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Doubt from moment of inertia

  1. Dec 18, 2015 #1
    in this video

    http://www.physicsgalaxy.com/lectures/1/44/234/Solved-Example-2#12(see only the question)

    the method illustrated is integration but i thought of an alternate method,

    moment of inertia of half disc with radius r2 is 1/2mr2^2 and that of half disc with radius r1 is 1/2mr1^2.so the moment of intertia of system must be equal to difference of these moment of inertia since half disc with radius r2 is cutoff from radius r1,so shouldn't moment of inertia be subtracted too?
  2. jcsd
  3. Dec 18, 2015 #2
    That is because you have have to be careful with the definition of the mass M.
    To use the subtraction method, we have to begin by considering a half disk of radius [itex]r_{2}[/itex] with mass [itex](\frac{1}{2}\rho \pi) r_{2}^{2}[/itex]. This can be decomposed into
    (1) A half disk of radius [itex]r_{1}[/itex] with mass [itex](\frac{1}{2}\rho \pi) r_{1}^{2}[/itex], and
    (2) The "cut-out" system of interest, which has mass M = [itex](\frac{1}{2}\rho \pi) (r_{2}^{2} - r_{1}^{2}) [/itex]

    Then the moment of inertia of our system is simply given by the difference of the moment of inertias of the two half disks,
    [tex]\frac{1}{2}\left(\frac{1}{2}\rho \pi r_{2}^{2}\right) r_{2}^{2} - \frac{1}{2}\left(\frac{1}{2}\rho \pi r_{1}^{2}\right) r_{1}^{2} [/tex]

    Finally, use [itex]M = (\frac{1}{2}\rho \pi) (r_{2}^{2} - r_{1}^{2}) [/itex] to recast the equation.
  4. Dec 18, 2015 #3
    great,thank you,completely cleared my doubt,i forgot about the mass.
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