Acceleration of a mass hanging from a pulley problem

  • Thread starter aftershock
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  • #1
aftershock
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EDIT: I just realized I have to use the tension as the force and not simply mg. It's been a while since I dealt with these problems and that slipped my mind. I think I get it now, sorry.

I can't figure out where I'm messing up on this. This is not a homework problem, bear with me as I try to explain this.

Imagine you have a pulley with a string attached to the edge of the pulley with a mass hanging off of it. You want to calculate the acceleration of that mass as it falls.

The force of the weight of the mass is producing the only torque. so torque = mgr where r is the radius of the pulley and m is the hanging mass.

The I of a pulley is (m'r^2)/2 where m' is the mass of the pulley. alpha is a/r where a is the linear acceleration at the edge of the pulley equal to the acceleration of the mass.

mgr = ((m'r^2)/2)*(a/r) ---> mg = (m'a)/2 ----> 2mg = m'a

so now to solve for a, the acceleration of the block...

(2mg)/m' = a

So what I'm getting from that formula is that if the mass of the pulley is less than twice that of the hanging mass it will accelerate with a magnitude larger than g.

This obviously makes no sense at all, where did I mess up?
 
Last edited:

Answers and Replies

  • #2
rcgldr
Homework Helper
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mg is the total force on the mass and pully system. This force is used to accelerate the mass and pulley, so after converting the pulley's inertia into it's equivalent linear inertia, the rate of acceleration will equal force / (mass of weight + equivalent of linear inertia of pulley).
 

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