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Angular velocity of a block of mass

  1. Apr 21, 2005 #1

    WY

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    Hey I was doing a question and I need someone to check if my method is ok - or stop me if i'm on completely the wrong track!!!

    A block of mass m is attached (with a massless string) to a wheel. Consider the bicycle wheel is not turning initially. The block is allowed to fall a distance of h. Assume that the wheel has a moment of inertia I about its rotation axis.

    Find the angular speed of the wheel after the block has fallen a distance of h in terms of m,g,h, r(of the wheel) and I

    I took the relative zero as where the mass is starting from:
    so the wheel has KE = 0 and the mass has KE=0 and PE=0

    Then after it has fallen a height of h:
    Wheel: KE = 1/2(I*omega^2)
    Mass: KE = 1/2mv^2 and PE=mgh

    so i came up with the conservation of energy equation to be:
    0 = 1/2(I*omega^2) + 1/2mv^2 + mgh
    the v the weight will be travelling at will be the same angular velocity the wheel is turning i substituted omega in for v
    the I rearranged it so that omega was the subject and got:
    omega = sqrt((-2mgh)/(Im))

    is this rite??? or ami completely wrong?? thanks in advance :)
     
  2. jcsd
  3. Apr 21, 2005 #2
    I don't see any radius r in your final answer. Did you express that:

    [tex]\omega = \frac{v}{r}[/tex] ?

    Also:

    [tex]0 = \frac{1}{2}I\omega ^2 + \frac{1}{2}mv^2 - mgh[/tex]

    So don't use (+) but (-) for mgh.
     
  4. Apr 22, 2005 #3

    WY

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    hahah right thanks for that!
     
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