Torque Calculation for Restoring Angular Speed of Two Disks

[amcavoy]

A uniform circular disk of radius R = 0.200 m and mass M = 1.00 kg rotates with angular speed wo = 10.0 radians/second on a frictionless pivot. A second disk, having half the radius of the first and made of the same material, is supported at rest a small distance above the first disk. When the small disk is dropped concentrically onto the larger disk, friction eventually causes the disks to reach a common angular speed.

What is the final angular speed?

I came up with 9.4 rad/s, which is correct.

What fraction of the initial rotational kinetic energy is converted to heat in the process?

I came up with .0588, which is also correct.

A motor must restore the angular speed of the combination to w_o in one revolution. What torque must the motor supply ?

I used the equation \tau=I\alpha here. I added each moment of inertia (for each disk individually):

I=I_1+I_2=.02125

The above I got using the standard formula for the moment of inertia for a disk (1/2)mr².

Now I found the acceleration:

10^2=9.4^2+2\alpha\left(2\pi\right)\implies\alpha=.926\text{rad}/\text{sec}^2

Then I multiplied the acceleration by the moment of inertia to come up with:

\tau=.0197\text{Nm}

...however the above is incorrect. Could someone please tell me why? I have gone through my steps many times, meaning that the only way this could be wrong is if I took the wrong steps; my arithmetic is fine. I'd appreciate any input on this.

Thank you very much.

 

[skeeter]

your procedure is correct ... your final solution for tau does have some round-off error in it from rounding the final omega from the 1st part of the problem ... 9.4 rad/s.

If you don't round, you should get tau = .0913092394 Nm

If this is an on-line problem that is machine graded, that could be the reason.

 

[amcavoy]

Thank you very much. It's nice to know my problem was in rounding digits rather than my procedure (it is web-based homework by the way).

Thanks again.