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String wrapped around a Disk with 4 Rods

  1. Nov 22, 2015 #1
    A string is wrapped around a uniform disk of mass M = 1.3 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 = 0.08 m, each with a small mass m = 0.3 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 = 33 N. At the instant when the center of the disk has moved a distance d = 0.034 m, a length w = 0.023 m of string has unwound off the disk.

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

    (b) At this instant, what is the angular speed of the apparatus?
    omega1 =

    (c) You keep pulling with constant force 33 N for an additional 0.039 s. Now what is the angular speed of the apparatus?
    omega2 =

    2. Relevant equations
    Idisk = 1/2 MR^2
    I = m1r1^2 + m2r2^2 ...

    torque = r x F

    3. The attempt at a solution
    I = 1/2 MR^2 + m1r1^2 + ... m4r4^2
    I = 1/2 (1.3)(0.05)^2 + 4*(0.3)(0.08)^2
    I = 0.009305 kg*m^2

    torque = r x f
    torque = 1.65 N*m

    I have no idea where to go from this.
     
  2. jcsd
  3. Nov 22, 2015 #2
    Was there no figure accompanying the problem?
     
  4. Nov 22, 2015 #3
    There was.
    c78475c2f9.jpg
    My bad.
     
  5. Nov 22, 2015 #4
    The fact that these two numbers are different tells you what?
     
  6. Nov 22, 2015 #5
    That the disk is rotating and the system is moving at different rates?
     
  7. Nov 22, 2015 #6
    Right. So from those two numbers you can tell how far it's moved (translated) and how many rotations it's made.
     
  8. Nov 22, 2015 #7
    I don't understand how to get velocity from that. Is there a formula?
     
  9. Nov 22, 2015 #8
    Have you tried using ##F=ma##?
     
  10. Nov 22, 2015 #9
    F = ma
    a = m/F
    a = 33 N / (1.3 kg + 0.3 kg + 0.3 kg + 0.3 kg + 0.3 kg)
    a = 13.2 m/s^2

    Maybe...
    0.034 * 13.2 = 0.4488 m^2 / s^2
    0.023 * 13.2 = 0.3036 m^2 / s^2
    0.4488 + 0.3036 = 0.7524 m^2 / s^2
    sqrt 0.7524 = 0.8674... m/s

    I'm just guessing through units, but is this it?
     
  11. Nov 22, 2015 #10
    Guessing is not a good idea. Don't you remember the formula that relates distance, velocity and acceleration? Distance times acceleration does give you something with units of ##v^2## but that doesn't mean it's equal to ##v^2##.
     
  12. Nov 22, 2015 #11
    I don't remember a formula like that at all.

    All I could find was:
    d = v1 + 1/2 at^2
    d = 1/2 at^2
    0.034 + 0.023 = 1/2 13.2 t^2
    0.057 = 6.6t^2
    0.008636364 = t^2
    0.092932038 = t

    t = (vf - vi) / a
    0.092932038 = vf / 13.2
    1.2267... = vf

    That doesn't look right though.
     
  13. Nov 22, 2015 #12
    Why did you add the two distances d and w together?
     
  14. Nov 22, 2015 #13
    Well, I tried calculating them separately using the same formula and got different times.

    But there can only be one time, so I thought it was wrong.
     
  15. Nov 22, 2015 #14
    Well, of course if you use two different distances d and w you get two different times.

    And if you use a third distance d+w you get a third time. If there's only one right answer then at least two of those three have to be wrong.

    You need to find a way to justify using one of them. Getting the right answer is not a justification. The justification needs to be based on sense-making, not on answer-making.
     
  16. Nov 22, 2015 #15
    Alright. So the first question is asking for the velocity of the system.

    The rotation of the disk is related to angular velocity, so that's not it.

    So the center of the system (center of the disk) has moved a distance of 0.034 m, which is d in the equation.
    d = v1 + 1/2 at^2
    d = 1/2 at^2
    0.034 = 1/2 13.2 t^2
    0.034 = 6.6t^2
    0.005151515 = t^2
    0.071774056 = t

    t = (vf - vi) / a
    0.071774056 = vf / 13.2
    0.947417543 = vf

    Does this make sense?
     
  17. Nov 22, 2015 #16
    Yes! If you remember to attach the units: 0.947 m/s.

    Next, repeat that same calculation, but this time you use the rotational motion versions of the linear motion equations.
     
  18. Nov 22, 2015 #17
    d = 1/2 at^2
    0.023 = 1/2 (13.2)t^2
    0.059032605 = t

    omegaf = omegai + RFt / I
    omegaf = (0.05)(33)(0.059032605) / 0.009305
    omegaf = 10.46789879 rad./s

    Okay, does that look correct?
     
  19. Nov 22, 2015 #18
    Is 13.2 m/s² the linear acceleration of the string? The reason I ask is because 0.023 m is the linear displacement of the string relative to the center of rotation.
     
  20. Nov 23, 2015 #19
    No, it's the acceleration of the system of rods and disk.

    Would the string's movement be measured in torques?
     
  21. Nov 23, 2015 #20
    The string moves a distance of 0.023 m relative to the center of rotation of the disk. What does this tell you about the disk's rotational motion?
     
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