# Angular Acceleration and Torque

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1. Dec 11, 2016

### jakec

1. The problem statement, all variables and given/known data
To lower himself from a balcony an 80kg stuntman grabs a rope connected to a 400kg cylinder with a 1.2m diameter that is free to rotate about its axis of symmetry. What is the stuntman's acceleration as he falls?

2. Relevant equations
I missed this on a homework assignment. I know the correct answer is 2.86m/s2 but I can't find what I did wrong.

3. The attempt at a solution
moment of inertia for a cylinder: I=1/2 Mr2

torque applied by stuntman = mgr = 80kg * 9.8m/s2 * 0.6m = 470.4Nm
Using t = Iα
470.4Nm = (0.5 * 400kg * 0.6m2
α = 470.4Nm / (0.5 * 400kg * 0.6m2) = 6.53rad/s2
a = αr = 6.53rad/s2 * 0.6m = 3.92m/s2

What am I missing?

2. Dec 11, 2016

### TSny

Hello. This is not the correct expression for the torque. The tension in the rope is not equal to the weight of the man. The tension is one of the unknowns. That means you'll need an additional equation besides $\tau = I \alpha$. Be sure to draw free body diagrams for both the cylinder and the man.

3. Dec 11, 2016

### Jonathan Scott

The force on the stuntman is accelerating the stuntman as well as turning the cylinder. You need to work out how to partition the force into those two parts.

4. Dec 11, 2016

### jakec

OK, so this is what I have now. I'm still a little off from the answer I should get. Any suggestions of what I can do better? (hopefully in time for finals tomorrow)

T= tension
ζ = torque
W = weight of stuntman

∑ζ = Iα
Since the tension in the cord is the only force providing torque:
Tr = Iα
Tr = (1/2 Mr2)(a/r)
T = 1/2 Ma

ΣF = ma
W - T = ma
Substituting from above:
W - (1/2 Ma) = ma
a = (2mg) / (2m +M)

a = (2 * 80kg * 9.8m/s2) / (2 * 80kg + 400kg) = 2.8 m/s2

5. Dec 11, 2016

### TSny

Your work looks correct to me. I also get 2.8 m/s2.

6. Dec 11, 2016

### jakec

Great! Thanks for the help!