# Rotational energy conservation

1. Jun 18, 2013

### 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.

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2. Jun 18, 2013

### 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.
$\frac{1}{2}I_0\omega^2 = \frac{1}{2}\left(I_{cm} + m\frac{L^2}{4}\right)\omega^2 = KE$

3. Jun 19, 2013

### dusurme

Could you also suggest me an alternative way of determining angular velocity?

4. Jun 19, 2013

### lightarrow

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).