Have you ever use your finger to press down one side of the ping-pong ball?If you press hard enough, you will find ping-pong ball running away from you.However, a few seconds later, the ping-pong starts to rolling back toward you. Why?

The ball starts with an initial velocity Vo, and an initial angular velocity w_o. Due to the friction force between ping-pong ball and the table
        F_u=u mg. (Blue arrow)

The center of mass of the ping-pong ball will slow down.
        V= Vo - a t

The friction force will produce a torque (ru mg, where r is the radius of the ball ) to change its rotating speed w = w_o -t * (r u mg)/I.

where I is the moment of inertia of the ball.

When the velocity of the contact point related to the table V - r w becomes zero, the friction force vanishes.

The motion become a free rolling (rolling without slipping).     V = r w

if V< 0. then ping-pong ball rolling backward (toward you). Is it passible to let the ball stop? (velocity V=0.) or moving forward(V>0.)?

Under what kinds of conditions? This java applet let you play with it. Enjoy!

The initial velocity of the ball is fixed. Vo= 200. cm/s. You can change different initial angular velocity w_o ( r*w_o was shown).

r*w_o<0 : counterclockwise,

r*w_o>0 : clockwise

Press start button to starts the java animation.

RightClick the mouse button to toggle the animation.

The white arrow below the ball represent the friction force.

Press Clear to clear the curves (Velocity V verses time t).

You can also change the friction coefficient u.

How to calculate?

1. What is the friction force between the ball and the table?
2. What is the acceleration of the ball？ How the speed changed?
3. What is the torque to the ball? ？How the angular velocity changed?
4. What is the velocity of the contact point relative to table?

Can you write down the equations of motion ? I did that already. It is your turn, now? （The momentum of initial for the hard sphere ball is I = (2/5) m r², I= (2/3)mr² for thin spherical shell )