Billiard Ball Center Of Percussion Question, Please

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SUMMARY

The discussion centers on the physics of a cue stick striking a billiard ball at its center of percussion (CP). It is established that when the cue strikes the ball at its CP, a horizontal force component facilitates pure rolling, while a vertical force component may push the ball into the surface. The conversation emphasizes that to achieve both linear and angular momentum, the cue must strike above the ball's center of mass. The moment of inertia of the ball is defined as (2/5) mR², and the relationship between rolling without slipping and angular velocity is clarified as R d(θ)/dt = v.

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Robert11
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Hello,

This isn't my area, so please bear with me a bit.

Am curious about what happens when a cue stick strikes a billiard ball at the ball's center of percussion. I understand center of percussion concept, I think, and have looked at the math and equations, and O.K.

But, am not sure how to think about, or handle this:

Assume the cue stick is horiz, and strikes the ball at the cp.
The only was a "force" can be transmitted to the ball, I think, would be on a line perpendicular
to a plane tangent to the ball at the cp. True ?

Therefore, there would be a horiz and a vert force component acting on the ball.

It's the vertical component that bothers me.

The horiz one would be responsible for the "pure rolling" the ball would experience as it is being struck at the its cp. True ?

But what about this vertical component ?
Does it tend to push the ball into the surface ?

How should one think about this vertical component, assuming it does exist ?

Or should one consider the cue stick as just putting an "Impulse" into the ball ?
If so, does the Impulse have a vertical component also ?

Much thanks,
Bob
 
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If the cue strikes the ball at its vertical center of mass, the ball will want to slide at first, and gain angular velocity only via friction with the table. The ball has to be struck above its center of mass to impart both the linear momentum and angular momemtum. The moment of inertia of the ball is (2/5) mR2. If it rolls w/o slipping, then R d(theta)/dt = v.
 

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