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A disk with a Spring around it

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A string is wrapped around a uniform disk of mass M = 2.5 kg and radius R = 0.05 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.09 m, each with a small mass m = 0.7 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 27 N. At the instant when the center of the disk has moved a distance d = 0.038 m, a length w = 0.020 m of string has unwound off the disk.


    (a) At this instant, what is the speed of the center of the apparatus?
    v = m/s

    (b) At this instant, what is the angular speed of the apparatus?

    (c) You keep pulling with constant force 27 N for an additional 0.035 s. Now what is the angular speed of the apparatus?



    I have no idea how to approach the problem. any help please
     
  2. jcsd
  3. Dec 6, 2008 #2

    Hootenanny

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    Well, a good place to start would be to determine the moment of inertia of the object.
     
  4. Dec 6, 2008 #3
    ok so the moment of inertia of the object is 1/2Mr^2 so = 1/2*2.5*(0.05)^2 = 0.003125
     
  5. Dec 7, 2008 #4

    Hootenanny

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    That we be the moment of inertia of the disk. What about the the small masses?
     
  6. Dec 7, 2008 #5
    ok since we have 4 masses ; m1r^2 + m2r^2 + m3r^2 + m4r^2

    so 0.2268
     
  7. Dec 7, 2008 #6

    Hootenanny

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    Correct. So the total moment of inertia is the sum of this and the moment of inertia of the disc.

    What do you suppose the next step will be?
     
  8. Dec 7, 2008 #7
    umm am assuming i have to find the angular speed cause it would one one way to find V, but i know the Lrot = IW but i dont have Lrot
     
  9. Dec 7, 2008 #8

    Hootenanny

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    Consider the force and resultant accelerations of the body.
     
  10. Dec 7, 2008 #9
    so i can use F = mxa?
     
  11. Dec 7, 2008 #10

    Hootenanny

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    Yes, but you also need to consider the torque on the body.
     
  12. Dec 7, 2008 #11
    ok..so i can find the torque on the body using r x Fnet, now is r the distance the center of the disk has moved? so 0.038 x 27 = 1.026? and F = mxa so a = f/m = 27/2.5 = 10.8
     
  13. Dec 7, 2008 #12

    Hootenanny

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    No, r is the [perpendicular] distance from the point of action of the force causing the torque and the centre of rotation.
     
  14. Dec 7, 2008 #13
    ok so i would use 0.05 x 27 = 1.35 and F = mxa so a = f/m = 27/2.5 = 10.8
     
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