Finding the angular velocity of a pendulum

[I_Try_Math]

Homework Statement

A pendulum consists of a rod of length 2 m and mass 3 kg with a solid sphere of mass 1 kg and radius 0.3 m attached at one end. The axis of rotation is as shown below. What is the angular velocity of the pendulum at its lowest point if it is released from rest at an angle of 30 degrees?

Relevant Equations

$I_{sphere} = \frac{2}{5}MR^2 + Md^2$
$I_{rod} = \frac{1}{3}ML^2 + Md^2$

My answer disagrees with the textbook and I have a feeling it may be due to how I calculated the moment of inertia. Is there anything obviously wrong with my calculation? Any help is appreciated.

$I_{sphere} = \frac{2}{5}MR^2 + Md^2$
$I_{rod} = \frac{1}{3}ML^2 + Md^2$

$I_{sphere} = \frac{2}{5}(1)(0.3)^2 + (1)(0.3)^2 = 0.126$
$I_{rod} = \frac{1}{3}(3)(2)^2 + 3(0.6)^2 = 5.08$

$I = I_{sphere} + I_{rod} = 5.21$

 

[kuruman]

$I_{rod} = \frac{1}{3}ML^2 + Md^2$
$I_{rod} = \frac{1}{3}(3)(2)^2 + 3(0.6)^2 = 5.08$

Why?
What is the equation for the parallel axis theorem?
What is $I_{\text{cm}}$ for the rod?
What is the distance of the CM of the rod from the axis of rotation?
Also, if you put numbers down remember to include the appropriate units.

 

[I_Try_Math]

Why?
What is the equation for the parallel axis theorem?
What is $I_{\text{cm}}$ for the rod?
What is the distance of the CM of the rod from the axis of rotation?
Also, if you put numbers down remember to include the appropriate units.

Ah I think I see where I was going wrong. I believe this should be the correct moment of inertia for the rod.
$I_{rod} = \frac{1}{12}ML^2 + Md^2$
$I_{rod} = \frac{1}{12}(3 kg)(2 m)^2 + (3 kg)(1.6 m)^2$

 

[kuruman]

That looks fine.