Determining $L_{o}$: Finding Angular Momentum of System

In summary, In this conversation, the concept of angular momentum is discussed in relation to a system of two disks rotating about an axis passing through O and perpendicular to the plane of the disk. The reasoning is that since there is no torque acting on the system, the only necessary factors to determine the angular momentum are the moment of inertia of the smaller disk about the axis through O and the angular velocity of the system about O. The system is given an initial angular velocity about a vertical axis passing through O', and both disks are considered to be a part of the system. However, the setup of the system is unclear and not fully defined. The original question from the exam is also confusing and unclear, making it difficult to determine the exact parameters and variables
  • #1
vcsharp2003
897
177
Homework Statement
The problem diagram is as given below.

A small disk with center ##O^{'}## is kept on a larger disk with center ##O## as shown in diagram.
The system is given an angular velocity ##\omega_ {0}## about a vertical axis passing through O'. The friction between disks is negligible and smaller disk is free to rotate about O'. Find the angular momentum of system at ##t = \frac {\pi } {\omega_0}## about a vertical axis passing through O.
Relevant Equations
##I_{o'} = I_{o} + m r^2## (parallel axis theorem)
##\vec {L} = I \vec {\omega}## (angular momentum)
IMG-20220528-WA0029~3.jpg


I think the the time given doesn't matter since no torque is acting on the system, but not sure. Therefore, all we need is to determine the angular momentum about the axis passing through O and perpendicular to the plane of disk. This will involve finding the moment of inertia of smaller disk about the axis through O by using the parallel axis theorem and also the angular velocity of the system about O. Does this reasoning sound correct or maybe I am missing something?
$$L_{o} \text { stands for the angular momentum of system} $$
$$L_{o} = L_{so} + L_{lo} \text { ,where s stands for smaller disk and l for larger disk} $$
$$L_{o} = I_{so} ~ \omega_{1} + I_{lo} ~ \omega_{1} \text { ,where } \omega_{1} \text { stands for angular velocity about O} $$

Also, the angular velocity about the axis through O should be different from angular velocity ##\omega_{0} ## about the axis through O'. I get ##\omega_{1}## in terms of ##\omega_{0}## as below, after equating the linear velocity of a point along OO' and on the edge of larger disk.
$$\omega_{0} ~ R = \omega_{1} ~ 2R$$
$$\therefore \omega_{1} = \frac {\omega_{0}} {2}$$
 
Last edited:
  • Informative
Likes Delta2
Physics news on Phys.org
  • #2
When you say
vcsharp2003 said:
The system is given an angular velocity ##\omega_ {0}## about a vertical axis passing through O'.
does the "system" include the disk?
 
  • #3
kuruman said:
does the "system" include the disk?
Yes, both the disks form the system.
 
  • #4
The setup is unclear to me. Are the disks allowed to move freely in the plane (apart from being linked)? Is the large disk fixed to O? If so how does the system rotate about O’ initially? Etc
 
  • Like
Likes nasu, Delta2 and Lnewqban
  • #5
The diagram suggests that angular velocity is for the big disc, while the text shows it to be “about a vertical axis passing through O', which is represented as the axis for the small disc.
As “friction between disks is negligible”, the point of application for the force or moment inducing the angular velocity is key, but not defined.
 
  • Like
Likes vcsharp2003
  • #6
The question is not making sense to me either. Is the system floating in space or mounted on a fixed axle at O or O'? Is it saying that each disk is given angular velocity ##\omega_0## about O'?
Please check that you have posted the whole question verbatim.
 
  • Like
Likes Delta2
  • #7
Orodruin said:
The setup is unclear to me. Are the disks allowed to move freely in the plane (apart from being linked)? Is the large disk fixed to O? If so how does the system rotate about O’ initially? Etc
The question was confusing me too. The actual question in printed form is as below.

