(adsbygoogle = window.adsbygoogle || []).push({}); A uniform rod of mass 0.62 kg is 6 m long.

The rod is pivoted about a horizontal, fric-

tionless pin at the end of a thin extension (of

negligible mass) a distance 6 m from the cen-

ter of mass of the rod. Initially the rod makes

an angle of 48◦ with the horizontal. The rod

is released from rest at an angle of 48◦ with

the horizontal, as shown in the figure.

What is the angular speed of the rod at

the instant the rod is in a horizontal position?

The acceleration of gravity is 9.8m/s2 and the

moment of inertia of the rod about its center

of mass is I = (1/12)mL^{2}

Answer in units of rad/s.

2. Relevant equations

U = mgh

KE = .5Iw^{2}

3. The attempt at a solution

So, I set U = KE

mgh = .5Iw^{2}

Then I plugged in the moment of inertia

mgh = .5(1/12)(mass of rod)(length of rod)^{2}w^{2}

Cancelled out the m and solved for the height of the center of mass

g(6sin(48degrees)) = .5(1/12)mL^{2}w^{2}

and finally, I plugged in values and solved.

(9.8)(6sin(48 degrees)) = .5(1/12)(6)^{2}w^{2}

And solved for w, which I found to equal 5.397339822.

This is wrong, but I have no idea where my error is. Can you guys help me spot it??

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# Homework Help: Angular velocity of a falling rod?

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