Is Angular momentum conserved in Impure rolling

In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $romega$, is angular momentum conserved about a point on the ground.

What confuses me is that friction decelerates the disk to make $v_com = romega$ and in this process some velocity is lost so according to formula of angular momentum $L=mvr$,about a point on the ground $L$ decreases as $v$ decreases.

But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.

edit:this question is about applying conservation of angular momentum in rolling with slipping.

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

1

$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
Mar 12 at 8:08

$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
Mar 12 at 14:46

add a comment |

In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $romega$, is angular momentum conserved about a point on the ground.

But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.

edit:this question is about applying conservation of angular momentum in rolling with slipping.

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

1

$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
Mar 12 at 8:08

$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
Mar 12 at 14:46

add a comment |

In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $romega$, is angular momentum conserved about a point on the ground.

But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.

edit:this question is about applying conservation of angular momentum in rolling with slipping.

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $romega$, is angular momentum conserved about a point on the ground.

But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.

edit:this question is about applying conservation of angular momentum in rolling with slipping.

newtonian-mechanics angular-momentum rotational-dynamics

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

edited Mar 12 at 11:45

Qmechanic♦

107k121991240

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

asked Mar 12 at 7:52

Lelouche Lamperouge

1125

1

$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
Mar 12 at 8:08

$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
Mar 12 at 14:46

add a comment |

1

$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
Mar 12 at 8:08

$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
Mar 12 at 14:46

Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?

– Mick
Mar 12 at 8:08

Only if the sum of the external torque is zero the angular momentum is conserved

– Eli
Mar 12 at 14:46

add a comment |

2 Answers
2

active

oldest

votes

It is easy to show that total angular momentum is conserved in this case. The case is for slipping where the friction force is retarding the translational velocity of the disk while increasing the rotation of the disk.

For disk of radius $R$, mass $m$, moment of inertia $I$ and velocity $v$, friction force $f$.

The angular momentum about a point on the ground is:

$L1=m v R$

The rate of change due to the friction force:

$fracdL1dt=-mfracdvdtR=-f R$

since the friction force $f$ is negative to the velocity.

Additionally the disk's spin angular momentum is increasing due to the torque created by $f$.
The spin angular momentum:

$L2=I omega$

and the rate of change:

$fracdL2dt=Ifrac d omega dt=tau =f R$

where $tau$ is the torque due to the friction force.

So we end up with

$fracdL1dt+fracdL2dt=0$

and the total angular momentum is constant.

Treating the whole system this way makes the friction force an internal force rather than an external one.

answered Mar 20 at 6:46

Bill Watts

39217

$begingroup$
How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too.
$endgroup$
– Vishal Jain
Mar 20 at 8:11

$begingroup$
So the answer by @Farcher is wrong?
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:49

$begingroup$
@BillWatts What is your “whole system”?
$endgroup$
– Farcher
Mar 20 at 17:20

$begingroup$
For pure rolling $omega$ and $v$ are constant. In this case, $v$ is decreasing and $omega$ is increasing due to slipping. By the whole system I mean the ground and the slipping disk.
$endgroup$
– Bill Watts
Mar 20 at 17:31

$begingroup$
BillWatts Is the pseudo force concept by @Farcher wrong? Please confirm, i dont know who is right.
$endgroup$
– Lelouche Lamperouge
Mar 21 at 6:45

|
show 4 more comments

If there is slipping and friction you now have to deal with an accelerating (non-inertial) frame of reference and axis of rotation.

To use Newton's laws of motion in the non-inertial frame of reference a pseudo-force has to be introduced which acts at the centre of mass, has a magnitude equal to the frictional force and acts in the opposite direction to the frictional force.

That pseudo-force produces a torque about the point of contact with the ground which changes the angular momentum of the disc about that point.

answered Mar 12 at 10:10

Farcher

52k340109

$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
Mar 12 at 10:17

$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
Mar 12 at 13:51

$begingroup$
@Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work.
$endgroup$
– Farcher
Mar 12 at 13:57

$begingroup$
@Farcher please review the 2nd answer. It seems to be opposite to your answer.
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:55

$begingroup$
@LeloucheLamperouge I answer your question But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change. You subsequently added an edit which has been answered by BillWatts.
$endgroup$
– Farcher
Mar 21 at 6:59

|
show 2 more comments

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2 Answers
2

active

oldest

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2 Answers
2

active

oldest

votes

For disk of radius $R$, mass $m$, moment of inertia $I$ and velocity $v$, friction force $f$.

The angular momentum about a point on the ground is:

$L1=m v R$

The rate of change due to the friction force:

$fracdL1dt=-mfracdvdtR=-f R$

since the friction force $f$ is negative to the velocity.

