# Angular Momentum of a Spinning Coin

1. Mar 30, 2004

### cpuwildman

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 $$L=\frac{1}{4}MR^2\omega$$. The angular momentum about a point away from the coin I thought would be $$L=\frac{1}{4}MR^2\omega + Mh^2\omega$$ where $$h$$ 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 $$MvR$$. 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
2. Mar 31, 2004

### Staff: Mentor

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):
$$L=\frac{1}{4}MR^2\omega$$
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?

3. Mar 31, 2004

### cpuwildman

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.)

4. Mar 31, 2004

### Staff: Mentor

Much clearer.

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)

5. Jul 20, 2004

### ChanDdoi

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

6. Jul 21, 2004

### Staff: Mentor

Welcome to PF, ChanDdoi.

In general, if an object rotates about an axis, the magnitude of the angular momentum will be $L = I\omega$, where I is the rotational inertia about that axis and $\omega$ is the angular speed.

7. Jul 21, 2004

### ChanDdoi

Well, why would the inertia of the coin be $$\frac{1}{4}MR^2\$$ i just can't seem to figure it out *may be cuz i'm slow*
The book said if you treat it as a disk then the inertia would be $$\frac{1}{2}MR^2\$$ 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 $$\frac{1}{4}MR^2\$$

btw, thanks for welcoming me to PF

8. Jul 21, 2004

### Staff: Mentor

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:
$$\frac{1}{2}MR^2$$
is the rotational inertia of a disk rotating about a perpendicular axis through its center.
$$\frac{1}{4}MR^2$$
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