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Moment of inertia of a wheel: linear to angular motion

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data

    You are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.600 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an 8.20-kg block of wood from the free end of the rope. You release the system from rest and find that the block descends 12.0 m in 4.00 s.

    2. Relevant equations

    K1 + U1 = K2 + U2
    K = 1/2*m*v2
    K = 1/2*I*omega2
    v2 = v02+2a(x-x0)
    v = r*omega

    3. The attempt at a solution

    d = 0.600 m
    r = 0.300 m
    m = 8.20 kg
    v0 = 0 m/s
    omega0 = 0 rad/s
    change in x = 12.0 m
    t = 4.00s

    The initial kinetic energy is zero and the final gravitational PE is 0, so we use the equation U1 = K2, which is:

    mgh = 1/2*m*v2 + 1/2*I*omega2
    mgh - 1/2*m*v2 = 1/2*I*omega2
    (mgh - 1/2*m*v2)/(1/2*omega2) = I

    To solve, we need the linear velocity. I used a kinematic equation to calculate the velocity.

    v2 = v02+2a(x-x0)
    v2 = 0 + 2(9.81 m/s2)(12.0 m)
    v2 = 235.44 m2/s2
    v = 15.34 m/s

    Then, v = r*omega
    omega = v/r
    omega = (15.34 m/s)/(0.300 m)
    omega = 51.15 rad/s

    Now back to the energy equation.

    I = (mgh - 1/2*m*v2)/(1/2*omega2)

    I = ((8.20kg)(9.81 m/s2)(12.0 m) - 1/2*(8.20 kg)*(15.34 m/s)2)/(1/2*(51.15 rad/s)2)

    I = 5.35*10-6

    But my homework site says this is wrong.
  2. jcsd
  3. Nov 16, 2016 #2

    Simon Bridge

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    Homework Helper

    ... you have an unjustified assumption - the acceleration cannot be 1g.
  4. Nov 16, 2016 #3
    But why not? There is no friction on the wheel from its bearing or mounting, and so the only external force acting on the block is gravity...?
  5. Nov 16, 2016 #4
    check this equation .....the mass is hanging ut your equation is representing free fall!
    v is related to w (omega) try to find the angular velocity ?
  6. Nov 16, 2016 #5
    Ah, I see what you mean.

    Now I've solved it. And as a courtesy to future answer-seekers, here's how.

    Use the kinematic equation delta-x = 0.5(v0-v)t

    12.0 m = 0.5*v*4.0 s
    v = 6.00 m/s

    then, omega = 20.0 rad/s

    Plugged into the energy equation above, I = 4.09 kg*m2
  7. Nov 16, 2016 #6

    Simon Bridge

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    Well done.
    The explanation you are looking for is that gravity does not just act on the hanging mass alone - it also has to accelerate the flywheel (via the string).
    It's just like when you pull a block across a table by attaching a hanging mass - you work it out by free body diagrams.

    You were halfway there with the energy description:
    You had already written that some of the change in PE gets stored as rotation in the flywheel and some as kinetic energy in the falling weight.
    If the weight were in freefall, all of the change in PE would go for it's kinetic energy ... so, with the flywheel attached, it should be slower.

    You can also check experimentally that a weight off a wheel falls slower than unattached - or that a yoyo falls slower when unwinding than it does if you just drop it.
  8. Nov 16, 2016 #7
    Thank you for the explanation. Your second paragraph especially helped!
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