponds to a distance  x = 12.4 m, the ball rolls without slipping, and frictional forces no longer enter the problem.

3. Cue Balls - Qualitative y R F f a)If the cue stick hits the ball at the height of the center of mass, the ball will initially translate without rotating. After that, it becomes like the bowling ball. b)If the cue stick hits the ball below the center of mass, there will be an initial backspin, along with translation. Motion will be rotation with slipping. c)At some height y above the center of mass height, it will be possible to hit the ball so that it starts translating and rotating without slipping.

4. Cue Balls - Quantitative Newton’s 2 nd Law for translation and rotation gives Given our goal of finding the position y at which we obtain rolling without slipping, along with the result of the bowling ball problem that f is irrelevant, i.e. can be set to zero, for non-slip rolling, we can solve the above to find

5. Cue Balls - Continued Let’s look at this problem in a slightly different way. We write down Newton’s 2 nd Law equations again: Assumptions: If I start by saying there is rolling without slipping, and now solve these two equations to eliminate f, I will arrive at the following result: This form is interesting because it effectively gives me a force acting through a lever arm of y + R and on the right-hand-side I have a moment of inertia calculated with respect to the contact point on the surface. For any object, the moment of inertia about a given axis is I = I cm = mh 2 where h is the separation between the axis passing through the center of mass and the new rotation axis. (“Parallel axis theorem;” see Tipler, p. 265) I have assumed here, however, that there is a real force acting along the plane.

6. Rolling on an Incline  Newton’s 2 nd Law: FnFn mg f =  F n We have done this problem before. Use the torque equation to eliminate the frictional force in the first relation, then solve for the center-of-mass acceleration. The key to solving this problem was to assume a torque about the center of mass due to the (static ? kinetic ? ) frictional force.

7. Rolling on an Incline, revisited Starting with the same equations, let’s concentrate on the angular acceleration, using the first equation To eliminate f in the torque equation. This gives where the non-slip condition has been used again. I chose to write this result in this way because it illustrates that the problem could have been approached from a different point of view. The first term in parentheses is a force (the weight), the second term is a lever arm through which the weight operates with respect to the point of contact on the plane. On the right-hand side we have the moment of inertia about the contact point, given again by the “parallel axis” theorem. Using the above result, we arrive at the same result as previously for the center of mass acceleration. The point: I could have started with the drawing on the previous slide and used it to calculate the angular acceleration without ever referring to the friction at all!

8. Rolling on an Incline, revisited Does the previous result mean that friction is irrelevant? No! I can use Newton’s 2 nd Law to find the frictional force explicitly. The result is (again, assuming rolling without slipping). Note that this  (defined by everything in square brackets) goes to zero as the Incline angle goes to zero. (Agreeing with our bowling ball example?)

9. The Wooden Yo-yo  T R1R1 R2R2 f Newton’s 2 nd Law Equations: Qualitatively, we saw in class that for small angles, the yo-yo will roll to the left, while larger angles result in a center-of-mass motion to the right.

10. The Wooden Yo-yo, cont’d. The first question is, for what angle does the translational acceleration become zero? If we assume the non-slip condition a = acm/R we can solve these equations to eliminate T. The result is Although this looks complicated at first, we see that a cm = 0 if the square bracket on the rhs is zero, which leads to For an inner axle with half the radius of the wheel, we find  = 30˚.

11. Real Frictional Forces Taking a look back at the result for the cue ball in which the torque about the point of contact was calculated, we see after staring at the equations for awhile that I have effectively assumed that there is a frictional force acting all the time. Of course, we would assume that there is such a force in a real-life system. The same is true for the yo-yo problem; with the results calculated thus far, it is possible, for example, to plot the center-of- mass acceleration as a function of the angle at which the string is pulled:  = 30˚ R 1 / R 2 = ½  = 0.1 Question: What is this force that now seems to be acting between the rolling object and the surface?

12. Real Frictional Forces, cont’d Surface deformation and “plowing” will take place in any real situation, at least to some extent. To the extent that not all of the normal forces act radially inward on the wheel, there will be a net force, and therefore a net torque that acts to reduce the angular speed of the wheel: left to roll on its own, the wheel will come to a stop. I will leave the story here, however. F net