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Moment of Inertia

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data

    A cylinder of radius R, length h, and mass m2 has two ends of mass m1 and radius R. Inside the cylinder is liquid which can be modelled as a solid cylinder of mass M, radius R and length h.

    This cylinder is placed onto a routegh surfaced conveyer belt which is stationary. When switched on, the belt has acceleration f. The time from which the cylinder starts to move is t, \theta is the angle through which the cylinder rotates and x is the distance moved by the centre of mass of the cylinder due to the rotation \theta.

    2. Relevant equations

    Rolling condition = R\theata = x
    I = Moment of inertia = 1/2MR^2
    second law:F = ma (where m is the total mass of the system)
    m = (M + 2m1 + m2)
    v = dx/dt
    w = -d^2x/dt^2
    a = f + w

    a) Use the rolling condition to show that the acceleration of COM of the cylinder is (f - Rw).

    b) Draw a force diagram, express forces in equation form.

    c) Appliy newtons second law, show the equation of motion of the COM of the cylinder

    d) Find the torque relative to the COM associated with each force identified in part b.

    e) Show that the moment of inertia is (1/2M +m1 + m2)R^2

    3. The attempt at a solution

    a) When the can rolls anticlockwise, the distance rolled is x, therfore R\theta = x. Relative to the origin, the position vector of the COM is Rj - xi, therfore the velocity is -dx/dt (since R is constant), therefore the acceleration is -d^2x/dt^2 = w. Therefore, the total acceleration of the COM is f - Rw as given in the question.

    b) Should there be 3 forces? W, N (normal reaction) and F (force due to the conveyor belt moving)?

    c) F = ma = (M + 2m1 +m2)(f + w)

    d) Not attempted this one yet (but any assistance appreciated)!

    e) I = 1/2mR^2 = 1/2(M + 2m1 + m2)R^2 = (1/2M + m1 + 1/2m2)R^2. Uhm... why is this wrong?

    Any hints/tips/help appreciated.
  2. jcsd
  3. Sep 17, 2007 #2


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    Are you mixing up [tex]\alpha[/tex] which is angular acceleration and [tex]\omega[/tex] which is angular velocity?

    Part a) looks good to me, except for the use of w instead of alpha... and -d^2x/dt^2 = -R*alpha.

    - R*alpha is the acceleration of the center of mass relative to the conveyer belt... then you add f to get the net acceleration of the center of mass which is f - R * alpha...

    For part c)

    You calculated the the acceleration from part a... why are you using a different acceleration here?

    There is no acceleration of the center of mass vertically... So the weight and normal must cancel vertically.

    d) Does the normal force exert a torque about the center of mass, what about the weight? What is the torque of the force by the conveyor belt about the center of mass

    e) What is the moment of inertia of each mass about the central axis... you just add all of them to get the total moment of inertia...
  4. Sep 18, 2007 #3
    Thanks for that.

    part a) and c) your correct, I made a mistake in the post.

    part d), there would be no torque about the COM from the weight (W) or the normal force (N), but there would from the force applied by the belt (F)? Is this understnding correct?

    part e) We are actually given the answer, it is (1/2M + m1 + m2)R^2, however, for each mass, the moment of inertia is:

    so for the two ends, 1/2(2m1)R^2 = m1R^2
    for the cyliner shell, 1/2m2R^2
    for the liquid contents, 1/2MR^2

    adding gives:
    (1/2M + 1/2m2 + m1)R^2

    where am I gong wrong?

  5. Sep 18, 2007 #4


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    Yes, there'd only be torque due to the force applied at the conveyer belt.

    The cylinder with mass m2 is a hollow cylinder. hollow cylinder's have a moment of inertia of mr^2
  6. Sep 18, 2007 #5
    DOH! You are correct, jeez, I'm stupid at times.

    For the torque at the COM due to the force applied by the belt, is force * position of COM?

    Given that the COM will move horizontally a distance x from the origin (O) and vertically a distance of R. Using vectors of i for horizontal and j for vertical, the position of the COM is xi + Rj.

    Therefore, the torque at the COM due to the force of the belt is f(xi + Rj) - does this seem correct?

    Many thanks for you input, your helping a great deal in making math 'click' and applicable to the real world.
  7. Sep 18, 2007 #6


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    The question is asking for the torque about the center of mass... the force applied by the belt is being applied at a point straight down right below the center of mass. if the COM is located at xi + Rj, then the force is being applied at xi. ie the torque will be the force * R counterclockwise (assuming the belt is moving rightward...)

    Also, the force in part b) Fnet = Fbelt = (M + 2m1 +m2)*acceleration = [tex](M+2m1+m2)\times(f-R\alpha)[/tex]

    So the torque would be R times that force.

    You're welcome. :smile: Glad to help.
  8. Sep 18, 2007 #7
    thankyou very much, appreciate it.
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