Additionally the disk's spin angular momentum is increasing due to the torque created by $f$.
The spin angular momentum:

$L2=I omega$

and the rate of change:

$fracdL2dt=Ifrac d omega dt=tau =f R$

where $tau$ is the torque due to the friction force.

So we end up with

$fracdL1dt+fracdL2dt=0$

and the total angular momentum is constant.

Treating the whole system this way makes the friction force an internal force rather than an external one.

answered Mar 20 at 6:46

Bill Watts

39217

$begingroup$
How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too.
$endgroup$
– Vishal Jain
Mar 20 at 8:11

$begingroup$
So the answer by @Farcher is wrong?
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:49

$begingroup$
@BillWatts What is your “whole system”?
$endgroup$
– Farcher
Mar 20 at 17:20

$begingroup$
For pure rolling $omega$ and $v$ are constant. In this case, $v$ is decreasing and $omega$ is increasing due to slipping. By the whole system I mean the ground and the slipping disk.
$endgroup$
– Bill Watts
Mar 20 at 17:31

$begingroup$
BillWatts Is the pseudo force concept by @Farcher wrong? Please confirm, i dont know who is right.
$endgroup$
– Lelouche Lamperouge
Mar 21 at 6:45

|
show 4 more comments

For disk of radius $R$, mass $m$, moment of inertia $I$ and velocity $v$, friction force $f$.

The angular momentum about a point on the ground is:

$L1=m v R$

The rate of change due to the friction force:

$fracdL1dt=-mfracdvdtR=-f R$

since the friction force $f$ is negative to the velocity.

Additionally the disk's spin angular momentum is increasing due to the torque created by $f$.
The spin angular momentum:

$L2=I omega$

and the rate of change:

$fracdL2dt=Ifrac d omega dt=tau =f R$

where $tau$ is the torque due to the friction force.

So we end up with

$fracdL1dt+fracdL2dt=0$

and the total angular momentum is constant.

Treating the whole system this way makes the friction force an internal force rather than an external one.

answered Mar 20 at 6:46

Bill Watts

39217

$begingroup$
How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too.
$endgroup$
– Vishal Jain
Mar 20 at 8:11

$begingroup$
So the answer by @Farcher is wrong?
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:49

$begingroup$
@BillWatts What is your “whole system”?
$endgroup$
– Farcher
Mar 20 at 17:20

$begingroup$
For pure rolling $omega$ and $v$ are constant. In this case, $v$ is decreasing and $omega$ is increasing due to slipping. By the whole system I mean the ground and the slipping disk.
$endgroup$
– Bill Watts
Mar 20 at 17:31

$begingroup$
BillWatts Is the pseudo force concept by @Farcher wrong? Please confirm, i dont know who is right.
$endgroup$
– Lelouche Lamperouge
Mar 21 at 6:45

|
show 4 more comments

For disk of radius $R$, mass $m$, moment of inertia $I$ and velocity $v$, friction force $f$.

The angular momentum about a point on the ground is:

$L1=m v R$

The rate of change due to the friction force:

$fracdL1dt=-mfracdvdtR=-f R$

since the friction force $f$ is negative to the velocity.

Additionally the disk's spin angular momentum is increasing due to the torque created by $f$.
The spin angular momentum:

$L2=I omega$

and the rate of change:

$fracdL2dt=Ifrac d omega dt=tau =f R$

where $tau$ is the torque due to the friction force.

So we end up with

$fracdL1dt+fracdL2dt=0$

and the total angular momentum is constant.

Treating the whole system this way makes the friction force an internal force rather than an external one.

answered Mar 20 at 6:46

Bill Watts

39217

For disk of radius $R$, mass $m$, moment of inertia $I$ and velocity $v$, friction force $f$.

The angular momentum about a point on the ground is:

$L1=m v R$

The rate of change due to the friction force:

$fracdL1dt=-mfracdvdtR=-f R$

since the friction force $f$ is negative to the velocity.

Additionally the disk's spin angular momentum is increasing due to the torque created by $f$.
The spin angular momentum:

$L2=I omega$

and the rate of change:

$fracdL2dt=Ifrac d omega dt=tau =f R$

where $tau$ is the torque due to the friction force.

So we end up with

$fracdL1dt+fracdL2dt=0$

and the total angular momentum is constant.

Treating the whole system this way makes the friction force an internal force rather than an external one.

answered Mar 20 at 6:46

Bill Watts

39217

answered Mar 20 at 6:46

Bill Watts

39217

answered Mar 20 at 6:46

Bill Watts

39217

answered Mar 20 at 6:46

Bill Watts

39217

$begingroup$
How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too.
$endgroup$
– Vishal Jain
Mar 20 at 8:11

$begingroup$
So the answer by @Farcher is wrong?
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:49

$begingroup$
@BillWatts What is your “whole system”?
$endgroup$
– