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Homework Help: Angular velocity of rigid object

  1. Jul 6, 2008 #1
    Hi I am having trouble with a problem that involves angular velocity. The problem states:

    The rigid object shown is rotated about an axis perpendicular to the paper and through point P. The total kinetic energy of the object as it rotates is equal to 1.4 J. If M = 1.3 kg and L = 0.50 m, what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles.

    There is an illustration that I tried to insert below...if it does not show, here is a description:

    The object has a total of 4 spheres. The first 2 spheres are each of mass M and are attached to opposite ends of a rod whose length is 2L. The other 2 spheres are each of mass 2M and are attached to opposite ends of a rod whose length is L. The rods cross one another at their midpoints.

    https://my.usf.edu/courses/1/PHY2048.801C08/ppg/examview/Chpater_img/mc025-1.jpg [Broken]

    The equations I was trying to use are:

    KE=(1/2)IW^2 (I'm using W for angular velocity)
    I also know KE=(1/2)mv^2 (for linear values) and v=WR.

    I don't know where to go from here. Could someone help me?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jul 6, 2008 #2
    Use the first equation you provided KE=(1/2)IW^2. You know net KE = 1.4, so you have to figure out I (which = MR^2) for each sphere, sum up what you get and solve for W.
  4. Jul 6, 2008 #3


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    Hi jhoffma4! :smile:

    (have a squared: ² and an omega: ω)

    You don't need I … the masses are particles, and you're not asked for the angular momentum anyway.

    You only need the angular velocity to calculate the actual velocities. :smile:
  5. Jul 6, 2008 #4
    I think I did what you suggested to do correctly...

    To find I(net),

    --> I(net)= I1+I2+I3+I4
    = 2[(M)(L²)] + 2[(2M)(L/2)²]

    and then I plugged I(net) into the KE equation to solve for ω...I got 1.7.

    Thank you!
  6. Jul 6, 2008 #5
    thx for the symbols tinytim :)
  7. Jul 6, 2008 #6
    hm looks right to me, I'm not sure off the top of my head how you can do this without I as tiny-tim said.
  8. Jul 6, 2008 #7


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    Hi cryptoguy! :smile:

    Because the energy, which is given as 1.4J, is ∑(1/2)mv², = ∑(1/2)mω²r². :smile:

    I don't understand how you get an I (unless you use the parallel axis theorem, with an I0 of 0).
  9. Jul 7, 2008 #8
    well the moment of inertia of a point mass is mr² so in effect, ∑(1/2)mω²r² = ∑(1/2)Iω²
  10. Jul 7, 2008 #9


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    But why bother, when ∑(1/2)mω²r² is already what you want? :smile:
  11. Oct 26, 2009 #10
    Given ∑(1/2)mω²r². Does that mean you solve the problem by setting it equal to Ke and solving for omega?
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