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Angular Momentum of a coin

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A 11 g coin of diameter 1.9 cm is spinning at 15 rev/s about a vertical diameter at a fixed point on a tabletop. A coin is a solid cylinder of length L and radius R, where L is negligible compared to R. Its moment of inertia is 1/4MR^2.

    (A)What is its angular momentum about a point on the table 10 cm from the coin?

    For the following questions, assume the coin spins about a vertical diameter at 15 rev/s while its center of mass travels in a straight line across the tabletop at 5 cm/s.

    (B)What is the angular momentum of the coin about a point on the line of motion of the center of mass?

    (C)What is the angular momentum of the coin about a point 10 cm from the line of motion?


    2. Relevant equations

    L=r*p
    L=I *omega



    3. The attempt at a solution

    I tried L= r*p, where p equals m*v, and r was .1m, m was .011kg, and v was omega times the radius of the coin, which gave me an angular momentum of 9.85e-4. I know this isn't right.

    For B should I add the angular momentum and the linear momentum?

    And for C, if I can figure out A, I might be able to do that with C.
     
  2. jcsd
  3. Oct 16, 2007 #2

    Doc Al

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    Staff: Mentor

    That's the formula for the angular momentum of a point mass. It should read:
    [tex]\vec{L} = \vec{r} \times \vec{p}[/tex]

    But that's not what you need for A. Use [itex]L = I \omega[/itex].

    You need to add the angular momentum about the center of mass (the result of A) and the angular moment of the center of mass (using the linear momentum of the coin, as described by the first equation above). Be sensitive to the difference between B and C.
     
  4. Oct 16, 2007 #3
    [itex]L = I \omega[/itex] just gives me the angular momentum about the center of mass though right? How do I account for a distance 10cm from that point?
     
  5. Oct 16, 2007 #4

    Doc Al

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    Staff: Mentor

    Right.
    By calculating the angular momentum of the center of mass using the equation I gave:
    [tex]\vec{L} = \vec{r} \times \vec{p}[/tex]
     
  6. Oct 16, 2007 #5
    Ok, now I'm really confused about this. For part A, I need to use [tex]\vec{L} = \vec{r} \times \vec{p}[/tex] where p is mass times velocity?
     
  7. Oct 16, 2007 #6

    Doc Al

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    Staff: Mentor

    No, that's not needed for part A. All you need is [itex]L = I \omega[/itex].
     
  8. Oct 16, 2007 #7
    ok I got you now. Now part B, I need to add part A to [tex]\vec{L} = \vec{r} \times \vec{p}[/tex] ?
     
  9. Oct 16, 2007 #8

    Doc Al

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    Staff: Mentor

    Exactly.
     
  10. Oct 16, 2007 #9
    I'm confused as to which r, I should use for each part. In the the r X p part, the r is .05m right? So the total momentum is:

    I(omega) + r X p

    so I got 2.34e-5 + .05m X (.011)([tex]\omega[/tex]*(.019/2) ?
     
  11. Oct 16, 2007 #10

    Doc Al

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    Staff: Mentor

    The r to use in the r X p part is specified as 10 cm in parts a and c, the distance to the coin. Note that they don't bother to specify a distance in part b--there's a good reason for that. Also, the p is the linear momentum of the coin. Note also that r X p is a vector cross product--direction counts.
     
  12. Oct 16, 2007 #11
    Ok. I think I might have it. So for C, I need to add what I got in part B to the cross product of v and p, and subtract it, cause there should be two answers. Is that right? Well, I'm kind of confused about how to exactly do the cross product, we didn't really go over that.


    Edit: Never mind the above part. I got it!

    You're a genius
     
    Last edited: Oct 16, 2007
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