Angular Acceleration and Torque

[jakec]

Homework Statement

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?

Homework 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.

The Attempt at a Solution

moment of inertia for a cylinder: I=1/2 Mr²
radius of cylinder = 0.6m

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

 

[TSny]

torque applied by stuntman = mgr = 80kg * 9.8m/s² * 0.6m = 470.4Nm

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.

 

[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.

 

[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

 

[TSny]

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

 

[jakec]

Great! Thanks for the help!