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Angular momentum dilemma

  1. Sep 25, 2011 #1
    I am attempting to verify the solution to a variable angular inertia problem. Initially, the problem is set up this way:

    Given a sphere with a diameter of 0.125 meter and a mass msfw=0.997 kg on a massless rod with a perpendicular axle such that the sphere can rotate about the axle at a distance from the center of rotation, r , of 15 inches and can increase continuously to 20 inches. The period to increase is 0 to 10 seconds, t. Thus, r as a function of t is: r(t)= 15…20 inches.

    The moment of inertia is calculated using the parallel-axis theorem (Steiner’s Theorem):

    Ivr(t) = 2/5*msfw*R2 + msfw*r(t)2 where R is the radius of the sphere

    The angular velocity is [itex]\omega[/itex] and is to remain constant at 1000 rpm. The question to answer is: What is the torque ([itex]\tau[/itex]) input on the axle required to maintain a constant angular velocity ([itex]\omega[/itex])?
    Since torque equals dL/dt the solution is simple: [itex]\tau[/itex](t) =[ (Ivr(ti)*w)-(Ivr(te)*[itex]\omega[/itex])]/t = 1.226 N*m

    To check this answer, I attempted to solve the same physical set up using linear momentum. Treating the sphere as a point-mass, I calculated the linear velocity based on the angular velocity (1000 rpm). Then calculating the linear velocity of the mass at 15 in. radius and also at 20 inch radius, I then used the two different velocities to calculate the beginning momentum and the ending momentum. Then using dp/dt to find the resultant force on the mass and lever arm (rod), torque equals force times r.

    The result I get is .703 N*m vs. 1.226 N*m using angular momentum. Can anyone offer an explanation why this is so?
  2. jcsd
  3. Sep 25, 2011 #2


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    I'm not sure exactly what assumption you are supposed to make here.

    Your post says the radius increases continuously but you then seem to be assuming it increases linearly. That is not necessarily the same thing.

    If r does increase linearly, the torque to maintain a constant RPM will not be constant, so the answer would be to give the torque as function of time, not a constant value.

    On the other hand, if the torque and rpm are both constant, the radius will not increase linearly, but you can find the torque from the change of energy over the 10 seconds without knowing exactly how the radius varies with time. You only need the radius at times 0 and 10.

    The moment of inertia of the sphere about its own center of mass is irrelevant if the RPM is constant, since its rotational energy about its own CM doesn't change.
  4. Sep 25, 2011 #3
    Sorry - I did leave out some key info. The radius increases as a function of time and both increase linearly. The torque does increase as the radius increases and I am only interested in the torque at te (time ending or after 10 s), which is at its maximum since the radius is at its maximum. In the delta L equation, the first term is the momentum at the 15 inch radius and the second is at the 20 inch radius (while angular velocity remains constant.
  5. Sep 26, 2011 #4
    After a long night of head scratching I uncovered an apparent error in my original math:
    In order to calculate torque using the linear approach I first calculated the force as

    f(t)=(pout-pin)/t and then torque as: tor(t)=f*r(t)

    In this case, the force on the lever arm r was incorrect. Using this equation yields the correct answer (that is, the same value as derived from an angular velocity method):

    tor(t)=[(pout * rout) - (pin * rin)]/t
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