Since there is no friction involved so we can say that there is no torque on any disk except the torque that was initially applied to achieve initial angular velocity. In absence of any torque the rotation about O' continues at ##\omega_{o}## angular velocity.

IMG-20220528-WA0029~2.jpg
 
  • #8
haruspex said:
The question is not making sense to me either. Is the system floating in space or mounted on a fixed axle at O or O'? Is it saying that each disk is given angular velocity ##\omega_0## about O'?
Please check that you have posted the whole question verbatim.
The question in printed form is as given in my recent post#7. It's confusing me too. I guessed that the entire system is given an angular velocity about O'. So, it's like the bigger and smaller disk rotating about O' together. As far as how they are mounted is not given in the problem statement which you can see in post#7.
 
  • #9
vcsharp2003 said:
Since there is no friction involved so we can say that there is no torque on any disk except the torque that was initially applied to achieve initial angular velocity
This is not true. It depends on which axis you consider as there can indeed be a force between the disks at O’.
 
  • #10
Orodruin said:
This is not true. It depends on which axis you consider as there can indeed be a force between the disks at O’.
What other force could be there causing a torque on the smaller disk? The friction on smaller disk from the larger disk should have caused a torque on smaller disk about O'.

If friction is absent between the two disks then the smaller disk would not rotate with the larger disk.
 
Last edited:
  • #11
vcsharp2003 said:
What other force could be there causing a torque on the smaller disk? The friction on smaller disk from the larger disk should have caused a torque on smaller disk about O'.
Torque from a local force is relative to an axis. A force through O' cannot have a torque about O', but it can have a torque about O.
 
  • Like
Likes Orodruin and vcsharp2003
  • #12
Out of curiosity I located the original question. It is Q13 here:
https://www.fiitjeenorthwest.com/admin/upload/AITS-1819-OT-JEEA-PAPER-2_3-2-19.pdf

The question appears (as already noted by others) to be ill-posed, making it (IMO) unanswerable.

Off-topic, it is interesting to note that the paper is for a 3 hour exam covering physics, chemistry and maths. There are 54 questions and it appears impossible to attempt all of them in the available time. What is required by the examiners is a mystery!
 
  • Like
Likes PhDeezNutz, Lnewqban and vcsharp2003
  • #13
Just reading through and parsing the instructions correctly will take half an hour … 😛
 
  • Like
Likes PhDeezNutz
  • #14

Steve4Physics said:
Out of curiosity I located the original question. It is Q13 here:
https://www.fiitjeenorthwest.com/admin/upload/AITS-1819-OT-JEEA-PAPER-2_3-2-19.pdf

The question appears (as already noted by others) to be ill-posed, making it (IMO) unanswerable.

Off-topic, it is interesting to note that the paper is for a 3 hour exam covering physics, chemistry and maths. There are 54 questions and it appears impossible to attempt all of them in the available time. What is required by the examiners is a mystery!
I also think that this is an impossible problem that is not properly framed. I tried an online Google search for this problem, but couldn't find any. I wonder how you found it.
 
  • #15
vcsharp2003 said:
I tried an online Google search for this problem, but couldn't find any. I wonder how you found it.
I used Google with text from original question:
"A disc of mass m and radius 2R has another disc of mass 2m"

Note the quotation marks. This tells Google to try to find an exact match.

Then picked the FIITJEE site.

Is this an exam you are studying for? I don't understanstand why there are so many (often long) questions in a 3 hour paper. Can you offer any insight?
 
  • #16
Steve4Physics said:
I used Google with text from original question:
"A disc of mass m and radius 2R has another disc of mass 2m"

Note the quotation marks. This tells Google to try to find an exact match.

Then picked the FIITJEE site.

Is this an exam you are studying for? I don't understanstand why there are so many (often long) questions in a 3 hour paper. Can you offer any insight?
I see. That's a nice trick in Google search that I never knew about.

