Energy conservation for a falling pin-ended rod

[dusurme]

Hi,
I am a little confused about the energy conservation of a pin ended free falling rod.
When i try to derive energy conservation equation i am not sure including angular and linear velocity at the same time. I try to visualize the problem in the attached picture and put my derivation also.
Any explanation will be appretiated.
Thanks.

 

[MisterX]

This is the kinetic energy in general:

$KE = \frac{1}{2}m v_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$


for this example the velocity of the center of mass
$v_{cm} = \frac{L}{2}\omega$

$KE = \frac{1}{2}m \frac{L^2}{4}\omega^2 + \frac{1}{2}I_{cm}\omega^2$

Notice if the center of rotation is at the end of the rod, the angular velocity $\omega$ is the same as for the center of rotation being the center of mass.

The moment of inertia changes by the parallel axis theorem
$I_0 = I_{cm} + m\left(\frac{L}{2}\right)^2 = I_{cm} + m\frac{L^2}{4}$

So we find that the rotational kinetic energy about the end of the rod is equal to the total kinetic energy from before.
$\frac{1}{2}I_0\omega^2 = \frac{1}{2}\left(I_{cm} + m\frac{L^2}{4}\right)\omega^2 = KE$

 

[dusurme]

MisterX thanks for the answer.
Could you also suggest me an alternative way of determining angular velocity?

 

[lightarrow]

MisterX thanks for the answer.
Could you also suggest me an alternative way of determining angular velocity?

No, he's saying that your solution is incorrect: either you use the kinetic energy of rotation with respect to the centre of mass (and then the moment of inertia is not the one you wrote) *and* the centre of mass' kinetic energy, or the kinetic energy of rotation with respect to the pin, *only* (with the moment of inertia you wrote).