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Yet another moment of intertia problem and such

  1. Jan 31, 2006 #1
    Hello again, got some rather interesting questions for those who have been kind enough to put up with my ridiculous free response questions over the past 3 weeks. Here it goes:

    A solid disk (of Radius R) has four holes cut out of it, each of which are centered at R/2. The holes have a radius of R/4 and are equidistance from the center of the disk, which has a mass M. I wish I had a picture but just imagine 4 holes of the size given. The center of the disk lies in the origin of the x-y plane, and the holes (at least the center of each) lie on the x and y axis, two in the positive and one in the negative. Very symmetrical indeed.

    The Moment of inertia of the disk is CMR^2, where C is some constant.

    a) The disk rolls down an incline of height H. In terms of C and H, what is the velocity at the bottom of the incline?

    b) A small pin is mounted at the top of the disk at (0,R) and is pointed in the z plane. The disk is displaced by a small angle and begins to oscillate. What is the period, in terms of C and R?

    c) What is the value of C?

    Okay, so part a and b are rather simple. Conservation of energy and the formula for a physical Pendelum are totally at hand here. The problem lay in part c.

    I know that some hidden math is involved. The use of the parallel axis theorem and "negative densities". That's not the issue. The only issue that I have is what do I need to equal each other in order to solve for C and what is the right setup to determine the "missing" mass of each of the holes in terms of M.

    ANY HELP WOULD BE SUPER APPRECIATED!!! If y'all need my work, just let me know.
     
  2. jcsd
  3. Jan 31, 2006 #2
    I think you can calculate each cut disk's(radii R/4) inertia about the axis. And
    [tex] I = I_{whole} - 4 I_{cut disk} [/tex]
     
    Last edited: Feb 1, 2006
  4. Feb 1, 2006 #3
    yeah, that's the right idea, where I=CMR^2 and 4Idisk is the "negative" moments of inertia about the center of mass. No Biggie. The problem I still have is with the Iwhole, Because the mass M is of the CUT disk, not of the whole disk. The way that the whole mass has to be extrapolated is from a careful tuning of a mass/area, but I'm not sure how to properly set it up. Problem is, when I use the ratio to sove for the "missing" mass in terms of the given mass, I get a coefficient that is greater than 1, which is definately not right. So yeah...any clues as to what I need to be doing? Thanks again.
     
  5. Feb 1, 2006 #4
    Since M is the mass of the cut disk, then

    [tex] M_{whole}=\frac{4}{3} M [/tex]

    [tex] M_{hole}= \frac{1}{16} \times \frac{4}{3} M [/tex]
     
    Last edited: Feb 2, 2006
  6. Feb 2, 2006 #5
    Hmm, seems right and logical, what with the whole number ratios and such. Thanks a whole bunch. Could you just quickly show me what your ratios where exactly, in the unsimplified form...I think I know where you got your fractions but I just want to make sure. Thanks a bunch dude.
     
  7. Feb 3, 2006 #6
    If it's a uniform disk, we can know that mass is proportioned to area, hence

    [tex] m \propto r^2 [/tex]

    I believe that you can go from here.:smile:
     
    Last edited: Feb 3, 2006
  8. Feb 8, 2006 #7
    Hmm...I see, setting the mass densities of the holes and the mass equal to each other.....excellent.

    I got 55/96. How did I do?
     
  9. Feb 8, 2006 #8
    I have the same answer as yours
     
  10. Feb 10, 2006 #9
    Awesome, thanks a bunch. Just in time for my huge physics exam....round one of the international physics olympiad screening exam.
     
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