How Much Torque Is Required to Stop the Rotating Rod System in 8.53 Seconds?

[Prophet029]

Homework Statement

A 1.02 kg ball and a 2.06 kg ball are connected by a 0.95 m long rigid, massless rod. The rod is rotating clockwise about its center of mass at 38 rpm. What torque will bring the balls to a halt in 8.53 s? (Give an absolute value of torque.)

Homework Equations

torque = F * r sin theta
(rpm/60)*2pi= angular velocity (int)
anglv fin = anglv int + anglaccel * delta t
accel = anglaccel * r
F=ma

The Attempt at a Solution

First I found the angular velocity initial of the system using the (38rpm/60)*2pi
then I found angular acceleration need to reach angular velocity final = 0 in 8.53s by
0 = (38rpm/60)*2pi + angular accel * 8.53s
I solved for accel by multiplying the answer found for angular accel by radius which is 0.95/2m.
Then i found the torques (F=ma so mass of ball 1 and 2 * accel * radius) and added them together since, I assumed that both forces were in the same direction. I thought that these torques added together would give me my answer. But I'm getting it wrong. Can anyone help? did I overlook some element to this problem.

 

[rock.freak667]

Well when you find the angular deceleration,\alpha. I believe you can just use \tau = I \alpha, where I is the moment of inertia.

 

[Prophet029]

Well when you find the angular deceleration,\alpha. I believe you can just use \tau = I \alpha, where I is the moment of inertia.

How would i solve for the moment of inertia? Is it I=1/12*M*L^2 for a rod with axis at center? what would the M be?

 

[rock.freak667]

How would i solve for the moment of inertia? Is it 1/12*M*L^2 for a rod with axis at center? what would the M be?

I guess since they said the mass is negligible you can just ignore it and just use I=mr² for the two masses

 

[Prophet029]

That did it. Thanks