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**Pivot point on a Uniform Rod and Acceleration**

I'm stuck on this question, which seems like it should be fairly simple:

A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle theta = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's CM.

theres a graphic of it here:

http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob28_1012rod.gif

ok so what I've done so far is:

Rmg(sinθ) = Iα

Rmg(sinθ) = (1/3)mL^2α

getting the expression I = (1/3)mL^2 from my text book for a uniform rod with a pivot at its end.

working that out I get:

(0.575)(9.8)(sin21) = (1/3)(1.15)^2 α

α = 4.58 rad/s

then converting α to a with:

a = Rα

a = (.575)(4.58)

a = 2.63 m/s^2

but apparently this isn't right.

can anyone help me out?

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