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Angular speed wheel problem

  1. Jun 17, 2009 #1

    If a person rides on a wheel that has the same speed throughout rotation. Does this mean that magnitude of angular velocity is same throughout rotation? I think if the wheel has the same speed, then the centripetal force is equivalent throughout rotation.

    Now, the person decides to carry a spring to weigh himself. The maximum of the spring reads X, and the minimum reads Y.

    So, wouldn't I have:

    w = angular velocity
    N = normal
    R = radius of wheel
    -N - mg = -mw^2*R (at top)
    N - mg = mw^2*R (at bottom)

    So, N top = mw^2*R - mg, N bottom = mw^2*R + mg? And N top = Y, and N bottom = X. I am supposed to find m, but somehow there's something wrong with this equation. Thanks.
  2. jcsd
  3. Jun 17, 2009 #2


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    Is that the problem as stated?
  4. Jun 18, 2009 #3
    Yes it is. I think mv^2/R would be the same both top and bottom because your speed is the same, no? Thanks.
  5. Jun 18, 2009 #4


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    If the wheel is rotating at the same speed, then yes, that is a question that answers itself from what you've given.
    If your problem is to find an expression for m ...

    Then you also need to consider that F = k*x such that

    kΔX = ΔF = mg + m*ω2r - (mg - m*ω2r)

    Then express as m?
  6. Jun 19, 2009 #5
    I'm sorry, but what is F= kx? I thought that was the force that causes a displacement of a spring? Thanks.
  7. Jun 19, 2009 #6


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    It is. The displacement of the spring at the bottom is X, at the top it is Y. The Δdisplacement of the spring reading they give you then is (X - Y). And this means the ΔForce is the Force at the Bottom minus the Force it reads at the Top.

    You can write then

    ΔF = kΔX = mg + m*ω2r - (mg - m*ω2r) = 2*m*ω2r = k*(X - Y)

    You can rearrange for m.
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