Relation between cue ball and cue speed

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SUMMARY

The relationship between cue ball speed and cue stick speed is primarily governed by the masses of both objects and the coefficient of friction. The formula to determine the ratio of cue ball speed to cue stick speed is given by {v_{\rm ball}\over v_{\rm cue}}={2K\over K+1}, where K represents the mass ratio of the cue stick to the cue ball, typically around 3. This results in a speed ratio of approximately 1.5. The collision between the cue stick and the cue ball is nearly elastic, meaning that energy is conserved during the impact, allowing for efficient transfer of momentum.

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  • Basic understanding of friction and its effects on motion
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gkusser
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Hey,
I am currently designing a pool playing robot and have a problem with the physics behind it.
I need the relation between the speed of the cue ball and the speed of the cue just before it hits the ball. In other words: How fast has the cue move in order to reach a certain speed of the cue ball.

Thanks for your help,

Georg
 
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Well, that's not too hard. You just need to figure out:
1)The mass of the cue stick
2)Mass of the cue ball
3)Coefficient of friction

And you'll be able to figure out how much acceleration you need to move the cue at to get the ball to move at a certain deceleration.

Just to warn you, you'll have to work with acceleration more than velocity. The only velocity you have will be instantaneous.
 
I would expect that the speed of the cue ball is pretty close to the speed of the stick. With cork for a tip, there isn't much elasticity (but there will be a little), so the ball won't go much faster than the cue stick. The cue stick will lose some speed, but again, not a whole lot because you're still applying a force to it and you have momentum both in the stick and your arm. That all probably washes out and makes the ball pretty close to the speed of the stick.
 
This is one of the simplest problems in collision physics. Two masses collide elastically (or nearly so) in one dimension. So both energy and momentum are conserved. The result is that the ratio of cue ball speed to initial cue speed is

{v_{\rm ball}\over v_{\rm cue}}={2K\over K+1}

where K is the ratio of cue mass to ball mass. I think K is around 3, so the ratio of speeds is 1.5.

BTW, about elasticity. I disagree with Russ. If the collisions were inelastic, that would mean the cue tip absorbs the energy, and after a few hard "breaks" it would quickly be broken.
 
Originally posted by krab
BTW, about elasticity. I disagree with Russ. If the collisions were inelastic, that would mean the cue tip absorbs the energy, and after a few hard "breaks" it would quickly be broken.
Why would it break? Why would it not dissipate the energy as heat like everything else?
 
Cue tips have very little elasticity. That's all I have to say about that.
 
Just a word of explanation here: The term "elastic" in physics means that no energy is dissipated, so that all the energy remains kinetic. It is not the same as the common definition as in "elastic band". For example, it would not occur to a (non-physicist) person to describe billiard balls as elastic balls; however, when one billiard ball strikes another, the collision is almost perfectly elastic in the physics sense. By comparison, when a hammer hits a nail on the head, driving it partway into wood, the hammer does not bounce off, and that collision is almost perfectly inelastic; all the kinetic energy being dissipated into heat.

Other examples:

A hammer hitting a steel block resting on a large mass of concrete will bounce back -> elastic.

The same hammer hitting a block of lead on concrete won't bounce back -> inelastic.

In the last example, the lead is deformed significantly; that's how the energy is absorbed. Getting back to the pool cue example, it hits the cue ball and the pad at the end is negligibly deformed. That's what makes me think the collision is elastic.
 
Alright, thanks. I guess it's very elastic then. That was a nice definition, thanks:).
 

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