# Falling rod angular speed

A thin 1.0-m rod pivoted at one end falls (rotates) frictionlessly from a vertical position, starting from rest. What is the angular speed of the rod when it is horizontal? [Hint: Consider the center of mass and use the conservation of mechanical energy.]

So far I'm thinking that intial potential energy = final kinetic energy, so
mgh = K, where h = 0.5 m (the center of mass)

I can't seem to use K = 1/2 * Icm * w2 + 1/2 * M * vcm2
because I don't know vcm... or can I find it?

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berkeman
Mentor
Can you express w(cm) in terms of v(cm) and then solve the equation?

hmm.. i'll express v in terms of w instead...

mgh=.5Iw2 + .5M(rw)2
2mgh=Iw2 + Mr2w2
w2=(2mgh)/(I + Mr2)

h = 0.5m, I = 1/2 ML2, where L = 0.5m, r = 0.5m

so cancelling out M,
w2 = (2gh)/(1/2 * 0.52 + 0.52)

Doc Al
Mentor
Recheck your expression for the moment of inertia of a rod about its center of mass.

Realize that you can also treat the rod as being in pure rotation about the pivot, so:
$$m g h = 1/2 I \omega^2$$
(where I is the moment of inertia about the pivot)

Doc Al said:
Recheck your expression for the moment of inertia of a rod about its center of mass.
I should have used I = 1/3 ML2

thanks! I got the right answer now.