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Rotational kinetic energy of a solid disk

  1. Aug 21, 2007 #1
    on the net I found some equation, and would like to know if they are correct ones.


    Stored energy = sum of kinetic energy of individual mass elements that comprise the flywheel
    Kinetic Energy = 1/2*I*w2 , where
    I = moment of inertia (ability of an obeject to resist changes in its rotational velocity)
    w = rotational velocity (rpm)
    I = k*M*R2 (M=mass; R=radius); k = inertial constant (depends on shape)
    Inertial constants for different shapes:
    Wheel loaded at rim (bike tire); k = 1
    solid disk of uniform thickness; k = 1/2
    solid sphere; k = 2/5
    spherical shell; k = 2/3
    thin rectangular rod; k = 1/2

    But I searched for that "I" (inertia momentum) on the Wiki http://en.wikipedia.org/wiki/Moment_of_inertia and there is not like it say above "I=k*m*w2"
  2. jcsd
  3. Aug 21, 2007 #2


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    Nowhere "above" does it say "I=k*m*w2."

    It says I=k*m*R2, apparently meaning
    where m is the total mass of the object and R its size, with k being different for differently shaped objects.

    There is a more general definition of moment of inertia which you may have found on Wikipedia in terms of a "Moment of inertia tensor." The values of I listed above are the diagonal entries of that tensor.
  4. Aug 21, 2007 #3
    I can't find the formula you refer to on Wikipedia, but the formulae you quote from the other websites are indeed correct.
  5. Aug 21, 2007 #4


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    I would be cautious about saying that, for example

    Kinetic Energy = 1/2*I*w2

    was correct, as a reader (or perhaps even the original poster might) confuse w2 with w*2 as opposed to w^2, i.e. w*w.

    Presumably this was a cut and paste job, and the '^' didn't cut and paste. Still, it's a potential source of confusion.

    In any event, one can conclude, correctly, that for an object of fixed shape rotating around a fixed axis, rotational kinetic energy is proportional to w^2, where w is the angular frequency, which seems to be the point of the question if I'm understanding it correctly.
  6. Aug 21, 2007 #5
    yes, my mistake I was in a hurry!

    so if I am calculating the flywheel kinetic energy this formula will be correct;
  7. Aug 21, 2007 #6


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    The kinetic energy in the flywheel will be

    E = (some constant) * (rpm)^2

    But it will take more work to get "some constant" right.


    says that

    E = (1/2) m r^2 * w^2 when w is measured in radians per second.

    If you are using MKS (standard metric) units, E will be in joules, m will be the mass of the flywheel in kg, r will be the radius of the flywheel in meters, and w will be the angular velocity in radians/second.

    To convert rpm to radians per second, you'd take

    (revolutions / minute) x (2 pi radians / revolution) x (1 minute) / (60 seconds)

    i.e w (rad/sec) = (rpm)*(2*pi) / (60)
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