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Angular Momentum of a Spinning Coin

  1. Mar 30, 2004 #1
    I'm trying to calculate the angular momentum of a coin spinning about a vertical diameter about its center of mass. Given is mass, diameter, and angular velocity. I thought it would be [tex]L=\frac{1}{4}MR^2\omega[/tex]. The angular momentum about a point away from the coin I thought would be [tex]L=\frac{1}{4}MR^2\omega + Mh^2\omega[/tex] where [tex]h[/tex] is the distance from the axis containing the diameter of the coin to the point away from the coin. And if the coin's center of mass were moving in a straight line while doing the two above things, then I thought that to each of those would be added [tex]MvR[/tex]. I'm guessing that it is just something simple that I missed in the beginning which propagated the error through the other parts of the problem. Help would be greatly appreciated.
     
    Last edited: Mar 30, 2004
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  3. Mar 31, 2004 #2

    Doc Al

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    What exactly is the motion of the coin? If it's just spinning about a vertical diameter, you would be correct that the angular momentum is (treating the coin as a thin disk):
    [tex]L=\frac{1}{4}MR^2\omega[/tex]
    I do not understand the second term in your second equation. Is the coin also revolving about some point with the same angular speed as it is rotating?
     
  4. Mar 31, 2004 #3
    For the sake of clarity, I will just post the whole question.

    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.)

    I appreciate your help.
     
  5. Mar 31, 2004 #4

    Doc Al

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    Much clearer. :smile:

    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)
     
  6. Jul 20, 2004 #5
    hi, sorry but I am a little slow
    Can you explain to me how did you get the formula for the angular momentum please
    Thanks for the help
     
  7. Jul 21, 2004 #6

    Doc Al

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    Welcome to PF, ChanDdoi.

    I'm not sure I understand your question. Which formula are you asking about?

    In general, if an object rotates about an axis, the magnitude of the angular momentum will be [itex]L = I\omega[/itex], where I is the rotational inertia about that axis and [itex]\omega[/itex] is the angular speed.
     
  8. Jul 21, 2004 #7
    Well, why would the inertia of the coin be [tex]\frac{1}{4}MR^2\[/tex] i just can't seem to figure it out *may be cuz i'm slow*
    The book said if you treat it as a disk then the inertia would be [tex]\frac{1}{2}MR^2\[/tex] and if I try to treat it as a sum of bars since it is spinning on a vertical axis then I got lost with the integrals and stuff . Either way I just cannot seem to get that [tex]\frac{1}{4}MR^2\[/tex]

    btw, thanks for welcoming me to PF
     
  9. Jul 21, 2004 #8

    Doc Al

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    rotational inertia of disk about diameter

    Now I see your question: How to find the rotational inertia of a disk spinning about a diameter. If you would like help with the integral, post what you've got and someone will give you a hand.

    But the smart way is to use the perpendicular axis theorem.

    Note:
    [tex]\frac{1}{2}MR^2[/tex]
    is the rotational inertia of a disk rotating about a perpendicular axis through its center.
    [tex]\frac{1}{4}MR^2[/tex]
    is the rotational inertia of a disk rotating about a diameter. The perpendicular axis theorem gives a simple way to calculate one from the other. Here's a link that explains it: http://hyperphysics.phy-astr.gsu.edu/hbase/perpx.html
     
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