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Yoyo calculation

by drPaZQaL
Tags: calculation, yoyo
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drPaZQaL
#1
Jan21-04, 09:49 AM
P: 4
I need to do a practical about a yoyo.

We made a video (somebody keeps te jojo from teh ground a certain height and let it fall, after a certain time it doesn't move anymore and stays at the bottom) in which you can see the motion and with a programma we can see the falling speed etc. But we can't measure the rotation speed, we need to derive it:


I found out that the energy comparison=

m*g*h = 1/2*m*v^2 + Rotational kinetic energy

normally rotational kinetic energy = 1/2 * I * w^2

But our teachers says that this is the comparison for a cylinder, I need to derive a comparison myself, it should be possible with a integral calculus.

Can somebody help me?

Thanks,
dr. PaZQaL
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Doc Al
#2
Jan21-04, 10:08 AM
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Originally posted by drPaZQaL
m*g*h = 1/2*m*v^2 + Rotational kinetic energy

normally rotational kinetic energy = 1/2 * I * w^2

But our teachers says that this is the comparison for a cylinder, I need to derive a comparison myself, it should be possible with a integral calculus.
I'm not sure what you are asking. I don't see anything in your two equations relating to a cylinder. Your first equation is just energy conservation; the second is the general definition of rotational KE.

Also, I see no big problem in modeling the yo-yo as two disks (essentially a cylinder) in order to calculate its moment of inertia.
drPaZQaL
#3
Jan21-04, 10:51 AM
P: 4
ok, my explanation was not very clear...

The problem is what is "I"? (normally: 0,5 * m * r^2)

This can be calculated with a certain integral calculus, like:

http://scienceworld.wolfram.com/phys...aCylinder.html

Because the jojo is not a cylinder i need some formula to calculate I..., this should be possible with a calculation that looks like a calculation of the centre of gravity.. (like link)

Doc Al
#4
Jan21-04, 11:39 AM
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Yoyo calculation

Originally posted by drPaZQaL
Because the jojo is not a cylinder i need some formula to calculate I..., this should be possible with a calculation that looks like a calculation of the centre of gravity.. (like link)
Model the yoyo as two disks. Of course a disk is just a cylinder, so the formula to calculate the moment of inertia is the same: [itex]I=\frac{1}{2}MR^2[/itex].
ophecleide
#5
Jan21-04, 10:25 PM
P: 34
I'm not sure how you could get a more accurate value for this than the two-cylinders model without measuring it directly. The problem is that all yoyo's are different and many of them are not the type of thing which will follow 'nice' equations like that of a cylinder. Unless the yoyo is shaped like a cylinder, I would recommend measuring the moment of inertia directly.
drPaZQaL
#6
Jan22-04, 10:02 AM
P: 4
Unless the yoyo is shaped like a cylinder, I would recommend measuring the moment of inertia directly.
How do I measure it?
ophecleide
#7
Jan23-04, 11:20 AM
P: 34
To be honest, it might not be all that easy. I would recommend building some type of apparatus to allow the yoyo to roll freely in place (without moving anything else), then attach a weight to the string and measure the angular acceleration as the weight falls. Depending on how much precision you're going for, it may be better to just make an estimate based on what you do know about the yoyo, but if you want me to go into more detail of what I'm thinking, I can.
Doc Al
#8
Jan23-04, 11:37 AM
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Before you get all crazy trying to measure the rotational inertia of the yoyo, I would suggest you speak with your instructor. Your first email said you were to derive the angular speed, not measure it. Also that you should be able to calculate the rotational inertia using calculus. Ask your instructor what model you should use for the yoyo. (I still think that modeling it as two uniform disks would be close enough. )
ophecleide
#9
Jan23-04, 02:28 PM
P: 34
If you model the yoyo as two uniform disks, you wouldn't need calculus if you could use the I=1/2 MR^2 formula. If, however, you assume the yoyo to have, say, twice the density on the outer part as the inner, it would be useful. My yoyo, for example, has, I'm guessing, half of its mass on an outer ring which would certainly change the moment of inertia. Then again, if you're only going for an order of magnitude, here, you'd be fine treating it as two cylinders.
drPaZQaL
#10
Jan24-04, 11:09 AM
P: 4
but if you want me to go into more detail of what I'm thinking, I can.
Could you tell me more about your idea of measuring it..., sounds usefull


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