What torque will bring the balls to a halt in 5.6s?

[aligass2004]

Homework Statement

A 1.2kg ball and a 2.2kg ball are connected by a 1.0m long rigid, massless rod. The rod is rotating clockwise about its center of mass at 26rpm. What torque will bring the balls to a halt in 5.6s?

Homework Equations

T = I(alpha)
Wf = Wi + alpha (delta t)
I = m1(r1^2) + m2(r2^2)

The Attempt at a Solution

I tried finding I = (1.2kg)(.5m)^2 + (2.2kg)(.5m)^2 = .85. Then I tried using Wf = Wi + alpha(delta t) to find alpha, which I got to be -.829. Then I tried using T = I(alpha) and I got -7.047, but it was wrong.

 

[hage567]

You need to find the centre of mass of the system. Since the balls are different masses, the axis of rotation will not be in the middle of the rod. Once you have the position of the centre of mass, you can find I.

 

[aligass2004]

The center of mass = (x1m1 + x2m2)/(m1 + m2), but I'm not sure what the positions would be.

 

[hage567]

Think of it like trying to balance this system on a fulcrum. You need to figure out where to set it so the system doesn't rotate. So the sum of the torques (rxF) will be zero. You can express x1 in terms of x2 since you've been given the total length of the rod. That let's you reduce your equation to one unknown variable (x2), that you can solve for and therefore determine the positions of both masses.

 

[aligass2004]

So it would be the Xc = (1-x2)m1 + x2m2/(m1 + m2). I'm still not sure how I would solve for x2. I mean, if I set that equation to zero I get 1.201, but that can't be right because it's bigger than the whole length.

 

[hage567]

For equilibrium: m1(1-x2) - m2x2 = 0 (summing the torques) The negative sign is what will make it work. (since one of the masses lies in the negative direction of your reference point)

I hope I'm making some sense, I'm short on time.

 

[aligass2004]

So x2 would be -1.2?