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Στ = Iα - A Physics Pulley Problem?
I cannot figure this out for the life of me. Any help is appreciated:
A pulley has a radius of 2.70 m and a moment of inertia of 39.0 kg·m^2. A hanging mass is 4.20 kg and it exerts a force tangent to the edge of the pulley. What is the angular acceleration of the pulley?
The answer is 1.60 rad/s^2
I am in the dark here - I know that torque = F x F_t in this case, where F_t = tension, but how to find that, I do not know.
1 Answer
- MadhukarLv 71 decade agoFavorite Answer
Let T = tension in the rope
and a = downward acceleration of the hanging mass
=>
mg - T = ma = mRα ... ( 1 ) for the hanging mass
T*R = Iα
=> T = Iα/R ... ( 2 ) for the pulley
Adding equations ( 1 ) and ( 2 ),
mg = mRα + Iα/R
=> α
= (mg) / (mR + I/R)
= (4.2 * 9.81) / (4.2*2.7 + 39/2.7)
= 1.6 rad/s^2.
