# Tension in the string of a rotating, massive pulley

**1. The problem statement, all variables and given/known data**

Hi, everyone. Me again. This time I’m here to ask about the tension in the string on a massive pulley.

We have a wheel of radius R with a moment of inertia about its principal axis of 2/3 MR^{2} , suspended above the ground on frictionless bearings. A massless string of length L is wound about the wheel, one end of which is attached to a block of mass m, which falls a distance h to the ground (where h is less than L.) We’re asked what speed the block reaches before it hits the ground, and what the tension in the string during the fall is.

**2. Relevant equations**

Conservation of energy:

1/2 I ω^{2} + 1/2 m v^{2} = mgh

Torque G = I [itex]\ddot{θ}[/itex]

mg – T = ma

**3. The attempt at a solution**

I think I got the first bit okay just by using conservation of energy – I said that the string wrapped around the pulley constrained the mass such that its speed was always equal to [itex] R \dot{θ}[/itex] where θ is the angle that the pulley has turned through. We can then express the rotational kinetic energy and the KE of the block in terms of [itex] R \dot{θ}[/itex] and solve for the final velocity:

mgh = 1/2 m ([itex] R \dot{θ}[/itex])^{2} + 1/3 M ([itex] R \dot{θ}[/itex])^{2}

which gives V = [itex]\sqrt{6mgh/(3m+2M}[/itex]

Now the tension in the string I was not so sure about, specifically how exactly it relates to the acceleration of the pulley. Is the tension multiplied by the pulley radius the torque that acts on the pulley during the fall? I’m also not sure whether the tension is constant (i.e. the block accelerates at a constant rate) – it seems like it should be this way as the rate of conversion of gravitational potential energy into kinetic energy has the same dependence as it would if the wheel wasn’t there (just a different constant of proportionality) but again, I’m unsure.

If we can say that the tension is constant and that the torque on the wheel is just T * R, then I think I can solve it: final angular velocity is V/R, and so the final angular momentum is IV/R and this implies that TR * t = IV/R where t is the total time the block was falling for. We then also have (g – T/m) is the acceleration of the block, and we can use V / (g – T/m) = time taken to fall and consequently solve these two equations for the tension – I’m just not sure if my approach is at all correct in the physics. Thanks in advance.

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