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Centre of mass of a pendulum

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Why is the attachment true? (It's from my notes)

    2. Relevant equations

    General centre of mass equations

    3. The attempt at a solution

    Really not sure how the lecturer did this. It's quoted as if it's obvious but I don't see it at all, this has been annoying me for a while.

    M(L-l) is obviously the moment of the ball about the center of masss.. I'm not sure where the other two terms come from really though.
     

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  3. Apr 4, 2012 #2

    tms

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    The two sides of your equation don't look equal to me; you are missing two terms.
     
  4. Apr 4, 2012 #3

    RK7

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    I'm not sure what you mean.. the equation given in the image is the equation the lecturer claims the distance l must satisfy. He then rearranged for l and uses it later in the notes. I don't really know how he got what's in the attached image at all. What do you think it should be?
     
  5. Apr 4, 2012 #4

    tms

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    I mean what I said. You have an equation. The two sides of the equation are not equal.

    I think something is missing, such as the length of the rod (or whatever [itex]m[/itex] is).
     
  6. Apr 4, 2012 #5

    RK7

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    They aren't equal in general, only for a specific value of l which is found by rearranging what's in the attached image to give [itex]l=\frac{(2M+m)L}{2(M+m)}[/itex]
    i.e. from this equation which the lecturer has said must hold, we can find the position of the centre of mass of the system.

    My problem is that I'm not sure why the equation given must hold since I can only see where one term is from.

    To be clear:
    [itex]M[/itex] is the mass of the disc on the end of the rod
    [itex]m[/itex] is the mass of the rod
    [itex]L[/itex] is the distance from the pivot to the centre of mass of the disc
    [itex]l[/itex] is the distance from the pivot to the centre of mass of the combined system

    [itex](L-l)[/itex] is the distance from the COM of the disc to the COM of the system so it looks like the [itex]M(L-l)[/itex] term is related to the moment of the disc about the centre of mass.
     
  7. Apr 4, 2012 #6

    tms

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    Sorry, I got a sign wrong. Ignore what I said.
     
  8. Apr 5, 2012 #7

    tms

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    You know that for each mass, the rod and the disk, the mass times the distance to the center of mass is the same:
    [tex]M(L - \ell) = m(\ell - x),[/tex] where [itex]x[/itex] is the (unknown) location of [itex]m[/itex]'s center of mass. Now take the equation in your graphic, collect all the terms in [itex]m[/itex] on the right, and factor out the [itex]m[/itex]. Set what is left on the right equal to [itex]\ell - x[/itex], and solve for [itex]x[/itex]. The result contradicts the drawing as I understand it, but the drawing may be wrong, or the exact problem may have been stated differently. I don't know where that stuff on the right comes from, but it leads to a correct result, modulo the drawing.
     
  9. Apr 5, 2012 #8

    RK7

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    In the end I got it. I have no idea what the lecturer was doing but I get the equivalent result by balancing the moments:
    [itex]M(L-l)=m(\frac{L}{2}-(L-l))
    \Rightarrow
    l = \frac{L}{2} . \frac{2M+m}{M+m}[/itex]
    where [itex]M(L-l)[/itex] is the ball's moment about the COM and [itex]\frac{L}{2}-(L-l)[/itex] is the distance from the COM of the rod to the COM of the combined body.

    Thanks
     
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