Potter's wheel

tomwilliam

1. The problem statement, all variables and given/known data
find the moment of inertia of a disc of diameter D that spins freely at 40 revs/min, then reduces to 30 revs/min when a mass m is added to the disc at x from the centre. I know m, x, both angular speeds, and the diameter of the disc.


2. Relevant equations
L = Iw
I = (mr^2)/2 (moment of inertia of a disc)
I = mr^2 (moment of inertia of a particle a distance r from the axis)

3. The attempt at a solution
I think this is a question about conservation of angular momentum. I've been assuming that because angular momentum (L) is the same before and after the mass falling, I can state:
I1w1 = (I1+I2)w2, where I1 is moment of inertia of the disc alone, w1 is angular speed before and I2 is the moment of inertia of the mass and w2 angular speed after the mass falls onto the wheel. I can calculate I2 = mr^2, so therefore calculate I1. Just to confirm, I have the angular speed before and after, I have the distance x and the mass m, I have the diameter of the disc also.
Can someone confirm that this is the right approach? I don't know how the diameter (or radius) of the disc comes into it, as it seems that this approach will calculate the value of I, so I don't need to find the mass and therefore find mr^2 to find I1. I also find that the values I've calculated so far seem pretty low, so I'm thinking I might have the wrong approach. Help appreciated - but please don't solve it for me! Just some pointers...
Thanks in advance

Delphi51

Welcome to PF, TomWilliam.
Your approach looks good to me.
It would be interesting to use conservation of energy as a check.

tomwilliam

Thanks for that.
The idea of checking with rotational energy is a good one. I'm not sure how to do it though. I don't have the mass of the disc - so I would have to use the calculated value of I1 to check using Erot = 1/2 I w^2.
Also, I don't know about conservation of energy in these situations, because I don't know how much energy the (unknown) additional mass on the disc will add to the system. I'll have a go at calculating this, but if it were possible to solve the problem using conservation of energy, I would have thought it would be much easier!