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

  • B
  • Thread starter Joe seki
  • Start date
  • Tags
    Ball
In summary, the force required to make a pool ball reach the pocket depends on the size of the ball and the properties of the table. If the ball's diameter is doubled, the mass and moment of inertia will also increase, requiring more than twice the force to reach the pocket. The number of balls in play is irrelevant. This is due to the counter-intuitive concept that the acceleration of a rolling object is determined by its mass, not its size. Additional factors such as friction and the conversion of slipping to rolling on the table's surface may also play a role.
  • #1
Joe seki
7
0
TL;DR Summary
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
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Delta2, sophiecentaur, Bystander and 2 others
  • #3
jbriggs444 said:
In what way are they not ideal? You tell us.
Exceptionally well said. :smile:
 
  • Like
Likes Delta2 and sophiecentaur
  • #4
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.
 
  • Like
Likes sophiecentaur
  • #5
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.
 
  • Like
Likes Delta2, sophiecentaur and PeroK
  • #6
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.
 
Last edited:
  • Like
Likes sophiecentaur
  • #7
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.
 
  • Like
Likes jbriggs444
  • #8
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.
 
  • #9
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"
 
  • Like
Likes Drakkith
  • #10
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.)
 
  • #11
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!
 
  • #12
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.
 
  • Like
Likes jbriggs444

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

How many pool balls are used in a game of pool?

There are a total of 16 pool balls used in a game of pool. This includes 1 cue ball, 7 solid-colored balls, 7 striped balls, and 1 black 8 ball.

What is the standard size of a pool ball?

The standard size of a pool ball is 2 1/4 inches in diameter. This is the size used in most professional and recreational games of pool.

What are the different colors of pool balls and what do they represent?

The colors of pool balls are typically red, yellow, blue, purple, orange, green, and brown. In a game of pool, the solid-colored balls represent the "low" numbered balls (1-7) and the striped balls represent the "high" numbered balls (9-15).

How much does a pool ball weigh?

The weight of a pool ball can vary, but the standard weight is around 5.5 to 6 ounces. This weight is important for maintaining the proper speed and trajectory of the ball during gameplay.

Can pool balls be used on any type of pool table?

Yes, pool balls can be used on any type of pool table as long as the table is the correct size and has a smooth, flat surface. However, different types of pool tables may have slightly different playing characteristics.

Back
Top