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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?
nicksauce said:Can you maybe clarify your question? Neither MV nor 1/2MV^2 represents a force in any way.
The correct definition of force is F = dp/dt.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.
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)
xArcherx said:Just though I would throw in the more accurate formula for momentum (or so it seems)...
mtworkowski@o said:But when looking at a trade off between speed and weight, we use MV and not 1/2MV^2?
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.
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.
Why would you do that? It adds a totally unnecessary complication for no good reason other than to make you look smart.
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,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?
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.
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.
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.
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.
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.
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.
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]
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.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...
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...
The weight of a bowling ball directly affects the force it exerts. The heavier the ball, the more force it will exert when it hits the pins. This is because the force of an object is directly proportional to its mass, according to Newton's Second Law of Motion.
Force and momentum are related concepts, but they are not the same thing. Force is the push or pull that an object exerts on another object, while momentum is the product of an object's mass and velocity. In bowling, the force of the ball hitting the pins causes a change in their momentum, resulting in them moving and potentially knocking down other pins.
There is no one ideal weight for a bowling ball to maximize force. The weight of a bowling ball should be chosen based on the individual's strength and bowling style. A heavier ball may exert more force, but it may also be more difficult to control and may cause strain on the bowler's arm.
The velocity of a bowling ball also plays a role in the force it exerts. As the velocity of the ball increases, so does the force it exerts. This is because force is also directly proportional to an object's velocity, according to Newton's Second Law of Motion.
Kinetic energy and force are also related in bowling. Kinetic energy is the energy an object possesses due to its motion, and it is directly proportional to an object's mass and the square of its velocity. This means that as the kinetic energy of a bowling ball increases, so does the force it exerts on the pins upon impact.