Understanding the Impact of Bowling Ball Weight on Force: MV vs. 1/2MV^2

  • Thread starter mtworkowski@o
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In summary, because a lighter bowling ball can be thrown faster, the force that the ball has is altered. Neither MV nor 1/2MV^2 represent a force in any way.
  • #1
mtworkowski@o
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My experience is that because a lighter bowling ball can be thrown faster, the force that the ball has is altered. Should I use MV or 1/2MV^2?
 
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  • #2
Neither. Force is not described by 1/2mv^2 or mv.
 
  • #3
Can you maybe clarify your question? Neither MV nor 1/2MV^2 represents a force in any way.
 
  • #4
MV = P which is Momentum.
Ke = 1/2 mv^2 which is Kinetic Energy which can be exchanged for Potential energy.

Force is typically defined as , F=ma.
 
  • #5
nicksauce said:
Can you maybe clarify your question? Neither MV nor 1/2MV^2 represents a force in any way.

Yes. In the context of trying to deliver energy to the pins, I'm thinking M and V are trade offs as far as energy is concerned. A heavier ball at slower speed or a somewhat lighter ball at higher V.
 
  • #6
razored said:
MV = P which is Momentum.
Ke = 1/2 mv^2 which is Kinetic Energy which can be exchanged for Potential energy.

Force is typically defined as , F=ma.
The correct definition of force is F = dp/dt.

Pete
 
  • #7
Just though I would throw in the more accurate formula for momentum (or so it seems)...

p = ɣmv
ɣ = (1-v^2/c^2)^(-1/2)
 
  • #8
xArcherx said:
Just though I would throw in the more accurate formula for momentum (or so it seems)...

p = ɣmv
ɣ = (1-v^2/c^2)^(-1/2)

I'm getting the dp/dt thing from calc 1 97 years ago. thank you Pete. But I missed the thing from xArcherX. Sorry. I'm rusty as hell. Break it down, as they say, please. thanks.
 
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  • #9
xArcherx's perspective is essentially the same as what you've been learning, but in general form. In classical physics, we mostly deal with what we can experience, which is one of the things that makes modern physics different (it tells us we've only been seeing part of the picture, and that phenomenon start acting differently over different scales).

If you've noticed, there is a c in the equations, which is the speed of light. Compared to v (the velocity of the bowling ball), its probably much bigger, so the formula for momentum can more or less be regarded as p = mv for this situation with the bowling ball.
 
  • #10
xArcherx said:
Just though I would throw in the more accurate formula for momentum (or so it seems)...

Why would you do that? It adds a totally unnecessary complication for no good reason other than to make you look smart. (And I'm afraid adding totally unnecessary complications doesn't make one look very smart)

A bowling ball might travel at 20 mph. That means we are talking about a correction in the 15th decimal place. Not only is this totally unmeasurable, it's tiny compared to the correction to the mass of the bowling ball as it travels and picks up and drops off various bits of this and that. So not only is it an unmeasurable correction, it's not even the biggest unmeasurable correction.
 
  • #11
But when looking at a trade off between speed and weight, we use MV and not 1/2MV^2?
 
  • #12
mtworkowski@o said:
But when looking at a trade off between speed and weight, we use MV and not 1/2MV^2?

What are you actually trying to calculate? It would be a lot easier if we had a well-defined problem.

Note that the two things you have formulae for are momentum and kinetic energy, respectively.
 
  • #13
cristo said:
What are you actually trying to calculate? It would be a lot easier if we had a well-defined problem.

Note that the two things you have formulae for are momentum and kinetic energy, respectively.

which is better, the 16lb ball or a 15. More detail, if I or anybody can throw a 16 lb ball at the pins and deliver a certain amount of energy(let's say that's as fast as he can comfortably throw the ball) than would a lighter ball ( assuming he can throw it a wee bit faster) deliver more energy. I don't know if its momentum, kinetic energy or if either one will work.
 
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  • #14
Firstly, it is both conservation of momentum and conservation of energy that dictate what happens. I good starting point is to assume perfectly elastic collisions. Together, conservation of momentum and conservation of energy tell what happens.

Secondly, your physiology rather than physics is the determining factor here. You are missing a key factor, which is throwing a ball accurately. Suppose that you can throw a 16 lb ball fast enough to get more pin action than you can get with a 15 lb ball. That does not necessarily mean you should prefer the heavier ball. If the heavier ball makes you less accurate you might want to use the lighter ball.
 
  • #15
D H said:
Firstly, it is both conservation of momentum and conservation of energy that dictate what happens. I good starting point is to assume perfectly elastic collisions. Together, conservation of momentum and conservation of energy tell what happens.

Secondly, your physiology rather than physics is the determining factor here. You are missing a key factor, which is throwing a ball accurately. Suppose that you can throw a 16 lb ball fast enough to get more pin action than you can get with a 15 lb ball. That does not necessarily mean you should prefer the heavier ball. If the heavier ball makes you less accurate you might want to use the lighter ball.

Thanks DH...I did better with the 15. But this is more of an academic question I guess. Compare the energy delivery of both balls when an equal force is used in each case. Maybe that says it better. Sorry.
 
