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## Homework Statement

A uniform rod of mass 3 kg is 17 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 17 m from the center of mass of the rod. Initially the rod makes an angle of 66◦ with the horizontal. The rod is released from rest at an angle of 66◦ with the horizontal, as shown in the figure below The acceleration of gravity is 9.8 m/s2.

Hint: The moment of inertia of the rod about its center-of-mass is Icm = 1/12mL^2.

What is the angular speed of the rod at the instant the rod is in a horizontal position?

Answer in units of rad/s.

## Homework Equations

U=mgh

K=1/2Iw^2

## The Attempt at a Solution

The moment of inertia I got for the whole system was:

I=M(L/2)^2+1/12ML^2

I've managed to get:

w=[tex]\sqrt{\frac{6gh}{L^2}}}[/tex], with w being the angular speed

But I am unsure if h is L*sin(theta) or (3/2L)*sin(theta), since the rod is off the pivot.

Any ideas?