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My teacher gave me this exercise:

Consider a billiard table with balls of radius $r$. How high should be the edge of the table to not allow undesirable pressures (and thus slips of the ball)? I.e. if there was no gravity, the billiard ball would still bounce off parallelly to the table plane.

However I do not even understand what exactly it says. My answer would be $r$ because otherwise I can see forces that push the ball off the table (in the upward direction). But that seems like a too easy answer for a calculus class.

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  • $\begingroup$ Without further information i would be with you... The table edges usually have an angle, so the ball only "sees" one height and not a wall. If its <r, you get upward forces. If its >r, you get downward forces which translate into upward forces by bouncing off the table itself. $\endgroup$
    – Anedar
    Commented Nov 22, 2017 at 22:34
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    $\begingroup$ Related: physics.stackexchange.com/questions/62619/… There is some very relevant discussion in the comments to the question. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 22, 2017 at 22:51

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This is a well known problem. Since this is obviously a homework problem, I will not give you the full solution (but it's easy enough to find online). Instead I will take you through some of the steps:

When the billiard ball is rolling (this is important) towards the edge with velocity $v$, it has both linear and angular momentum. Linear momentum is $mv$, angular momentum is $I\omega = \frac25 m r^2 \omega = \frac25 m v r$. The impact with the wall will impart an impulse $F\Delta t$. This reverses the direction of both linear and angular momentum. The change in angular momentum is given by $Fh\Delta t $ where $h$ is the height of the bumper.

Now you should be able to solve the problem. If you are still stuck, see this

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  • $\begingroup$ Thank you. I am sorry if this is stupid, but the last time I've taken any lessons on physics was on high school (I had to read a lot of articles on wiki just to understand a bit of your comment). And it is not a graded homework, more like an exercise on integration :) So if $L = I\omega$: $\Delta L = Fh\Delta t $. $\Delta L = \frac25 m \Delta v r$ and $F = ma$. I can see I can get rid off $m$, but what is $a$, the ball is certainly not accelerating? $\endgroup$
    – SlowerPhoton
    Commented Nov 23, 2017 at 8:57
  • $\begingroup$ EDIT: It just ocurred to me: isn't $a = \Delta v$? $\endgroup$
    – SlowerPhoton
    Commented Nov 23, 2017 at 9:06
  • $\begingroup$ Use $m\Delta v = F \Delta t$ instead - you don't need to know the acceleration, only the integral of acceleration with time (change in velocity). $\endgroup$
    – Floris
    Commented Nov 23, 2017 at 12:00
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Your answer (that the edge should be at height r) sounds right to me. That is the only height that would cause the ball to bounce off in the same plane as the table. Of course in practice the bumper is somewhat elastic, so when the ball hits it the bumper will deform a little, and the area of contact will be a small area and not a single point.

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