  • #16
Why would you do that? It adds a totally unnecessary complication for no good reason other than to make you look smart.

Yeah it is pretty much unnecessary at speeds that are not relativistic. Although, personally, I like to work with as close to exacts as I can. I also like to try and look smart :biggrin:, even if it does backfire on me (something that happens way too often):tongue:
 
  • #17
to all. you know my question is perfectly clear. If you can't answer it than I'm out. you can just debate it for the next thousand years, which I'm sure you will. By The idea of using relativistic physics on a bowling bowl. Ha Ha Hal. What a joke.
 
  • #18
mtworkowski@o said:
My experience is that because a lighter bowling ball can be thrown faster, the force that the ball has is altered. Should I use MV or 1/2MV^2?
If we assume that you want to accelerate each bowling ball to the same speed, because you only have one swing to do it and want the same final speed, then you can assume,
[tex]F\propto m[/tex].
Here's why.
[tex]\frac{1}{2}mv^2=W=\int_0^d ma dx=ma\int_0^d dx=mad[/tex]
So...
[tex]\frac{1}{2}mv^2=Fd[/tex]
and finally,
[tex]F=\frac{1}{2d}mv^2[/tex]
with d and v constants,
[tex]F\propto m[/tex].
 
  • #19
mtworkowski@o said:
Yes. In the context of trying to deliver energy to the pins, I'm thinking M and V are trade offs as far as energy is concerned. A heavier ball at slower speed or a somewhat lighter ball at higher V.

Assuming that you use the same length swing, and are able to exert the same force throughout the swing, then the energy delivered to the ball is found by

[tex]E= fd[/tex]

IOW, the ball will deliver the same energy regardless of its mass.
 
  • #20
Janus said:
Assuming that you use the same length swing, and are able to exert the same force throughout the swing, then the energy delivered to the ball is found by

[tex]E= fd[/tex]

IOW, the ball will deliver the same energy regardless of its mass.

So are you saying it's a one to one trade off between mass and velocity? So I don't use 1/2MV^2. That was my original question, I think.
 
  • #21
Janus said:
Assuming that you use the same length swing, and are able to exert the same force throughout the swing, then the energy delivered to the ball is found by

[tex]E= fd[/tex]

IOW, the ball will deliver the same energy regardless of its mass.

You know, I'm reading your remarks again and I think adding swing length is not a good thing to do. I'm iterested in the energy that can be delivered to the pins. Now that is either kinetic energy or momentum or both that we're talking about. In one formula V is squared in the other it's not. I though if you trade mass for V and V is squared, than you have an advantage in using a lighter ball and throwing it faster with the same force applied from your end. I hope that makes the question clearer. Sorry.
 
  • #22
Janus said:
Assuming that you use the same length swing, and are able to exert the same force throughout the swing, then the energy delivered to the ball is found by

[tex]E= fd[/tex]

IOW, the ball will deliver the same energy regardless of its mass.

PS. E=fd.....? I remember W=fd. I'm a bit confused by this E thingy.
 
  • #23
mtworkowski@o said:
You know, I'm reading your remarks again and I think adding swing length is not a good thing to do. I'm iterested in the energy that can be delivered to the pins. Now that is either kinetic energy or momentum or both that we're talking about. In one formula V is squared in the other it's not. I though if you trade mass for V and V is squared, than you have an advantage in using a lighter ball and throwing it faster with the same force applied from your end. I hope that makes the question clearer. Sorry.

Swing length determines how much energy you impart to the ball, ergo how much energy the ball has to impart to the pin.

So, assuming that you use the same swing to roll the lighter ball as you do the heavier ball, and assuming that you can apply the same amount force throughout the swing, the smaller ball will have a greater velocity when you release it, But it will have the same energy as the heavier ball would.

For instance, you can roll a 12lb ball faster than a 16 lb ball, but not [itex]1 \frac{1}{3} [/itex] times faster but only [itex]\sqrt{1 \frac{1}{3}}[/itex] times faster. As far as the energy of the ball goes, the gain of velocities effect is matched by the reduced masses effect.

BTW E stands for energy, which is the same as work, which the W stands for.
IOW E=W
 
  • #24
Could anyone tell me if MV is the formula or is it 1/2MV^2. Read my original question. This is a joke. Now I know whyu the Japanese are ahead of us. Answer my freakin' question please!
 
  • #25
You have to use both momentum and energy, just like I said in post #14.
 
  • #26
D H said:
Firstly, it is both conservation of momentum and conservation of energy that dictate what happens. I good starting point is to assume perfectly elastic collisions. Together, conservation of momentum and conservation of energy tell what happens.

Secondly, your physiology rather than physics is the determining factor here. You are missing a key factor, which is throwing a ball accurately. Suppose that you can throw a 16 lb ball fast enough to get more pin action than you can get with a 15 lb ball. That does not necessarily mean you should prefer the heavier ball. If the heavier ball makes you less accurate you might want to use the lighter ball.

When you said conservation of energy and conservation of momentum, you were telling me which formula to use?......OK? Soooooooooooooooooo, Which, oh forget it.
 
