# Torque change of a rod with unequal masses

1. Jan 27, 2008

### Metalsonic75

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s?

I used (m1x1+m2x2)/(m1+m2) to solve for the center of mass (2/3 meters from the 1kg weight). Then I used that to solve for the moment of inertia, using I= m1r1^2+m2r2^2, and I got I=1. I know $$\alpha$$=$$\tau$$/I, and $$\alpha$$=($$\omega$$_final - $$\omega$$_initial) / time. I plugged in my known values (0 rps, 1/3rps, and 5 seconds, respectively) and got -0.1333 for $$\alpha$$. Then I multiplied $$\alpha$$ by I and got -0.1333, which is wrong. I don't know where I screwed up. Any help would be greatly appreciated. Thanks

Last edited: Jan 27, 2008
2. Jan 27, 2008

### cepheid

Staff Emeritus
In the calculation for the moment of inertia, you used the wrong radius for the wrong mass. The (2/3 m)^2 should be multiplied by 1.0 kg.

Last edited: Jan 27, 2008
3. Jan 27, 2008

### Metalsonic75

It doesn't seem to work... I took 1*(2/3)^2 + 2*(1/3)^2 and got (2/3). Then I multiplied -0.1333 by (2/3) (angular acceleration times I) and got 0.0888, which is still wrong.

4. Jan 27, 2008

### Hootenanny

Staff Emeritus
That is not the angular velocity, that is simply the frequencey, you need the angular frequency/velcoity.

5. Jan 27, 2008

### cepheid

Staff Emeritus
Oh crap...how did I miss that? I didn't understand what Hootenanny was saying until I realized that his / was just a slash, and not a division sign. Anyway to the OP, yes it's true...you're out by a factor of 2*pi as far as your angular velocity goes.

6. Jan 27, 2008

### Metalsonic75

Ah, yes. I multiplied 1/3rps by 2*pi and everything worked. Thank you!