# Angular velocity of a falling rod?

spursfan2110
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)mL2

U = mgh
KE = .5Iw2

## The Attempt at a Solution

So, I set U = KE

mgh = .5Iw2

Then I plugged in the moment of inertia

mgh = .5(1/12)(mass of rod)(length of rod)2w2

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

g(6sin(48degrees)) = .5(1/12)mL2w2

and finally, I plugged in values and solved.

(9.8)(6sin(48 degrees)) = .5(1/12)(6)2w2

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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## Answers and Replies

Gold Member
parallel axis theorem, since the system is not rotating about its center of mass

Redsummers
Yeah, you have to use Steiner's theorem to work out the inertia at that point.

spursfan2110
So if I read my stuff right, instead of inserting (1/12)mL2 for momentum, I would insert (1/12)mL2 + md2?