# Bowling ball rolling with slipping at the instant it stops

I have done lots of algebra working for this, but I've decided to not include it because I would prefer an explanation in words. There are some questions on here similar to this, but please may I ask that you make sure they address the exact doubt here before marking as a duplicate.

Suppose I have (uniform) ball rolling on a rough surface with backspin, where we model friction with a coefficient of sliding friction $$\mu$$. The friction will produce a forward angular rotation(i.e. the initial backspin decreases as the ball rolls). We can choose the spin to be large enough such that when the translation velocity is zero(i.e. translation KE lost to friction), the angular velocity is nonzero. I would like to know what happens at this instant. Does friction point towards or away from the initial position? I am told that it will begin moving towards its initial position, but I don't understand how this can happen as friction will be pointing in the same direction as velocity.

My thoughts where that it would come to rest, and perhaps spin on the spot for a second or two, but some other neglected force of rotational friction will stop it. However, after reading this thread, where a live experiment is suggested(basically do the same thing with a hula hoop), it did return instead of going to rest.

It is also said in the linked thread that the backspin determines the direction of friction, not the velocity, which is new to me ..

• Have you ever seen low english (aka backspin) put on a cue ball while playing billiards? Jul 25, 2021 at 22:01
• Friction resists relative motion on two sliding surfaces, not the absolute motion of the center of mass. Jul 26, 2021 at 13:23

The direction of friction does not flip when the ball stops rolling forwards and reverses direction.

The figure below shows a progression with time from right to left. At each stage, it is important to consider where the center of rotation is (blue circle). Above the rotation center, the velocity vectors point to the right, and below the velocity vectors point to the left.

1. Initially the ball moves to the left with a large amount of backspin (clockwise rotation). The slip velocity at the contact contains the effects of the forward motion as well as the rotation. The velocity on the top is small at this stage.

2. As friction slows down the velocity of the ball significantly, the rotation center moves downwards reducing the slip velocity and increasing the velocity on top.

3. At this point the ball has reversed the direction of travel, but the rotation is still in the same sense (clockwise). The rotation center is now below the center of the ball and the contact is still sliding to the left, while most of the ball is moving to the right.

4. At this point the sliding motion ceases and the ball is under pure rolling to the right, with rotation still clockwise. The rotation center is at the contact point (no-slip condition).

For cases 0 and 1, friction force opposes the forward motion of the ball. But for case 2, friction is acting in the same direction as the motion of the ball. This is where you think there is a problem. But in reality, this is not an issue, as friction still opposes the relative motion at the contact.

Doing a quick simulation I found the initial conditions for this to happen. Below the dot is at coordinates $$(v,\omega R)$$ and as the simulation progresses it moves down and to the left with a slope equaling $$I/m R^2$$ where $$I$$ is the mass moment of inertia, $$m$$ is the mass and $$R$$ is the radius.

If the graph crosses the x-axis and the ball still has some spin, it will start rolling backward (stages 2 and 3 above). But if the initial spin is not high enough, then the graph is going to cross the x-axis with still some forward speed.

The requirement is thus $$\omega > \frac{m\,v R}{I}$$ or for a solid ball with $$I=\frac{2}{5} m R^2$$ $$\omega > \frac{5}{2} \frac{v}{R}$$

Kinetic friction points in the direction that opposes the relative motion between the surfaces. So if the ball is still spinning while its center of mass is at rest, then based on how I understand your description the friction force will continue to point towards the initial position at that instant.

• Ah I think I see what you're suggesting! The velocity of the centre of mass is irrelevant, its the velocity of the point at the bottom of the bowling ball? Jul 25, 2021 at 17:58
• @user950314 Yes, that is correct. The velocity of that point relative to the surface it is moving along. Jul 25, 2021 at 18:00

Quoting your word: "but I don't understand how this can happen as friction will be pointing in the same direction as velocity."

The bold sentence is not correct. The kinetic friction is opposite to your initial velocity, also the torque is also opposite to your initial back spin. Because at the contact point, the relative velocity to ground is

\begin{align*} V_{contact} &= V_c \text{ (forward) } + R \omega \text{ (forward) };\\ f_k &\text{ in backward direction.} \end{align*}

Therefore, during the initial stage of motion, your velocity is de-acceleration, and the backspin slowing down.

At reached the condition that the forward speed is vanished (due to the kinetic friction), but the large backspin survived.

\begin{align*} V_{contact} &= 0 + R \omega \text{ (forward) };\\ f_k &\text{ still in the backward direction.} \end{align*}

This backward kinetic frictional force renders the ball to roll back. It is just like you throw a back-spin basket ball, it will bounce back to you.