Torque on a CD accelerating to 450 rev/min in 3.0 revolutions

[jbgibson]

The problem states: when the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius of 6.0-cm and a mass of 17-g, what is the torque exerted on it?

I used the the formula tau = I * alpha; I = .5(0.017)(0.06)^2 = 3.06E-5;
in order to calculate alpha, I used wf^2 = wo^2+2alpha(theta); from this I get alpha = square root (wf^2)/(2*theta). I realize some converting is necessary so I convert 450 rev/min into 47.124 rad/s, and 3.0 revolutions into 18.50 rad. Applying these figures to solve for alpha, I get 60.018 rad/s^2. Finally, tau = I*alpha = 0.00184 N*m.

This is incorrect. Can someone direct me in the right direction? Thanks in advance.

 

[KingOfTwilight]

You have two equations, omega = alpha * t, and fi = ½ * alpha * t^2, these equations are like v = at and s = ½at^2 in linear motion. You also have two unkwons, so you can solve both.

Then use M = J*alpha.

 

[lightgrav]

alpha = wf^2/(2 theta) , no sqrt.
you see Work and Energy hiding in this kinematic equation?

 

[jbgibson]

alpha = wf^2/(2 theta) , no sqrt.
you see Work and Energy hiding in this kinematic equation?

I still don't have a clue where to go with this. I was in error, there is no sqrt., but the solution I'm getting is incorrect. The answer to this problem is 0.0072-N*m. How is that? What am I doing wrong?

 

[andrevdh]

I get the same answer as you. Maybe the given answer is incorrect.