Tension in the string of a rotating, massive pulley

[raving_lunatic]

Homework Statement

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 MR² , 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.

Homework Equations

Conservation of energy:

1/2 I ω² + 1/2 m v² = mgh

Torque G = I $\ddot{θ}$

mg - T = ma

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 $R \dot{θ}$ 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 $R \dot{θ}$ and solve for the final velocity:

mgh = 1/2 m ($R \dot{θ}$)² + 1/3 M ($R \dot{θ}$)²

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


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.

 

[Tanya Sharma]

Is the tension multiplied by the pulley radius the torque that acts on the pulley during the fall?

Correct.

Apart from this you need two more equations .

1) Write ∑F = ma for the block in vertical direction .
2) Write no slip condition for the pulley i.e the tangential acceleration of the pulley at the point where the string leaves it is equal to the acceleration of the block .

These three equations will give you the value of tension .

 

[paisiello2]

You got everything right except finding T is much simpler than you propose. Your approach might work also but since above you already wrote the equation out why not just use that. If T is constant then it should be independent of time.

 

[haruspex]

Apart from this you need two more equations .

Why? r_l had to find the final velocity, and that has been done. The tension can now be deduced by considering the acceleration of either the wheel or the mass.
Of the two, I would have thought the mass a little simpler, which may be what paisiello2 is hinting at.
Also, it is not necessary to consider time if the appropriate SUVAT equation is used. r_l, you need the one that relates acceleration, initial speed, final speed and distance. Or, completely equivalently, consider force, energy change and distance. It's the same as using the SUVAT equation, but multiplying everything by the mass.

 

[HalfLostAndSearching]

Your approach to solving for the final velocity of the block using conservation of energy is correct. As for the tension in the string, it is not constant throughout the fall. Initially, the tension will be equal to the weight of the block (mg) as it accelerates downwards. However, as the block falls and the pulley rotates, the tension in the string will decrease due to the increase in tension on the other side of the pulley. This decrease in tension will slow down the rotation of the pulley. The tension in the string can be calculated using the equation T = ma + mg, where a is the acceleration of the block and g is the acceleration due to gravity. The torque acting on the pulley can be calculated using the equation T*R = I*α, where α is the angular acceleration of the pulley. As the pulley rotates at a constant rate, the torque on the pulley will also be constant. Therefore, your approach of equating the torque and angular momentum is correct. Overall, your approach to solving for the final velocity and tension in the string is correct, but the tension is not constant throughout the fall.