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Torque Required to bring a drum to rest

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    I have a problem involving a goods lift with a cage m1 = 1500kg which is raised at V = 4m/s via a winding drum with a radius of 0.6m with the help of a counter weight of m = 900kg. i am asked to find the braking torque required to bring it to rest in 1 1/3 seconds. Also i am to consider gravity playing its part.

    3. The attempt at a solution

    I have calculated a few things that i thought i would need

    inertia of the drum = 95 kg.m^2.
    angular acceleration = -5 rad/sec^2
    angular velocity= 6 2/3 rad/sec
    linear distance to stop = 8m

    My idea was to add the kinetic energy of m1, The drum and the potential energy of m2 together to get a total energy so i could use T= E/theta. But then i realized that m1 would contribute to the braking force.
    i have attached an image of what i am working with. Any assistance or guidance is much appreciated.

    Thanks
    Kane
     

    Attached Files:

  2. jcsd
  3. May 16, 2015 #2
    sorry about the image being 90 degrees off optimal! it was the right way when i viewed it earlier.
     
  4. May 16, 2015 #3

    haruspex

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    I don't get 8m to stop. Did you miss something out?
    You don't state a mass for the drum, so where does the 95kg m2 come from?
    Working entirely with energy should be ok. I see no mention of the change in PE of m1.
    If the braking torque is ##\tau##, what work will it do in the stopping distance?
    (The two pages you posted are identical. Should they be different?)
     
  5. May 16, 2015 #4
    Thanks for the input, the drum inertia was a given and the 8m is what i was using for potential energy(using kinematic equations solving for theta). Do i use the distance to stop when calculating the potential energies?

    I've tried again this morning using a different method

    method

    calculate the tension forces in each side and find the torque required my multiplying the total force by the radius.


    I tried again with the energy method. ive left out potential energies because they don't seem to contribute towards the braking torque, or do they? I've attached pictures of both, with very different results. where am i going wrong?

    Thanks

    Kane
     

    Attached Files:

    • PF2.JPG
      PF2.JPG
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  6. May 16, 2015 #5
    sorry, second pic is here..i'm having no luck with uploads here.. please excuse that
     

    Attached Files:

    • PF1.JPG
      PF1.JPG
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  7. May 16, 2015 #6

    haruspex

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    I forgot to mention: please do not post working as images, especially if you don't seem to be able to get them to appear right way up!
    You still have not explained how you get 8m. Please post your working for that.
    And yes, you do need to consider gravitational PE. Why do you think gravity doesn't affect the braking torque?
     
  8. May 16, 2015 #7
    s=ut+(1/2)at^2

    u= 4m/s; t= 1 1/3s; a=3/ms^2

    s=8m

    For m1, it is travelling vertically upward, contributing to the braking torque. i don't understand how the height that it is at effects the force it contributes to braking. or which height to use. The only height i have is the stopping distance. Also i don't think the kinetic energy would contribute anything, considering it is going up and slowing down.

    For m2 the force of tension takes into account mg and ma, so i don't know what i am missing there. I think the kinetic energy for this mass plays a big role. i just don't know how to apply the concept.

    Thanks for taking the time!
     
  9. May 16, 2015 #8

    haruspex

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    It's getting faster while braking?!
    The height doesn't affect the force, but you're mixing up the energy view with the force view.
    In the force view, the gravitational force affects the braking force needed.
    In the energy view, the change in height affects the quantity of energy the braking has to dispose of.
     
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