How Is Angular Momentum Calculated for a Spinning Coin?

[royguitarboy]

Homework Statement

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?

Homework Equations

L=r*p
L=I *omega

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.

 

[Doc Al]

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.

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

But that's not what you need for A. Use $L = I \omega$.

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

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.

 

[royguitarboy]

$L = I \omega$ just gives me the angular momentum about the center of mass though right? How do I account for a distance 10cm from that point?

 

[Doc Al]

$L = I \omega$ just gives me the angular momentum about the center of mass though right?

Right.

How do I account for a distance 10cm from that point?

By calculating the angular momentum of the center of mass using the equation I gave:
$$\vec{L} = \vec{r} \times \vec{p}$$

 

[royguitarboy]

Ok, now I'm really confused about this. For part A, I need to use $$\vec{L} = \vec{r} \times \vec{p}$$ where p is mass times velocity?

 

[Doc Al]

Ok, now I'm really confused about this. For part A, I need to use $$\vec{L} = \vec{r} \times \vec{p}$$ where p is mass times velocity?

No, that's not needed for part A. All you need is $L = I \omega$.

 

[royguitarboy]

ok I got you now. Now part B, I need to add part A to $$\vec{L} = \vec{r} \times \vec{p}$$ ?

 

[Doc Al]

Exactly.

 

[royguitarboy]

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)($$\omega$$*(.019/2) ?

 

[Doc Al]

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.

 

[royguitarboy]

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