Billiard Ball, Conservation of Angular Momentum, Force below CM point

[mancity]

Homework Statement

A billiard ball of mass m initially at rest is given a sharp blow by a cue stick. The force is horizontal and is applied at a distance 2R/3 below the center line as shown in the figure. The initial speed of the ball is v_o and the coefficient of kinetic friction is u_k.
(a) What is the initial angular speed w_o?
(b) What is the speed of the ball once it begins to roll without slipping?
(c) What is the initial kinetic energy of the ball?
(d) What is the energy loss to heat as it slides on the table?

Relevant Equations

L=angular momentum about CM + angular momentum due to translation

The solution is also attached. I understand everything here except for part (b). What I don't get - why is it that when the ball starts rolling without slipping, we use moment of inertia about the point of contact instead of about the center of mass; and why is it that when the ball starts rolling without slipping we have no more angular momentum due to translation (mv_0 r)? Thanks.

 

[kuruman]

When the ball rolls without slipping, the axis of rotation is the point of contact with the surface. No slipping means that the contact point P on the ball is instantaneously at rest relative to the corresponding contact point Q on the surface. Point P is the axis about which the CM instantaneously rotates with angular speed $\omega.$ Of course the angular momentum about point P is $$L_P=I_P\omega \tag{1}$$ where $I_P$ is the moment of inertia of the ball about the CM. It is given by the parallel axes theorem $$I_P=I_{CM}+mR^2$$ where $I_{CM}$ is the moment of inertia about the center of mass. Thus, we can see that the angular momentum of the ball in equation (1) can also be written as $$L_P=I_{CM}\omega +mR^2\omega. \tag{2}$$ The first term in equation (2) is the angular momentum of the ball due to its spin about the CM and the second term is the term that you call "angular momentum due to translation" that you get when you replace $\omega$ with $v/R.$

 

[haruspex]

What I don't get - why is it that when the ball starts rolling without slipping, we use moment of inertia about the point of contact

The reason for using it is that the friction has no moment about that point, so the moment of inertia about it is conserved.

 

[mancity]

When the ball rolls without slipping, the axis of rotation is the point of contact with the surface. No slipping means that the contact point P on the ball is instantaneously at rest relative to the corresponding contact point Q on the surface. Point P is the axis about which the CM instantaneously rotates with angular speed $\omega.$ Of course the angular momentum about point P is $$L_P=I_P\omega \tag{1}$$ where $I_P$ is the moment of inertia of the ball about the CM. It is given by the parallel axes theorem $$I_P=I_{CM}+mR^2$$ where $I_{CM}$ is the moment of inertia about the center of mass. Thus, we can see that the angular momentum of the ball in equation (1) can also be written as $$L_P=I_{CM}\omega +mR^2\omega. \tag{2}$$ The first term in equation (2) is the angular momentum of the ball due to its spin about the CM and the second term is the term that you call "angular momentum due to translation" that you get when you replace $\omega$ with $v/R.$

Thank you. To clarify: prior to when the object starts its rolling without slipping, the axis of rotation is the CM of the ball, but after rolling without slipping begins, the axis of rotation is (instantaneously) the point of contact between the ball and the floor?

 

[haruspex]

To clarify: prior to when the object starts its rolling without slipping, the axis of rotation is the CM of the ball

No, if that were true the ball would have no velocity. The instantaneous centre of rotation will, in this case, start above the height where the ball is struck (I make it $\frac 35R$ above the ball's centre, with the ball rotating backwards). Subsequently, the linear speed drops and the backwards rotation likewise.
The instantaneous centre of rotation rises. When it is at height y above the ball's centre, using the usual convention that up, right and anticlockwise are positive, $v-y\omega=0$. Eventually, the rotation falls to zero and the instantaneous centre passes through $+\infty$ to $-\infty$. The rotation is now forwards.
The forwards rotation increases as the instantaneous centre rises until it becomes the point of contact.

 

[kuruman]

It is usually easier to visualize the rolling ball (with or without slipping) as being the sum of two motions: (a) translation of the CM with velocity $V_{cm}$ and rotation about the CM with angular speed $\omega.$ Unless the ball translates on a frictionless surface, there is always rotation about the CM.

Consider a point at some distance $r$ on the radius joining the center and the point of contact P. The velocity relative to the surface of this point is in the horizontal direction and has magnitude $$v=V_{cm}-\omega~r.$$ The picture below shows the side and end views of a yo-yo that may roll (a) either in contact of its outer radius $R_P$ with a black surface or (b) in contact of its inner radius $R_Q$ with a red rail.

If (a), then point P is at rest relative to the black surface while point Q moves in the forward direction with velocity $v_Q=V_{cm}-\omega~R_Q.$

If (b), then point Q is at rest relative to the black surface while point P moves in the backward direction with velocity $v_P=V_{cm}-\omega~R_P.$ This is the case, for example, with rail car wheels riding on railroad tracks.

Also note that rolling with slipping occurs when $\omega~R_P >V_{cm}.$ In this case friction would normally reduce both $\omega$ and $V_{cm}$. Eventually the inequality becomes an equality and rolling without slipping begins.

However, in all cases and as long as $\omega \neq 0$, there will be rotation about the CM. You can easily see why that is. Imagine moving in a car with velocity $V_{cm}$ next to the wheel. You see the center of the wheel at rest with respect to you but you also see the wheel spin about its axis. Just match speeds with the car on the adjoining highway lane and you'll see what I'm talking about.

You may wish to read this article that I wrote a few years ago having in mind people with questions like yours.

 

[Lnewqban]