  • #27
MV is momentum (which is a measure of motion...or you could say inertia of motion). The more momentum something has, the more impulse is needed to stop it. Under inertial reference frames, the mass of a system can be considered conserved and so could velocity (inertial -- no acceleration)...which is why you can consider momentum a conserved quantity when there is no net force on the system.

1/2MV^2 is kinetic energy (energy of motion). If you've noticed, v is squared, so whatever change one does to the velocity has more of an effect than changing the mass.

If you're asking whether a lighter or heavier ball can be thrown faster: F = ma...when applying a force to something, you're transferring energy. Lighter and heavier refer to weight, and since we're assuming g (gravitational acceleration) as constant, heavier balls will have more mass compared to lighter balls. Let's say someone applies a certain amount of force to 2 balls (one heavier, m1, and one lighter, m2), in which the amount of force is equivalent...F = m1*a1 and F = m2*a2. Then F/m1 = a1 and F/m2 = a2. If m1 greater than m2, then a2 greater than a1 (and vice versa). So it is harder to accelerate heavier balls.
 
  • #28
In hopes of putting this thread to rest, here's the last and best explanation I can come up with...
The force you apply when throwing [itex]F_{applied}[/itex] times the distance you accelerate it with, using your hand [itex]d_{thrown}[/itex] is equal to the force the ball applies on the pins [itex]F_{pins}[/itex] times the distance it is slowed over by the pins, [itex]d_{slowed}[/itex]. So mathematically,
[itex]F_{applied}d_{thrown}=F_{pins}d_{slowed}[/itex].
There is no mass in either equation. There is no velocity in either equation, not so intuitive. For some more insight look to my other post. It contains the equation,
[tex]F=\frac{1}{2d}mv^2[/tex].
where,
-F is the force the ball applies to the pins,
-d is the distance the ball is slowed down over,
-m is the mass of the ball, and
-v is the initial velocity of the ball.

So if the force you apply to the ball and the distance you apply this force over is constant, the product [itex]Fd=\frac{1}{2}mv^2[/itex] is constant, thus only d, the distance the ball is slowed over, affects the force the ball applies on the pins. Mathematically,
[tex]F_{pins}\propto \frac{1}{d}[/tex]
 
  • #29
gamesguru said:
In hopes of putting this thread to rest, here's the last and best explanation I can come up with...
The force you apply when throwing [itex]F_{applied}[/itex] times the distance you accelerate it with, using your hand [itex]d_{thrown}[/itex] is equal to the force the ball applies on the pins [itex]F_{pins}[/itex] times the distance it is slowed over by the pins, [itex]d_{slowed}[/itex]. So mathematically,
[itex]F_{applied}d_{thrown}=F_{pins}d_{slowed}[/itex].
There is no mass in either equation. There is no velocity in either equation, not so intuitive. For some more insight look to my other post. It contains the equation,
[tex]F=\frac{1}{2d}mv^2[/tex].
where,
-F is the force the ball applies to the pins,
-d is the distance the ball is slowed down over,
-m is the mass of the ball, and
-v is the initial velocity of the ball.

So if the force you apply to the ball and the distance you apply this force over is constant, the product [itex]Fd=\frac{1}{2}mv^2[/itex] is constant, thus only d, the distance the ball is slowed over, affects the force the ball applies on the pins. Mathematically,
[tex]F_{pins}\propto \frac{1}{d}[/tex]

I don't think anyone can read. WHICH FORMULA DO I USE? I'M SICK OF ASKING. WHICH ONE A:MV B:1/2MV6^2 SAY A OR B. DON'T SAY ANYTHING ELSE. THANK YOU AND GOODBYE...
 
  • #30
When looking at a trade off between energy and mass, we use E = mc^2. But what do you mean MV or 1/2MV^2?...what exactly is your question (state it as precisely as possible so that we could sort of grasp what you're asking)?
 
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  • #31
mtworkowski@o said:
I don't think anyone can read. WHICH FORMULA DO I USE? I'M SICK OF ASKING. WHICH ONE A:MV B:1/2MV6^2 SAY A OR B. DON'T SAY ANYTHING ELSE. THANK YOU AND GOODBYE...
You don't "use" either formula. You're forcing us to answer an impossible question. If you want to know which is most closely related to the force the ball applies, it's 1/2mv^2. Don't get mad at us because you don't know how to ask a question right. You can't use either formula, it's dimensionally wrong. Instead divide 1/2mv^2 by d and you will get the force that it applies. End of story.
 
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  • #32
mtworkowski@o said:
I don't think anyone can read. WHICH FORMULA DO I USE? I'M SICK OF ASKING. WHICH ONE A:MV B:1/2MV6^2 SAY A OR B. DON'T SAY ANYTHING ELSE. THANK YOU AND GOODBYE...

Nice attitude. I think you need to spend more time trying to help people understand your poorly presented problem rather than throw a hissy fit and insult them.
Just use B, for what it's worth. Do you even know what to do with it to try to solve your "problem"?
Also, you might want to look a little harder at a physics textbook and get a basic understanding of things next time, or at least so you can get the terminology right (you know, before you go calling other people stupid).
 

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