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Bike Problem Help

  1. Nov 29, 2003 #1
    Moments of inertia Help

    A wheel is formed from a hoop and n equally spaced spokes extending from the center of the hoop to its rim. The mass of the hoop is M, and the radius of the hoop (and hence the length of each spoke) is R. The mass of each spoke is lower case m.

    A. Determine the moments of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel.

    From a table in my book the long thin rod with rotation axis through the center is Icm = (1/12)m(L)^2

    I think I have to do this problem in two steps, one for the hoop and one for the rods.
    (I think the moment of inertia is about the z axis through the origin)

    Iz = integral (r^2) dm = (R^2) integral dm =M(R^2)

    Ok, for the rods I have to apply the parallel-axis theorem
    I = Icm+M(D^2)
    I = (1/12)m(R^2) + m(R^2)
    I = n ((1/12)M(R^2) + m (R^2))

    So now I just add the two systems together to find the moment of inertia.

    I = n ((1/12)m(R^2) + m (R^2))+ M(R^2)

    Ok, here is what I am confused about: I am sure that the hoop passes through the z axis but I do not think the rods do. I think the rods pass through the y-axis because it says perpendicular to the wheel which is the hoop plus the rods. If that is true then, my answer changes to I = n(1/12)m(R^2)+ M(R^2). Can you actually have two objects passing thorough two different axis?

    B. Determine the moment of inertia of the wheel about an exis through its rim and perpendicular to the plane of the wheel.

    I still need to finish this one please do not reply to B until I put what I got. thanks
    Last edited: Nov 29, 2003
  2. jcsd
  3. Nov 29, 2003 #2


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    Science Advisor

    What do you mean by "passing through" an axis? The axis of rotation is perpendicular to the entire wheel. The rim of the wheel does not pass through the axis of rotation but the spokes do- exactly the opposite of what you say.
    As far as the x,y, z axes are concerned- you are free to set up the coordinate axes as you wish. I assume that you set up your coordinate system so that the wheel is in the xy-plane and the z-axis goes through the center of the wheel. In that case, yes, it is distance from the origin that is important.
  4. Nov 29, 2003 #3
    Here is my final answer for A.
    A. I = M(R^2)+N(( m(R^2) )/3)

    Here is my final answer for B.

    Irim= I + (M + nm)(R^2)
    Irim= M(R^2)+N(( m(R^2) )/3)+ (M + nm)(R^2)
    Last edited: Nov 30, 2003
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