# Angular acceleration of a falling arm

**1. The problem statement, all variables and given/known data**

Assume your arm is a solid rod, 0.740 m in length, free to rotate about your shoulder (so its

moment of inertia is (1/3)ML^2). If you hold your arm out sideways, and then completely relax

your muscles, what is the initial downward angular acceleration of your arm?

**2. Relevant equations**

τ=LF=Iα

F=mg

I=(1/3) ML^2

**3. The attempt at a solution**

I’m not given a specific formula to use, but I think the above formula for Torque applies. I’m not using conservation of energy because I dont have a value for the height to use for potential energy.

If I plug the formulas for inertia and Fg into the torque formula:

Lmg=(1/3)ML^2 α

α=3g/L = (3*9.81)/0.740 = 39.77 but the answer key says the answer is 19.9 (half).

What am I doing wrong here?

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