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Homework Help: Finding the change in Rotational Kinetic Energy

  1. Jun 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A sphere with a 3 kg * m^2 moment of inertia about an axis through its center has its angular velocity changed by 4 rad/s owing to a 2 N*m torque applied about that axis for 6s. If its initial angular velocity is 10 rad/s, the change in its rotational kinetic energy in J is...
    The answer according to the book is 144 J.

    2. Relevant equations
    1/2Iiwi^2 - 1/2Ifwf^2 = change in Ke

    3. The attempt at a solution
    The wording of the question is kind of throwing me off because i don't understand the part about the Torque.

  2. jcsd
  3. Jun 15, 2008 #2


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    It will be helpful to you to know that there are analogies in rotational mechanics to the definitions and equations you learned in linear mechanics.

    Back in linear mechanics, you have the definition of work:

    W = F · delta_x

    and the work-kinetic energy theorem:

    W = KE_f - KE_i .

    In rotational mechanics, the rotational work will now be

    W_rot = (torque) · (delta_theta) ,

    with delta_theta being the angle the object has turned through while the torque is applied.

    The work-kinetic energy theorem still applies,
    now with

    KE_rot = (1/2) · I · (w^2) .
  4. Jun 15, 2008 #3
    Ok so the initial KE_rot = 1/2 * I * (w^2)
    = 1/2 * 3 * (10^2) = 150
    I see what you are saying that W_rot = torque * delta theta
    but how would that apply I am not given an angle?
    and am I finding the W_rot and subtracting it from 150?
  5. Jun 15, 2008 #4


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    Hi crazyog! :smile:

    You don't need an angle or a torque or anything else for this part of the question …

    You know it goes from 10 rad/s to 14 rad/s … so what is the increase in rotational KE? :smile:
  6. Jun 15, 2008 #5
    Oh ok so when they say "changed by 4 rad/s" it means it is now 14 m/s
    so that would be

    1/2(3)(14^2) = 294 J
    294 - 150 = 144 J

    Thank you!
  7. Jun 15, 2008 #6


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    Oh, yes, quite so! Since they gave that information, I thought we'd need it somewhere. (Presumably, there are other parts to this problem...)

    I wonder how they expected the student to know that the angular speed was being increased. The problem states that the "angular velocity (is) changed by 4 rad/s". You get a different magnitude for the rotational KE change if the angular speed goes to 6 rad/sec...
  8. Jun 15, 2008 #7


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    yes … you're right! … :smile:

    Now I'm wondering why I assumed it was increasing! :redface:
  9. Jun 16, 2008 #8


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    And the change in angular momentum (rotational impulse) is

    delta_L = (torque)·(delta_t) .

    So the change in angular momentum is

    delta_L = +/- (2 N·m)·(6 sec) = +/-12 kg·(m^2)/sec ,

    so the new angular momentum would be either

    L' = (3 kg·m^2)·(10 rad/sec) + 12 kg·(m^2)/sec
    = 42 kg·(m^2)/sec


    L' = (3 kg·m^2)·(10 rad/sec) - 12 kg·(m^2)/sec
    = 18 kg·(m^2)/sec ,

    which are consistent with final angular speeds of 14 or 6 rad/sec .

    So the other information in the problem doesn't provide a clue either...

    So a second valid answer would be -96 J for the change in rotational KE. (I think the solver for the textbook must have assumed the speed was increased.)
    Last edited: Jun 16, 2008
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