What is Angular Momentum and How Do I Calculate It?

  • Thread starter acerulz54
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In summary: So even though the object is moving in a straight line, the angular momentum is still conserved.Thanks for the clarification! OK, I got part (c). But now I'm stumped on part (d). The line of motion moves in a straight line, but the center of mass (which I assume is the coin) is moving around. What do you mean by "around"? Thanks, and I'm glad a place like this exists for people like me!Hi Doc Al,In summary, the angular momentum of a spinning object is the sum of the angular momentum of the center of mass and the angular momentum about the center of mass. If the center of mass doesn't move, you can
  • #1
acerulz54
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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!
 
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  • #2
acerulz54 said:
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.
Hey, I was younger then!
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?
What do you need 1/3 M L^2 for? (Not sure I understand where you got that.)



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.
Read my response, which you quoted:
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
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
 
  • #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?
 
  • #4
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.
 

1. Who is Doc Al and what is "Blast from the Past"?

Doc Al is a renowned scientist and inventor who has created a time machine called "Blast from the Past." This machine allows people to travel back in time and experience historical events firsthand.

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Doc Al's time machine works by harnessing the power of quantum mechanics and wormhole theory. It creates a portal through which a person can enter and travel through time.

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Yes, "Blast from the Past" is completely safe to use. Doc Al has taken every precaution to ensure the safety of travelers, and the machine has been extensively tested and approved by scientific experts.

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While anyone can technically use the time machine, there are certain limitations. The machine requires a great deal of scientific knowledge and skill to operate, so only trained individuals are allowed to use it. Additionally, there are restrictions on what events can be visited to prevent altering the course of history.

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The main consequence of using "Blast from the Past" is the potential to alter the timeline and create a ripple effect that could drastically change the present. That's why strict guidelines and precautions are in place for the use of the time machine.

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