Angular acceleration of a falling arm

[usernamed2]

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?

Homework Equations

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 don't 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?

 

[SammyS]

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?

Homework Equations

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 don't 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 ?

 

[usernamed2]

L/2, the very center of the length?

 

[SammyS]

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 ?

 

[usernamed2]

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]

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 !

 

[usernamed2]

awesome! Thank you so much for the help!

 

[SammyS]

awesome! Thank you so much for the help!

You're welcome.

By The Way, welcome to PF !

 

[usernamed2]

Thanks!