Calculating Angular Speed of a Falling Rod

[endeavor]

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 * I_cm * w² + 1/2 * M * v_cm²
because I don't know v_cm... or can I find it?

 

[berkeman]

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

 

[endeavor]

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)
w = 5.1 rad/s

but the answer is supposed to be 5.4 rad/s...

 

[Doc Al]

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)

 

[endeavor]

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.