## head scratching momentum

A horizontal disk of diameter 12.0cm is spinning freely about a vertical axis through its centre at an angular speed of 72 revolutions per minute. A piece of putty of mass 5.0g drops on to and sticks to the disk a distance of 4.0cm from the centre. The angular speed reduces to 60 revolutions per minute.

1. I need to calculate the moment of inertia of the disc. I can assume that no external torques are applied to the system during this process.

2. A constant tangential force is now applied to the rim of the disk which brings it to rest in 6.0s. Calculate the magnitude of this force.

3. Calculate the rotational energy of the system before and after the putty is added to the disc.
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 wurz, we need to have you show some work or thought about the problem before helping out. I'll add a suggestion that maybe the first question is like a collision which sticks, where Mvi=(M+m)vf where vi and vf are initial and final velocities.
 I think you must use these equations for the questions: L=lw and you must know that I=mr^2. You must also use the analogous equations--similar to those in linear mechanics-- of rotational mechanics equations to solve the problem. Just plug in and solve.

## head scratching momentum

I understand Denverdoc, thanks for the lead... I think I've got a good answer using the conservation laws: m1v1 = m2v2, and I=mr^2
the answer is very small however, with I = 2.00x10-5kg m^2 does this sound anything like what you would expect??
 nice one Atomicbomb22, I got to that point as I read your reply. many thanks for re-inforcing what I had suspected...
 Wurx, uh, might have taken what I suggersted too literally, Its a rotational problem using angular quantities and conservation of these. Without doing the work looks ballpark where 72X=(x+2)60

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 Quote by Wurz I understand Denverdoc, thanks for the lead... I think I've got a good answer using the conservation laws: m1v1 = m2v2, and I=mr^2 the answer is very small however, with I = 2.00x10-5kg m^2 does this sound anything like what you would expect??
I think you are missing a factor of 2 in your answer. Use conservation of angular momentum, I'm sure that's what the question intended.

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 Quote by denverdoc Wurx, uh, might have taken what I suggersted too literally, Its a rotational problem using angular quantities and conservation of these. Without doing the work looks ballpark where 72X=(x+2)60
How did you get a 2 in there? I'm assuming X and x are representing the same thing.
 Yes, the moment of inertia of the disk. For that I recall being 1/2mr^2, while the point is mr^2

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