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Flywheel, rotational mechanics problem

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A flywheel is a homogeneous cylinder of mass M with a hole in the middle where the axle connects with it. The outer radius of the wheel is R. The radius of the axle is r, and the mass of the axle is m. A weight of mass m_w is suspended from a string wound around the axle. The axle is also a homogeneous cylinder.
    (a) Calculate the moment of inertia of the flywheel + axle combination in terms of the variables given in the problem. In what follows, call this I.
    (b) Express the kinetic energy of the flywheel, and the kinetic energy of the falling weight, in terms of the variables given in the problem, and the instantaneous angular speed ω.
    (c) Express ω as a function of h, the height that the falling weight drops, assuming that everything starts from rest. State any other assumptions you make, and why.

    [​IMG]

    2. Relevant equations

    I(cylinder) = 0.5MR^2 (in general)
    K(linear) = 0.5mv^2
    K(rot) = 0.5Iω^2

    3. The attempt at a solution

    a) I = I(axle) + I(wheel)
    I(axle) = 0.5mr^2
    I(wheel) = 0.5MR^2
    Therefore, I = 0.5(mr^2 + MR^2)

    I thought this one was pretty straightforward but somehow it doesn't look right, like I should somehow be compensating for the fact that the wheel is not a complete cylinder, but a cylinder with an inner cylinder removed. But if the total mass of the wheel is M, which is given, this should already be compensated for in the problem, no?

    b) The kinetic energy of the flywheel is easy enough to solve for, since K(rot) = 0.5Iω^2, and we don't need to simplify any further as we just solved for I. For the mass, I assumed the velocity of the block to be equal to ωr, as the motion of the weight must be the same as the motion of the rim of the axle, and from circular motion we know that v = ωr. So therefore the kinetic energy of the weight, assuming the string doesn't slip, is 0.5(m_w)(ωr)^2

    c) We know that the potential energy of the weight before it is dropped is equal to the kinetic energy at the instant before it finishes its drop, so we can say that U_i = K_f = m_w*g*h, or

    0.5m_w(ωr)^2 = m_w*g*h
    0.5(ωr)^2 = gh, h = 1/g * 0.5(ωr)^2

    This was a bit of a leap though, as it doesn't take into account the kinetic energy of the wheel itself, which it seems I should be doing. I'm just not sure how. Any thoughts on anything I've done? I can sort of grasp at the important concepts this question is trying to get me to think about but I seem to be tripping over some details here or there. Any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 12, 2007 #2

    andrevdh

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    Homework Helper

    (c) Assuming that no friction acts on the axle and not taking drag into consideration the mechanical energy of the system will be conserved.
     
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