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Moment of inertia and Parallel axis theorem

  1. Oct 26, 2009 #1
    1. Thehttp://nft01.nuk.edu.tw/lib/exam/97/master/97ap-master.pdf" [Broken] statement, all variables and given/known data
    A solid, uniform disk of mass M and radius R is oscillating about an axis through P. The axis is perpendicular to the plane of the disk. Suppose the friction at P can be ignored. The distance from P to the center, C, of the disk is b (see figure 1). The gravitational acceleration is g.

    What is the moment of inertia for rotation about the axis through P?

    3. The attempt at a solution

    I know the formula

    [tex] L = Iw [/tex]

    Moment of inertia is I, while rotation per second is w.
    The question asks to calculate the moment of inertia for rotation.
    This seems to be apparently "L".

    However, I am not sure, since this problem should be apparently solved with Parallel Axis theorem i.e.
    [tex] I = I_{cm} + MR^2 [/tex]
    where likely [tex] I_{cm} = .5 MR^2 [/tex] is the moment of inertia though the center of mass.
    However, I amn not sure, since this would imply

    [tex] I = 1.5 MR^2 [/tex]


    How find the moment of inertia for rotation about the axis though P?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 27, 2009 #2

    Andrew Mason

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    No. L is the angular momentum of the object when rotating at angular speed [itex]\omega[/itex].

    You have found it. Why do you think it is wrong?

    AM
     
    Last edited by a moderator: May 4, 2017
  4. Oct 27, 2009 #3
    I did not know that the same Steiner's formula applies for moment of inertia of rotating objects.
    There is no "w" in the Steiner's formula which made me to think that there is something wrong in the result.

    Thank you for your answer!
     
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