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Tension in the string of a rotating, massive pulley

  1. Apr 13, 2014 #1
    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 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.


    2. Relevant equations

    Conservation of energy:

    1/2 I ω2 + 1/2 m v2 = 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.
     
  2. jcsd
  3. Apr 13, 2014 #2
    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 .
     
  4. Apr 13, 2014 #3
    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.
     
  5. Apr 13, 2014 #4

    haruspex

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    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.
     
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