How does the size of pool balls affect the amount of force needed for a shot?

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SUMMARY

The discussion centers on the physics of pool shots, specifically how doubling the size of pool balls affects the force required for a shot. It is established that if the diameter of the balls is doubled, their mass increases by a factor of eight due to volume changes. Consequently, the force needed to achieve the same shot becomes significantly greater than merely doubling the initial force, as factors such as increased rolling resistance and moment of inertia must be considered. The conversation highlights the importance of impulse and momentum transfer in understanding the dynamics of the shot.

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  • Familiarity with the principles of rolling resistance and moment of inertia.
  • Knowledge of impulse and its role in momentum transfer.
  • Basic understanding of the geometry of spheres and their motion.
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  • Research the concept of impulse in physics and its applications in sports dynamics.
  • Study the effects of rolling resistance on different surfaces, particularly in billiards.
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Joe seki
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TL;DR
Can someone clear this up for me please
DB4983DA-05E6-43F3-9C45-421317641239.jpeg

A=Cueball start position
CP=Centre Point where balls collide
B=The pocketHey guys, I was asked this question that seems completely obvious but I keep second guessing myself over it, I have included a basic picture to explain what I’m about to describe

Say you were to take the above pool shot, at just enough force for the yellow ball to reach B and fall into the pocket

How many more times of that force would be required if the balls were double in size?

My brain says it’s obviously just twice the force but considering there are 2 balls that will be increased in size I’m having doubts lol

Thanks in advance and sorry if this is a silly question 🙃

Joe Seki
 
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Joe seki said:
How many more times of that force would be required if the balls were double in size?
It depends.

For ideal pool balls on an ideal table, the question does not make sense since there is no minimum force that will do the job. Any force, no matter how small will be enough to make a pool ball roll an infinite distance.

So we must be talking about balls and tables that are not ideal.

In what way are they not ideal? You tell us.

Probably it will come down to how one models rolling resistance on felt.
 
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jbriggs444 said:
In what way are they not ideal? You tell us.
Exceptionally well said. :smile:
 
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jbriggs444 said:
Probably it will come down to how one models rolling resistance on felt.
I would also expect some rolling with slipping before the collision.
 
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Joe seki said:
How many more times of that force would be required if the balls were double in size?

My brain says it’s obviously just twice the force but considering there are 2 balls that will be increased in size I’m having doubts lol
Assuming that doubling the size means to increase the diameter of the ball by 2x, then the mass of each ball would increase by a factor of 8 (volume increase from 4π/3 to 32π/3 if we set the initial radius to 1). Thus the mass is now 8x that of normal and far more than 2x force is needed. Especially if we look at the moment of inertia increase (formula below).

Assuming the balls increase in mass by 2x, and by size proportionally, the force necessary is probably somewhat more than 2x since friction on the balls has now increased due to the increased mass of the balls. Then there's the increased moment of inertia. So not only are the balls twice as hard to accelerate linearly, the are also harder to accelerate rotationally proportional to r2 since the moment of inertia of a solid sphere is given by l =2/5MR2.
 
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Drakkith said:
also harder to accelerate rotationally proportional to ##r^2## since the moment of inertia of a solid sphere is given by ##l =2/5MR^2##
That part cancels out neatly. You double the radius and quadruple the moment of inertia, thus quadrupling kinetic energy. But you halve the angular velocity which quarters the kinetic energy. For a given linear velocity, rotational kinetic energy ends up proportional to mass alone. The radius is irrelevant.

If you look at angular momentum, force and acceleration, the cancellation still happens. This time the moment of inertia quadruples and the angular rotation rate halves, so angular momentum rises proportional to radius. But the moment arm of the pool cue (or of the felt converting slipping to rolling) increases as well.
 
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Joe seki said:
My brain says it’s obviously just twice the force but considering there are 2 balls that will be increased in size I’m having doubts lol
The number of balls is irrelevant.
The cue ball will stop on contact. So only one ball is moving at one time for an inline cannon, or carom shot.
 
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jbriggs444 said:
That part cancels out neatly. You double the radius and quadruple the moment of inertia, thus quadrupling kinetic energy. But you halve the angular velocity which quarters the kinetic energy. For a given linear velocity, rotational kinetic energy ends up proportional to mass alone. The radius is irrelevant.
My mistake then. Been a while since I took my physics 101 course.
 
Drakkith said:
My mistake then. Been a while since I took my physics 101 course.
One of those counter-intuitive notions that one stores in the "cool things" corner of the brain.

"A sphere rolling down a slope will accelerate down at the same rate regardless of radius, but hollow it out and it'll accelerate at a different rate"
 
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jbriggs444 said:
"A sphere rolling down a slope will accelerate down at the same rate regardless of radius, but hollow it out and it'll accelerate at a different rate"
And a can of chicken noodle soup will accelerate slower than an identical can that has been frozen. (I used to do this demo.)
 
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jbriggs444 said:
One of those counter-intuitive notions that one stores in the "cool things" corner of the brain.

"A sphere rolling down a slope will accelerate down at the same rate regardless of radius, but hollow it out and it'll accelerate at a different rate"
Gravity is geometry in more ways than one!
 
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Joe seki said:
Summary: Can someone clear this up for me please

just enough force for the yellow ball to reach B and fall into the pocket
The thing that gives the cue ball forward momentum from the cue is best not described as Force. What counts is for how long the force is applied and the momentum transferred will depend on the elasticity of the cue and tip, its mass and the speed it's traveling on contact. Force times time applied is called Impulse and the great thing about it is that you can treat it as a quantity, rather than messing about trying to specify how long the contact time is.

Probably the OP didn't actually want a Physics Lesson but the question is one of 'that family' of questions like "What's the force of a punch" and "What's the force of a traffic collision". Neither of those has a satisfactory answer if Impulse (or sometimes Work Done) aren't considered. With pool balls of equal mass, they will approximately 'exchange velocities', leaving the cue ball stationary - but it's never as simple as that unless they are on an ice surface (for example).
I often watch World Championship Snooker and I notice that the commentators (all highly qualified ex-players) seem to make a point of avoiding pseudo Physics in their descriptions of what's going on. They are the angels who fear to tread.
 
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