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Suppose there is a snooker ball rolling on a table. The velocity of the center of the ball is Rω and its direction is horizontally right. I don't understand why the point right below the center on the ball has zero instantaneous velocity. Much obliged if anyone can help me.

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marked as duplicate by John Rennie, anna v, Kyle Kanos, ACuriousMind, Ryan Unger Sep 29 '15 at 19:23

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    $\begingroup$ Assume there is no sliding of the ball. It is because the table is at rest, and the point touching, to touch , it must be instantaneously at rest otherwise there would be sliding of the ball (as on ice) . $\endgroup$ – anna v Sep 29 '15 at 5:41
  • $\begingroup$ Thanks a lot. But how do I explain by considering the contact point velocity relative to the center of the ball, and the velocity of the center relative to the table? $\endgroup$ – Linkin Wai Sep 29 '15 at 5:51
  • $\begingroup$ The contact point will always move relative to the center of the ball. That's rotational velocity of sorts and with constant speed, that will remain unchanged when it touches the table. However the velocity relative to the center of the table will zero at the point of contact. Think of it as 2 velocities that can be added together, rotational velocity (the ball turns) and directional velocity (the ball moves). I can try to give an example if you want but there's already many examples in the related question. $\endgroup$ – userLTK Sep 29 '15 at 7:00
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The ball is rolling along positive x direction with velocity v. The points of the ball directly below the center are rotating around the center of the ball in the -x direction with instantaneous velocity u equal to the distance from the center of the ball times the angular velocity of the ball. Where v = u you have the condition you are looking for.
Truly there is a line below the center of the ball perpendicular to the direction of v where all points on the line have the same zero instantaneous velocity.

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