# Angular acceleration of a falling arm

## Homework Statement

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?

τ=LF=Iα
F=mg
I=(1/3) ML^2

## 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?

Last edited:

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SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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?

τ=LF=Iα
F=mg
I=(1/3) ML^2

## 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?
What is the location of the center of mass of the arm ?

L/2, the very center of the length?

SammyS
Staff Emeritus
Homework Helper
Gold Member
L/2, the very center of the length?
Yes.

That's where gravity effectively acts.

So, what is the torque produced by gravity about the shoulder ?

τ=(L/2)F=(L/2)mg
But the value of the length in the formula for inertia will remain the full length right??

SammyS
Staff Emeritus
Homework Helper
Gold Member
τ=(L/2)F=(L/2)mg
But the value of the length in the formula for inertia will remain the full length right??
Right !

awesome!! Thank you so much for the help!!

SammyS
Staff Emeritus