Pool Ball Terminal Velocity

Cornerman has this 100% correct. I am currently a tutor in physics and other things engineering and I graduate in December as a mechanical engineer. 8ballr you lost all credibility when you said that terminal velocity is 9.8 m/s^2. That Is acceleration and not velocity. Acceleration is the rate of change of velocity per unit of time. A velocity can never be an acceleration and vice versa.

Ya I was getting ahead of myself...didn't realize I would have to spoon feed you guys the obvious. Terminal velocity according to the op's question would be the same for all objects regardless of mass. Do I need to explain why or can you "mechanical engineers" figure it out?
 
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Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)

Not sure you understand what terminal velocity is. Without air resistance, I think most physicists would say the terminal velocity of a pool ball or anything else is the speed of light. Typically, the term "terminal velocity" refers to the maximum speed an object can reach under earth's gravity WITH air resistance. The air resistance is what imposes the limit on the speed of a falling object. A human being with clothing typical has a terminal velocity of anywhere from 120-140 mph. I would imagine a pool ball would be significantly faster than that. I'd take a stab in the dark guess that it is something like 200 mph. Hard to say though...I imagine a more dense object would likely have a higher terminal velocity than a less dense object, all other things being equal. It seems like a pool ball is more dense than a human being, although I may be wrong about this. It is surely harder on average, but maybe not more dense. A pool ball sized glob of fat probably weighs more than a pool ball. FOr some reason if I picture it falling at a fast speed, it seems the air would very much influence its fall. Maybe it doesn't have the density to fall as fast as a person, but it sure has a better shape I would think. But I would make a much more confident guess that it is MUCH MUCH faster than anyone can break.

Eager to read the thread and see what answers people give.

KMRUNOUT
 
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You're confused...nobody is talking about a vacuum here...only you. Terminal velocity is the same for most all falling objects of any mass...ie if you drop a golf ball and a bowling ball from a building they will hit the ground at the same time. The exception is if you drop a feather and a bowling ball...obviously the bowling ball will first...unless in a vacuum then they will hit at the same time...ie on the moon.

This is not at all correct. When you compare a golf ball and a bowling ball, you are selecting objects that share a number of qualities, and hence yield terminal velocities that are relatively close. However, you have already disproved your own claim with the example of the feather. Think about a ping pong ball, a beach ball, a 4x8 sheet of plywood, a lead ball bearing. They will all have different terminal velocities than a golf ball or a bowling ball.

It seems you are not understanding the term "terminal velocity".

Now a question to the people that actually understand physics: The "force of gravity" is the "same" for objects on earth because we are comparing objects that are of essentially negligible mass compared to the mass of the earth. But would it not be true that an *extremely* massive object would have a greater acceleration under gravity, in that the gravitational force exerted by the object on the earth would have an additive effect? I mean the whole m1m2/r^2...wouldn't an object that is 1/4 the mass of the earth "fall" with more acceleration? (Or rather "move towards the earth"). Is this right?

KMRUNOUT
 
lol dude...I hope you are better in pool than in physics.
http://www.physlink.com/education/askexperts/ae6.cfm

This is getting good. Few things entertain me more than this sort of thing. I believe the equation is E=WR^2, where E is entertainment value, W is the amount the person is wrong, and R is the amount they think they are right. 8ballr is really maxing out the value of E here.

Edit: bonus points for the fact that the link says exactly what Freddie was saying.

KMRUNOUT
 
Ya I was getting ahead of myself...didn't realize I would have to spoon feed you guys the obvious. Terminal velocity according to the op's question would be the same for all objects regardless of mass. Do I need to explain why or can you "mechanical engineers" figure it out?

Ok I get it. We have been trolled. It almost seemed like this was simply a very ignorant person wallowing in their own foolishness.

Good one man, you got us. Now I feel like the idiot for not getting this sooner.

KMRUNOUT
 
Ya I was getting ahead of myself...didn't realize I would have to spoon feed you guys the obvious. Terminal velocity according to the op's question would be the same for all objects regardless of mass. Do I need to explain why or can you "mechanical engineers" figure it out?

Lol. Because the OP stated "with no air resistance". But there is no terminal velocity without resistance. The object would continue to accelerate until it met the Earth. So I, and others, ignored the "no air resistance" in order to calculate a true terminal velocity. The real answer to the exact question would be: none - no terminal velocity
 
Thank you

How far must the average pool ball fall before Terminal velocity is reached?

Are you planning a fabulous trick shot? Lol. I'd say you'd have to drop the ball from a little more than 100m (about 350ft). It would take it about 5 seconds, and terminal velocity would be reached at about the same time the ball crashes into the table or ground. A more advanced physics guy can probably calculate the exact values.
 
Ask a sky diver, they will know terminal velocity. It is somewhere in the 125mph range near sea level, for a near-round object with no special wind resistance like a diver falling without a parachute. Feathers, lifting bodies, gliders, and parachutes of course can fall slower depending on wind resistance generated by the item's shape. Thicker air at sea level slows the terminal velocity.

The average break shot is well under 30 mph. I've heard some pro or pros can break 31-32 mph, but that would be a very fast break.

There is a cool smartphone app (free?) to show players the actual speed of their break. You put the cueball on the head spot - center of the table, opposite the second diamond - and rack normally with the lead ball on the foot spot, Set the phone on the rail, set the app, and fire away. The phone hears the first and second contact and calculates speed based on the known distance between the head spot and foot spot.

So far my best effort is 26-something mph, and if I don't concentrate my break speed falls to under 24 mph. I believe pros can break faster, but not twice as fast.


