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Doc Al -Blast from the Past

  1. Sep 17, 2007 #1
    Doc Al!!!--Blast from the Past

    Hi Doc Al,
    In 2004 you helped out a guy with this problem:

    A 10 g coin of diameter 1.3 cm is spinning at 16 rev/s about a vertical diameter at a fixed point on a tabletop.

    (a) What is the angular momentum of the coin about its center of mass?
    (b) What is its angular momentum about a point on the table 10 cm from the coin?
    (c) If the coin spins about a vertical diameter at 16 rev/s while its center of mass travels in a straight line across the tabletop at 5 cm/s, what is the angular momentum of the coin about a point on the line of motion?
    (d) What is the angular momentum of the coin about a point 10 cm from the line of motion? (There are two answers to this question.)

    You then responded:

    This may help you: the angular momentum of an object is the sum of:
    (1) the angular momentum of the center of mass
    (2) the angular momentum about the center of mass

    Thus:
    a: See previous post
    b: What's the movement of the center of mass?
    c: See above
    d: See above (there are two answers since you could be on either side of the line of motion)




    2. I know I need 1/4 MR^squared for part a, but do I need 1/3 ML^squared for either b, c, or d?



    3. I can handle part a, but but in part b I'm stuck: the center of mass is fixed there, but then in part c and d the cm travels in a straight line in the horizontal direction and I'm not sure how to take that into account. Can you help me? Thanks, and I'm glad a place like this exists for people like me!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2007 #2

    Doc Al

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    Staff: Mentor

    Hey, I was younger then!
    What do you need 1/3 M L^2 for? (Not sure I understand where you got that.)



    Read my response, which you quoted:
    If the center of mass doesn't move, you can skip part (1).

    If it does move, you have to add that additional angular momentum to the spinning part. What's the angular momentum of a moving point mass about some point? Hint: [tex]\vec{L} = \vec{r}\times\vec{p}[/tex], where p is the linear momentum of the mass. Read this: Angular Momentum of a Particle
     
  4. Sep 18, 2007 #3
    Thanks for answering! OK, I got part (b). Now I think that if [tex]\vec{L} = \vec{r}\times\vec{p}[/tex] is worked out it becomes mvrsin[tex]\theta[/tex]. But if that's the case, the object is moving in a straight line. Doesn't that make [tex]\theta[/tex] zero? And if so, isn't L now zero as well?
     
  5. Sep 18, 2007 #4

    Doc Al

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    Staff: Mentor

    The angular momentum contributed by the object's linear momentum depends on your reference point. If you compute the angular momentum about a point on the line of the object's motion, then theta is zero (or 180) and the linear momentum contributes nothing to the total angular momentum. (Read the link I gave in the last post.) But the spinning still contributes to the total angular momentum.
     
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