# Homework Help: Moment of inertia question

1. Oct 12, 2006

### John O' Meara

A cylinder of radius 20cm is mounted on an axle coincident with its axis so as to be free to rotate. A cord is wound on it and a 50-g mass is hung from it. If after being released, the mass drops 100cm in 12s, find the moment of inertia of the cylinder? Take down as positive.
Torque=I*(alpha), i.e., b*T=I*(alpha) => T=I*(alpha)/b ...(i)
m*g - T=m*a ...(ii), on adding i and ii, we get
I=m*b(g-a)/(alpha); now alpha=a/b, where a=acceleration, b=radius, T=tension, I=moment of inertia and s=1m.
v_avg=(v+u)/2=.04166 m/s: v^2=u^2+2*a*s therefore a=-u^2/2/s=-8.681*10^-4. Hence I=225.78 kg.m^2.
I also did the calculation from the energy point of view and got the same answer. Yet the answer in the book is 1.4kg.m^2. Can anyone show me where I went off track? Thanks very much.

2. Oct 12, 2006

### Noein

Looks like you used the average velocity for the final velocity in determining 'a.'

3. Oct 12, 2006

### OlderDan

Yes. And that calculated average velocity is wrong by a factor of 2, so the v used is off by a factor of 4.

4. Oct 14, 2006

### John O' Meara

I used the initial value for the velocity v, and now I get I=56.46kg.m^2; which is still the wrong answer. So I wounder what else is wrong? Any help would be welcome.

5. Oct 14, 2006

### OlderDan

What initial value would that be? The initial velocity is zero. The final velocity at the end of 12 seconds can be calculated from the average and initial velocities, and the average velocity can be calculated from the distance and the time. Show us the calculation you did using what you believe to be the correct velocity. Then we can find your mistake.

6. Oct 17, 2006

### John O' Meara

The torque=I*(alpha); b*T=I*(alpha) therefore T=I*(alpha)/b
m*g - T = m*a therefore I=m*b^2(g/a-1) where (alpha)=a/b; v_avg=1/12 = 8.333*10^-2; therefore v=2*v_avg=16.666*10^-2m/s;
a=16.666*10^-2/12s=1.388*10^-2m/s/s.
I=.04*.05(9.8/1.388*10^-2 - 1) = 1.41kg.m^2. Much thanks for your help OlderDan.

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