Weight on a pulley, torque, mechanical energy

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SUMMARY

The discussion focuses on calculating the speed of a block (m1) attached to a pulley (m2) after falling a height (h) using conservation of mechanical energy principles. The relevant equation derived is v = sqrt[(m1gh) / (0.5(m1) + 0.25(m2))], where the moment of inertia for the pulley is defined as (1/2)(m2)R^2. The solution confirms the correctness of the approach and the final formula for speed.

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  • Familiarity with angular velocity (w) and its relationship to linear velocity (v)
  • Basic algebra for simplifying equations
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Homework Statement


A block of mass (m1) is attached by a massless string to a pulley of mass (m2) with radius R. Starting from rest, what is the block's speed after it has fallen height h?

Homework Equations


I'll call angular velocity w.
I'm thinking conservation of mechanical energy, so (m1)gh = (1/2)(m1)v^2 + (1/2)Iw^2
Moment of inertia for the pulley would be (1/2)(m2)r^2

The Attempt at a Solution



(m1)gh = (1/2)(m1)v^2 + (1/2)[(1/2)(m2)r^2](v^2/r^2)
simplifying would give
v=sqrt[(m1gh) / (.5(m1) + .25(m2))]

Is this right?
 
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Yes.

ehild
 

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