# Angular acceleration problem - masses on massless rod

1. Apr 21, 2006

### FlipStyle1308

A 3 kg mass and a 4 kg mass are attached to either end of a 3 m long massless rod. If the system is rotated about the center of mass by a force of 7 N acting on the 4 kg mass, (perpendicular to the rod), what will the size of the angular acceleration of the system be?

I used angular acceleration = rF/mr^2, and got 1.3608 rad/s^2, but got the answer wrong. Anyone have any idea why?

2. Apr 21, 2006

### nrqed

What did you use for r? You must first find where the center of mass is. Did you do this?

3. Apr 21, 2006

### FlipStyle1308

For r I used 1.286. I got the center of mass is 15.4286 kgm^2, which is correct, according to WebAssign.

4. Apr 21, 2006

### nrqed

Ok (btw, the value you give is the moment of inertia, not the center of mass).

5. Apr 21, 2006

### FlipStyle1308

Whoops, you're right, I gave the rotational inertia for the center of the rod lol. Center of mass is 1.714 m, from the 3 kg mass. What did you plug in for that 0.5835?

6. Apr 21, 2006

### nrqed

rF/I = 1.286m * 7 N/15.428 kg m^2

7. Apr 21, 2006

### FlipStyle1308

Oh okay, thanks! So I was supposed to use the inertia of the center of mass. Interesting. Thanks!

8. Apr 21, 2006

### nrqed

I am not sure what you mean. But the r you use there is the distance between where the force is applied and the axis of rotation, as a general principle. And the I is always calculated with respect to where the axis of rotation is located.

9. Apr 21, 2006

### FlipStyle1308

Makes sense...I never thought of that. Thanks!