A uniform rod of mass 1.2 kg and length 1.8 m is pivoted in the horizontal position as shown (black point).
The rod is at rest and then released. The acceleration due to gravity is $g = 9.8 m/s^2$. 1) Find the rotational inertia of the rod relative to the axis perpendicular to the screen and passing through the pivot point. 2) What is the angular speed (in rad/s, but do not include units) of the rod as it passes through the vertical position (when end marked B is at the bottom)?
My attempt of solution:
To find moment of inertia, I use the moment of inertia of a uniform rod respect to it's center of mass (middle in this case), which is $$I_{cm}=mL^{2}/12$$ where $m$ is mass of the rod and $L$ its length. Then I use the parallel axis theorem to find the moment of inertia in this case: $$mL^{2}/12+0.45^{2}m=0.567$$
For the 2º part, since the net torque on the rod is not constant I use potential energy to find first the speed of center of mass at the vertical position. The initial height of center of mass is 0.45 m taking as reference point position of the middle point of the rod at vertical position. So: $$mg0.45=1/2mv_{cm}^{2}+1/2I\omega ^{2}=1/2mv_{cm}^{2}+1/2I(v_{cm}^2/0.45^{2})$$
From here I get $v_{cm}$ and later the angular velocity.
Where am I wrong?
