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Angular momentum of a purely rolling body

  • #1
Krushnaraj Pandya
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Homework Statement


A disk is undergoing pure rolling motion with speed v. The radius of the disk being R and mass M. Then the angular momentum of the disk about the
1)bottom most and
2)top most point

Homework Equations


1) L(orbital) = m*v*r where v is the velocity of cm which is perpendicular to the given axis and r is the perpendicular distance between vector mv and and the axis.
2) L(spin) = Iw where I is the moment of inertia about the axis of rotation

The Attempt at a Solution



I got the answer in the first case-Since bottom most point is stationary in pure rolling I took L(orbital) as mvr and L(spin) as I*w=(MR^2/2)*w= mvr/2. adding both since both are in a clockwise sense- I get the answer L=1.5MVR which is correct.
In the second case however- the topmost point has velocity 2v and therefore v(cm) is relatively going to v to the left w.r.t to that point. since mv is in a clockwise sense, I took L(orbital) as mvr again and Iw remains unchanged giving the same answer which is incorrect, please let me know where I'm wrong
 

Answers and Replies

  • #2
Nathanael
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You found the angular momentum in the frame where the top point is motionless, right? Perhaps they meant to find the angular momentum about that point in the ground frame. The question is ambiguous in that regard.
 
  • #3
Krushnaraj Pandya
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You found the angular momentum in the frame where the top point is motionless, right? Perhaps they meant to take the angular momentum about that point in the ground frame. The question is ambiguous in that regard.
yes I did....let me try it keeping that in mind
 
  • #4
Krushnaraj Pandya
Gold Member
697
71
You found the angular momentum in the frame where the top point is motionless, right? Perhaps they meant to find the angular momentum about that point in the ground frame. The question is ambiguous in that regard.
I'm getting Very confused changing frames, how do I take angular momentum about a moving point w.r.t to the ground frame
Ah! I think I have a considerable logic which gives the correct answer, please confirm. So about bottom most point, L is mvr + Iw =3/2mvr
about topmost point, relative to it v(cm) is -v therefore direction of L(orbital) is reversed while L(spin) is still mvr/2...then mvr/2-mvr would give the right answer but the only problem is...in the first case radius vector points upwards and velocity to the right, using right hand screw rule gives L inwards. In the second case radius vector points downwards and velocity to the left...so L is inwards...again and should give 3/2 mvr on adding with L(spin). Is my reasoning wrong anywhere here?
if yes...I need to consider the ground frame but I have no idea how
 
Last edited:
  • #5
Nathanael
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I'm getting Very confused changing frames, how do I take angular momentum about a moving point w.r.t to the ground frame
When I said “find the angular momentum about that point in the ground frame” what I mean is to take the origin to be (initially) at the top-most point but not moving with it.

If you imagine the origin as moving with the top point, then we would need to leave the ground frame to analyze it.
 
  • #6
Krushnaraj Pandya
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When I said “find the angular momentum about that point in the ground frame” what I mean is to take the origin to be (initially) at the top-most point but not moving with it.

If you imagine the origin as moving with the top point, then we would need to leave the ground frame to analyze it.
so the top most point at t=0 is origin, then the disc moves right on and that point is still our origin, right?
 
  • #7
Nathanael
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so the top most point at t=0 is origin, then the disc moves right on and that point is still our origin, right?
Right. So the origin will only coincide with the top most point for the initial instant.

Again this may not be what they meant... the question is unclear about it.
 
  • #8
Krushnaraj Pandya
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so the top most point at t=0 is origin, then the disc moves right on and that point is still our origin, right?
Ah! yes, then the radius vector points downwards and velocity to the right, giving L upwards (in the opposite direction to first case!)
so -mvr + mvr/2 gives us the right answer. I'm guessing vector L upwards as positive or negative is a matter of convention
 
  • #9
Krushnaraj Pandya
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Right. So the origin will only coincide with the top most point for the initial instant.

Again this may not be what they meant... the question is unclear about it.
the question is indeed unclear! but I'd never have thought of it this way...you've a very nice thinking process, thank you :D
 
  • #10
Nathanael
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Ah! yes, then the radius vector points downwards and velocity to the right, giving L upwards (in the opposite direction to first case!)
so -mvr + mvr/2 gives us the right answer. I'm guessing vector L upwards as positive or negative is a matter of convention
I just want to make sure you know that L is not “upwards” as in along gravity.. L is “coming out of the page.”

But yes, which way you say is positive is just convention. The common way is to follow the “right hand rule” but all your answers will come out the same if you use the “left hand rule” as long as you just stick to one or the other.
 
  • #11
Krushnaraj Pandya
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I just want to make sure you know that L is not “upwards” as in along gravity.. L is “coming out of the page.”

But yes, which way you say is positive is just convention. The common way is to follow the “right hand rule” but all your answers will come out the same if you use the “left hand rule” as long as you just stick to one or the other.
yes! I should have said outwards, I misspoke...Its clear now, thank you very much :D
I wanted to ask for some advice on a career in research but its late here and I have school tomorrow...so good night(/day), see you tomorrow(/today)
 
  • #12
Nathanael
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I wanted to ask for some advice on a career in research but its late here and I have school tomorrow...
I am also a student, so I can’t help with that. There is a sub forum here “career/academic guidance” where you will hopefully get some excellent advice.

good night(/day), see you tomorrow(/today)
:biggrin:
 
  • #13
Krushnaraj Pandya
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I am also a student, so I can’t help with that. There is a sub forum here “career/academic guidance” where you will hopefully get some excellent advice.


:biggrin:
how do you have such a good grasp of physics then (P.S I'm just in high school)
 
  • #14
Nathanael
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how do you have such a good grasp of physics then (P.S I'm just in high school)
There is a lot I don’t know; I just try to make sure that what I do know, I really know.
Also I am in college... in a few years when you are my age I’m sure you will be at least as good!
 

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