The OP wrote "Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)"

When he mentioned something he called Terminal Velocity 'with no air resistance', he showed no understanding of what Terminal Velocity actually is:
...the velocity at which a falling body moves through a medium, as air, when the force of resistance of the medium is equal in magnitude and opposite in direction to the force of gravity.

or, the maximum velocity of a body falling through a viscous fluid.

So the initial question is a non-question, or ignorant question, or a troll, pick one.
 
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Lol. Because the OP stated "with no air resistance". But there is no terminal velocity without resistance. The object would continue to accelerate until it met the Earth. So I, and others, ignored the "no air resistance" in order to calculate a true terminal velocity. The real answer to the exact question would be: none - no terminal velocity

The answer simply is less than c.

The same theory says that objects gain mass as they speed up, and that speeding up requires energy. The more mass, the more energy is required. By the time an object reached the speed of light, Einstein calculated, its mass would be infinite, and so would the amount of energy required to increase its speed

Reference https://www.google.ca/search?q=obje...ome..69i57.10694j0j7&sourceid=chrome&ie=UTF-8

BUT this is a moot point because...

g is only 9.81m/s/s on the Earth's surface. The inverse square law applies and, as you go further away, the acceleration decreases and decreases. At a 'great distance', the acceleration would be nearly zero!

Reference https://www.physicsforums.com/threa...um-at-the-acceleration-due-to-gravity.350292/
 

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Ya I was getting ahead of myself...didn't realize I would have to spoon feed you guys the obvious. Terminal velocity according to the op's question would be the same for all objects regardless of mass. Do I need to explain why or can you "mechanical engineers" figure it out?

We got that immediately. But you went with complete nonsense that Terminal Velocity = acceleration due to Earth gravity. The rest of assumed "no additional resistance," and four posts already told him the question wasn't worded correctly.

From there you showed how little understanding you had of anything. We saw it, don't worry. "A little ahead of yourself" doesn't remotely come close to the situation.
 
The answer simply is less than c.

The same theory says that objects gain mass as they speed up, and that speeding up requires energy. The more mass, the more energy is required. By the time an object reached the speed of light, Einstein calculated, its mass would be infinite, and so would the amount of energy required to increase its speed

This is moot.

If you are trying to think of terminal velocity as the speed at which an object dropped from a near infinite distance would strike the earth given that nothing else interfered (i.e. air, other planets or suns), this is simple to calculate and a small number relative to the speed of light, namely it is 11.2km/s, The speed of light is about 300,000km/s.

If I did my math right, that means the increase of mass from this falling-from-infinite-height is 0.00000007%. Good luck measuring that.

This is the escape velocity - they are an identical concept, basically one where all the energy from falling is converted into the equivalent potential energy, the latter is an integration from the earth's surface to infinite height where the gravitational force gets progressively weaker with height.

Slow college students do this derivation in first year physics; good ones do it in high school. A bit later they derive the special relativistic formula. (Well, they used to - now they learn to feel good about not being able to do math without asking Siri.)

We live today in the age of google -- people can find perfectly correct information that they can misapply and post authoritatively. :rolleyes:
 
We got that immediately. But you went with complete nonsense that Terminal Velocity = acceleration due to Earth gravity. The rest of assumed "no additional resistance," and four posts already told him the question wasn't worded correctly.

From there you showed how little understanding you had of anything. We saw it, don't worry. "A little ahead of yourself" doesn't remotely come close to the situation.

I answered the op's question as it was presented. You and others tried to change the parameters to fit your answers...that's called observation bias.
 
I answered the op's question as it was presented. You and others tried to change the parameters to fit your answers...that's called observation bias.

This from a guy who gave this gem:

Acceleration to a to terminal velocity...which for a cueball or a marble or just about any object in a free-fall position is about 195 km/h.

Try to spin that answer. Hint: it doesn't follow that you understood anything yet.
 
Good question! Are we in a vacuum?

Of more interest is that the interest in this topic has been dropping as well. I has halved at each new posting, and was not high to begin with.

As the Engineer said, paraphrased, "interest is close enough to zero for practical purposes".

---- caution politically insensitive matter below ---

For those not familiar with the joke it is a variant on zenos' paradox. There's a game show and they bring out a mathematician and an engineer, both male. They then open a curtain and a gorgeous lady stands before them, equal distances from each.

They both cheer up considerably.

The host says "I will ring a bell periodically, and each time, you will each move half-way to this beautiful lady".

The mathematician immediately looks distressed, and sits down and cries.

The host asks why -- he says "if I can only move half-way each time, at finite intervals, I can never quite get there. I quit".

The Engineer however is grinning and happy. The host now asks him "why are you so happy, did you not hear? Do you not understand math?"

The Engineer just smiles and says "but I can get close enough for all practical purposes".

--- we now return to this thread unfortunately probably still in progress, with interest at terminal velocity for sure ---
 
All I can imagine when asked about cueballs and terminal velocity. Is a published story tells of young men,in the quest of a bigger break made a interesting discovery. They removed the leadshot from a 12gauge shotgun shell,inserted the shell into the barrel chamber,stuck a modified shaft down the barrel,and let whitey loose. Wood splinters were everywhere. Whitey resembled humptydumpty and I believe paneling suffered schrapnel scarring! Perhaps a ear got pulled I just remember the breakshot.
 
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Yeah I goobed the question a little with the part about the air resistance. What I meant was, no additional wind or turbulence, not no air. Fortunately most of you understood my intent anyways.

So if a strong break can reach 30mph, approximately how high would I have to drop a ball from to reach 30mph?
 
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