No, I'm not studying for it. This is not the real JEE exam, which never gives these impossible problems. It's from a study center that helps students prepare for JEE exams. These study centers also called coaching centers make millions in helping students prepare for the real JEE exam.
The real JEE exam is very tough, which is taken by a huge number of students and I'm guessing there is relative marking done in these exams since most students cannot complete the test paper in alloted time due to the difficulty level of most questions.
 
  • Like
Likes Lnewqban and Steve4Physics
  • #17
vcsharp2003 said:
I see. That's a nice trick in Google search that I never knew about.

No, I'm not studying for it. This is not the real JEE exam, which never gives these impossible problems. It's from a study center that helps students prepare for JEE exams. These study centers also called coaching centers make millions in helping students prepare for the real JEE exam.
The real JEE exam is very tough, which is taken by a huge number of students and I'm guessing there is relative marking done in these exams since most students cannot complete the test paper in alloted time due to the difficulty level of most questions.
I wonder about the spread in scores in this sort of test. It seems to me that the purpose of something like this would be to rank the takers according to score from 100% to zero. Uncertainty in ranking becomes larger if the difference between the highest and lowest scoring paper is like 25% at whatever average because only one quarter of the full range is used. This looks like a low average - narrow spread test to me.
 
  • Like
Likes vcsharp2003
  • #18
kuruman said:
I wonder about the spread in scores in this sort of test. It seems to me that the purpose of something like this would be to rank the takers according to score from 100% to zero. Uncertainty in ranking becomes larger if the difference between the highest and lowest scoring paper is like 25% at whatever average because only one quarter of the full range is used. This looks like a low average - narrow spread test to me.
The Wiki article on the exam https://en.wikipedia.org/wiki/Joint_Entrance_Examination_–_Advanced makes quite interesting reading.

”It has a very low qualification rate [presumably that means pass rate] (about 9,369 in 479,651 in 2012; ~1.95%)[9] The qualification rate of the JEE-Advanced in 2017 was approximately 0.92% (about 11,000 out of 1,200,000 who applied for JEE-Main).[10] The level of Physics, Chemistry and Mathematics in the exam requires high analytical thinking, hence regarded as the second toughest examination of India, after UPSC Civil Services Examination and third in the world after China's Gaokao and India's UPSC.“
 
  • Informative
  • Like
Likes vcsharp2003 and kuruman
  • #19
kuruman said:
I wonder about the spread in scores in this sort of test. It seems to me that the purpose of something like this would be to rank the takers according to score from 100% to zero. Uncertainty in ranking becomes larger if the difference between the highest and lowest scoring paper is like 25% at whatever average because only one quarter of the full range is used. This looks like a low average - narrow spread test to me.
Yes, JEE gives a unique rank to each qualified student like AIR 1 or AIR 2 or AIR 5000. The questions are so difficult that it's generally not possible for anyone to get 100%. So, I think if someone gets even 60% correct then that student would get a very good rank.
 
  • #21
haruspex said:
Torque from a local force is relative to an axis. A force through O' cannot have a torque about O', but it can have a torque about O.
Is it possible that the small disk also rotates about O' with the same angular velocity as the larger disk ##\omega_{o}##? If small and large disks are somehow attached at O', then perhaps this could happen.
 
  • #22
vcsharp2003 said:
Is it possible that the small disk also rotates about O' with the same angular velocity as the larger disk ##\omega_{o}##? If small and large disks are somehow attached at O', then perhaps this could happen.
If both disks are mounted on a fixed axle at O' then there is no interaction between them. All gets rather boring.
Likewise if the larger disk is on a fixed axle at O.
 
  • Like
Likes vcsharp2003
  • #23
haruspex said:
If both disks are mounted on a fixed axle at O' then there is no interaction between them. All gets rather boring.
Likewise if the larger disk is on a fixed axle at O.
The problem statement says that small disk is free to rotate about O', so it must be attached to some axle. Now, if this axle is given a certain angular velocity then the disks connected to these axles must also rotate with the angular velocity of axle about O'.
 
