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Torque on a Pulley

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Two paint buckets, initially at rest, are connected by a lightwieght rope which is wrapped over a pulley which has a mass 2 kg and a radius of .5 m. Only conservative forces do work.
    Mass A is 4 kg and starts on the ground. Mass B is 6 kg and starts 5 m above the ground.
    a- Find the speed at which the top bucket hits the ground.
    b- What was the average torque on the pulley?

    2. Relevant equations



    3. The attempt at a solution
    a - 4.22 m/s
    b- Using torque = (tension 1 - tension 2) r, I get 9.8 N for net torque. Is this correct?
     
  2. jcsd
  3. Nov 11, 2008 #2

    LowlyPion

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    Because the pulley has mass and hence moment of inertia, your acceleration won't be entirely just the difference in the paint buckets accelerating just the paint buckets. There is also the increase in angular kinetic energy that goes into the pulley.
     
  4. Nov 11, 2008 #3
    So how do I account for that in the equation? Should I add 1/2 Iω^2?
     
  5. Nov 11, 2008 #4

    LowlyPion

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    Yes. You could approach it from the point of view of kinetic energy of the buckets and changes in potential energy and the change in kinetic energy of the pulley and noting along the way that ω is equal to v/r.
     
  6. Nov 11, 2008 #5
    or the acceleration = (driving force)/(Inertia) will work quite well, giving you the tension in both strings (which will be unequal, since the pulley has mass) :)
     
  7. Nov 11, 2008 #6
    I dont know how to use these equation in a system like this. For example, do I need to focus on 1 mass? In the term for initial potential, only one has initial potential. How do I find info about the pulley using this?

    Kf + Uf + RKf = Ki + Ui + Rki
     
  8. Nov 11, 2008 #7
    i think that applying force dynamics and rotational dynamics will make this problem easier.
     
  9. Nov 11, 2008 #8

    LowlyPion

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    Well that equation certainly would work.

    What is the initial KE then? 0 right? And same for Rotational KE. For PE the initial is the m*g*h of the 6kg bucket.

    For final KE you have the masses of the two buckets in motion at some velocity and your rotational KE related by v/r and you have the 4 kg bucket with its m*g*h when the heavier bucket touches down.
     
  10. Nov 11, 2008 #9
    So I dont consider the mass of the pulley in the Kf term? Should it just be .5 ( 6 + 4 kg) vf^2? And the info about the pulley is only used in the RKf term?
     
  11. Nov 11, 2008 #10

    LowlyPion

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    The pulley is the one rotating.

    The buckets are the ones with mv2/2.

    You just have to account for the energy wherever it is. That after all is what's conserved isn't it?
     
  12. Nov 11, 2008 #11
    OK, I plugged that al in and found ω = 8.49 m/s. Using I ω= τ , I end up with τ = 18.02. Does this look right?
     
    Last edited: Nov 11, 2008
  13. Nov 11, 2008 #12

    LowlyPion

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    They asked you for v in part a) didn't they? Not ω.

    As for Torque isn't that merely Net force * radius?
     
  14. Nov 11, 2008 #13
    So the answer is incorrect? I would rather use I ω= τ and not find all the seperate forces.
     
  15. Nov 11, 2008 #14

    LowlyPion

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    I ω= is not torque. That's Angular momentum sometimes written as L.

    I * α = τ
     
  16. Nov 11, 2008 #15

    LowlyPion

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  17. Nov 11, 2008 #16
    Argh! Thanks for being so patient. So I cant find α without time. I can't find time without rotations. So I do need to find all the forces. Would that be -pulley weight - mass1 weight - mass 2 weight + tension 1 + tension 2?
     
  18. Nov 11, 2008 #17

    LowlyPion

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    If you wanted to find α that's a/r and a can be found by V2 = 2*a*x = 2*5*a

    So take your answer from part a) square it and divide by 10 and that will be a.
     
  19. Nov 11, 2008 #18
    Oh! So I alpha = I (a/r) = torque
    3.56*.25 = .89

    torque = .89 Nm
     
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