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Moments of Inertia

  1. May 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A rod [pq] has lenght 4l and mass M.

    q is attatched to the circumfrence of a disc of mass M and radius l, in such a way that that p and q are colinear with the disc and the two pieces don't overlap.

    PART A

    Find the moment of intertia of the system about an axis through p perpendicular to the plain of the disc.

    PART B

    If the system is released from rest when q is vertically above p, find its angular speed and the speed at the centre of the disc when q is vertically below p.

    2. Relevant equations

    I don't actually know what I'm supposed to put here? I'm sure everything you guys need to help me is included in the scans though.

    3. The attempt at a solution

    I did the whole thing and got an answer. However, the answer is different that the one given at the back of the book. Did I make a mistake somewhere?

    I think all of Part A is correct. I must have done something wrong with Part B.

    I didn't want to type it all out again so I scanned my work. Is this okay? Sorry if my writing is hard to read. :redface:


  2. jcsd
  3. May 12, 2007 #2

    Part B
    If the system is released from rest
    "when q is vertically above p"
    , find its angular speed and the speed at the centre of the disc
    "when q is vertically below p".

    Notice the two conditions at the beginning and the final one.
    In your calculations about B, it seems not the conditions in the question.

    Best regards.
  4. May 12, 2007 #3
    Oh my God, how did I not see that? :surprised

    Why on earth did I have q to the right of p? :rofl:

    Thanks for pointing that out. I feel so dumb. :biggrin:
  5. May 12, 2007 #4
    Okay so I did it again. My answer now resembles the back of the book's answer more, but it's still not the same.

    Where is that 5l coming from? :redface:

  6. May 12, 2007 #5

    There are two quantities in the answer of Part B.

  7. May 12, 2007 #6
    I don't understand. Could you explain further? :confused:
  8. May 12, 2007 #7
    The angular speed is one of the answer, you have got this one, correct one.

    The other answer is more trivial if you know the tangent velocity: [tex]v=\omega r[/tex].
    and [tex]5L\times\omega[/tex] is the speed of the center of the disc.
  9. May 12, 2007 #8
    Ah. I've never heard of that before. Thanks! :smile:
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