  • #24
vcsharp2003 said:
The problem statement says that small disk is free to rotate, so it must be attached to some axle. Now, if this axle is given a certain angular velocity then the disks connected to these axles must also rotate with the angular velocity of axle about O'.
Oh yes, it is definitely mounted on an axle on the larger disk. The question is whether the system as a whole is floating in space or mounted on an externally fixed axle, and if so, where?
Odd that we are told the initial rotation is about a vertical axis, but not whether the disks are in a horizontal or vertical plane. The only clue we have is the curved arrow in the diagram.
 
  • Like
Likes vcsharp2003
  • #25
haruspex said:
Oh yes, it is definitely mounted on an axle on the larger disk. The question is whether the system as a whole is floating in space or mounted on an externally fixed axle, and if so, where?
Can't we make some reasonable assumptions here? Perhaps mounted in air with no solid surface beneath the larger disk. If this assumption is made, then we can say that as larger disk rotates about O', the smaller disk rotates also about O' with same angular velocity. Also, since there is no friction between the disks, so smaller disk cannot move with larger disk. Hence we have a scenario where the larger disk is rotating about O' and also the smaller disk is rotating about O'.
 
  • #26
I think this problem is solvable if we make some reasonable assumptions.
 
  • #27
vcsharp2003 said:
Can't we make some reasonable assumptions here? Perhaps mounted in air with no solid surface beneath the larger disk. If this assumption is made, then we can say that as larger disk rotates about O', the smaller disk rotates also about O' with same angular velocity. Also, since there is no friction between the disks, so smaller disk cannot move with larger disk. Hence we have a scenario where the larger disk is rotating about O' and also the smaller disk is rotating about O'.
If there is no fixed axle at O' then the larger disk won't continue to rotate about it.
Ignore the small disk for the moment. If the large disk is initially set to rotate about O' then we can resolve that into a rotation about its centre plus a linear motion of the centre. Both of those will continue, but the point of the disk that is its instantaneous centre of rotation keeps changing. The point in space that is its instantaneous centre of rotation will move within the linear velocity of the disk's centre.

But you are right that then introducing the small disk makes a worthwhile question. Have a go at it.
 
  • Like
Likes vcsharp2003
  • #28
haruspex said:
If there is no fixed axle at O' then the larger disk won't continue to rotate about it.
Ignore the small disk for the moment. If the large disk is initially set to rotate about O' then we can resolve that into a rotation about its centre plus a linear motion of the centre. Both of those will continue, but the point of the disk that is its instantaneous centre of rotation keeps changing. The point in space that is its instantaneous centre of rotation will move within the linear velocity of the disk's centre.

But you are right that then introducing the small disk makes a worthwhile question. Have a go at it.
If I focus only on the larger disk then following are the positions at ##t=0## and ##t=\frac {\pi}{\omega_{o}}## as it rotates about O' at an angular speed of ##{\omega_{o}}##. We want to find the angular momentum about an axis passing through O and perpendicular to the plane of paper, of the larger disk at ##t=\frac {\pi}{\omega_{o}}##.

Are the two positions I have shown correct?
If yes, then would it be correct to say that angular momentum about O axis is moment of inertia of disk about O axis X angular speed about O axis at given instant?
CamScanner 05-30-2022 15.24.jpg
 
  • #29
vcsharp2003 said:
If I focus only on the larger disk then following are the positions at ##t=0## and ##t=\frac {\pi}{\omega_{o}}## as it rotates about O' at an angular speed of ##{\omega_{o}}##. We want to find the angular momentum about an axis passing through O and perpendicular to the plane of paper, of the larger disk at ##t=\frac {\pi}{\omega_{o}}##.

Are the two positions I have shown correct?
If yes, then would it be correct to say that angular momentum about O axis is moment of inertia of disk about O axis X angular speed about O axis at given instant?View attachment 302116
You do not seem to have understood my post #27.
If there is no fixed axle at O' then the large disk cannot continue to rotate about it, with or without the smaller disk. There is nothing to provide the centripetal force.

