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Rotational Force

  1. Nov 15, 2006 #1
    Here is what I did initially:

    v = 3.77m/1.6s
    mgh = (1/2)mv^2 + (1/2)*I*omega^2
    omega = v/R
    (20kg)(9.8)(3.77m) = (1/2)(20kg)(2.36 m/s)^2 + (1/2)I((2.36)^2/(0.60)^2)
    I = 88.3 ... which is more than I = MR^2 = 25.2 ...

    So this is entirely wrong, because I know it should be less, since the mass is more evenly distributed.
    Second approach, I believe the drops 3.77 m in 1.6 s is still due to gravity, is not constant velocity. There must also be tension then since it is less than gravitational acceleration.

    mg - T = ma
    y=(1/2)a*t^2
    (3.77m) = (1/2)a(1.6)^2
    a= 2.95 m/s^2

    (20 kg)(9.8) - T = (20kg)(2.95)
    T = 137N

    torque = F*R
    Only T does something in relation to the rotation of the axis? So:
    torque = T*R = I * alpha
    alpha = a/R = 2.95m/s^2 / 0.60m = 4.92 rad/s^2

    137N * 0.60m = I * (4.92 rad/s^2)
    I = 16.6 N s

    Well I got the less inertia part now. But was that the right way to do it?
     
  2. jcsd
  3. Nov 15, 2006 #2

    OlderDan

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    Science Advisor
    Homework Helper

    Both ways should give you the same result. You made an error in your first calculation. A hint as to where that error is can be found in your second calculation (a = 2.95m/s²). How fast is the mass moving at the end of 1.6 seconds?
     
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