Angular Momentum of a Spinning Coin

In summary, the angular momentum of a coin spinning about a vertical diameter at a fixed point on a tabletop can be calculated using the formula L=\frac{1}{4}MR^2\omega, where M is the mass of the coin, R is the radius of the coin, and \omega is the angular velocity. This formula can also be used to calculate the angular momentum about a point 10 cm from the coin, as long as the coin is spinning without any additional motion. If the coin's center of mass is moving in a straight line while spinning, the angular momentum about a point on the line of motion can be calculated by adding the angular momentum of the center of mass and the angular momentum about the center of mass.
  • #1
cpuwildman
10
0
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.
 
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  • #2
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?
 
  • #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.
 
  • #4
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)
 
  • #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
 
  • #6
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.
 
  • #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 because 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
 
  • #8
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
 

What is angular momentum?

Angular momentum is a physical quantity that measures the amount of rotational motion an object has. It is a vector quantity, meaning it has both magnitude and direction.

How is angular momentum of a spinning coin calculated?

The angular momentum of a spinning coin is calculated by multiplying the moment of inertia (a measure of an object's resistance to rotational motion) by the angular velocity (rate of change of angular displacement).

What factors affect the angular momentum of a spinning coin?

The angular momentum of a spinning coin can be affected by its mass, moment of inertia, and angular velocity. Additionally, external forces such as friction or air resistance can also impact its angular momentum.

Can angular momentum of a spinning coin change?

Yes, the angular momentum of a spinning coin can change if there is a change in its mass, moment of inertia, or angular velocity. This can occur due to external forces acting on the coin or changes in the coin's spin.

How is the conservation of angular momentum applied to a spinning coin?

The conservation of angular momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. This means that if a spinning coin experiences no external torque, its angular momentum will remain constant.

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