Rotational energy conservation

[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).