Leaving out the smaller disk, we can think of the larger disk's initial motion as the sum of a rotation rate ##\omega_0## about its own centre and a linear velocity ##R\omega_0## in the negative y direction. And with no external forces, that is how it would continue.

Now adding in the smaller disk, we would need a centripetal force on that to keep the large disk rotating about its own centre, so something else must happen.

Any rotation the smaller disk has on its own axis is irrelevant since the centripetal force for that would be internal, so it seems to me we can treat the smaller disk as a point mass.
 
  • Like
Likes vcsharp2003
  • #30
haruspex said:
You do not seem to have understood my post #27.
If there is no fixed axle at O' then the large disk cannot continue to rotate about it, with or without the smaller disk. There is nothing to provide the centripetal force.

Leaving out the smaller disk, we can think of the larger disk's initial motion as the sum of a rotation rate ##\omega_0## about its own centre and a linear velocity ##R\omega_0## in the negative y direction. And with no external forces, that is how it would continue.

Now adding in the smaller disk, we would need a centripetal force on that to keep the large disk rotating about its own centre, so something else must happen.

Any rotation the smaller disk has on its own axis is irrelevant since the centripetal force for that would be internal, so it seems to me we can treat the smaller disk as a point mass.
Ok. I need to understand your previous post and what you stated in this post.

I was thinking that there is a fixed axis going through O', and another fixed axis going through O. And the larger disk is rotating about the O' axis. As the larger disk rotates, the O axis as well as O' axis remain fixed. Also while the O' axis has a fixed axle, the O axis doesn't have an axle

Perhaps, I am missing something. Let me read your last post again.
 
Last edited:
  • Like
Likes Delta2
  • #31
vcsharp2003 said:
I was thinking that there is a fixed axis going through O', and another fixed axis going through O. And the larger disk is rotating about the O' axis. As the larger disk rotates, the O axis as well as O' axis remain fixed.
I thought we agreed in posts #25 and #27 that there is no axle fixed in space. The two disk system is floating in space. All that is 'fixed' is that there is an axle mounted on the large disk, and the small disk rotates freely about that. And as I noted in post #29, this means the rotation of the smaller disk is irrelevant and we can replace it with a point mass fixed to the larger disk at O'.
 
  • Like
Likes Lnewqban and vcsharp2003
  • #32
haruspex said:
If the large disk is initially set to rotate about O' then we can resolve that into a rotation about its centre plus a linear motion of the centre.
Is this fact always applicable when a disk rotates about a vertical axis perpendicular to the disk passing through a point other than it's center?
If yes, then this is pretty advanced stuff for me.

I know that a rolling object on a horizontal surface without slippage, can have its motion resolved into a translational motion and a pure rotation about it's center as you have mentioned.
 
  • #33
vcsharp2003 said:
Is this fact always applicable when a disk rotates about a vertical axis perpendicular to the disk passing through a point other than it's center?
It is always valid to represent a motion as a sum of motions, and it often helps to represent a rotation about a point other than the CoM as a rotation about the CoM plus a linear motion of the CoM.
 
  • Like
Likes Lnewqban and vcsharp2003
  • #34
haruspex said:
plus a linear motion of the CoM.
And if the bodies in interest are rigid, a linear motion of the CoM implies the same linear motion for every point of the rigid bodies.
 
  • Like
Likes Lnewqban and vcsharp2003
  • #35
haruspex said:
It is always valid to represent a motion as a sum of motions, and it often helps to represent a rotation about a point other than the CoM as a rotation about the CoM plus a linear motion of the CoM.
Ok, that's a very powerful truth. I was used to thinking in these terms only for a rolling object with no slippage on a horizontal surface since that's the scenario textbooks normally explain.
 

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
831
Replies
13
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
10
Replies
335
Views
9K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
Back
Top