Solving Angular Acceleration of a Rod-Ball System

[ambient_88]

Homework Statement

A rod with mass M and length L is mounted on a central pivot. A ball is attached at the end of the rod. The rod initially makes an angle q = 27^{\circ} with the horizontal. The rod is released from rest. What is the angular acceleration immediately after the rod is released?2. The attempt at a solution
M = 1.4 kg
m = 0.7 kg
L = 1.2 m

\sum\tau = I\alpha

\alpha = \frac{\tau}{I}

\tau = F_{g}rcos(q) = (mg)(\frac{L}{2})cos(27^{\circ})

I = (\frac{1}{12})ML^{2} + (M(\frac{L}{2})^{2})

\alpha = \frac{(mg)(\frac{L}{2})cos(27^{\circ})}{(\frac{1}{12})ML^{2} + (M(\frac{L}{2})^{2})}

\alpha = \frac{(0.7kg * 9.81 m/s^{2})(\frac{1.2m}{2})cos(27^{\circ})}{(\frac{1}{12})(1.4 kg)(1.2 m)^{2} + (1.4kg(\frac{1.2 m}{2})^{2})} = \frac{3.67113 \frac{kgm^{2}}{s^{2}}}{0.672 kgm^{2}} = 5.46 \frac{rad}{s^{2}}

I believe I did everything correctly, however, the computer is saying I'm not. Would any of you please point out what's wrong with my solution?

Thanks!

 

[Hootenanny]

Welcome to Physics Forums,

You may want to recheck your moment of inertia, particularly the masses...

 

[ambient_88]

Thanks for the reply. Upon further inspection of the MOI, I noticed that I used the incorrect mass for the ball (M instead of m). I've corrected the issue and got the correct answer.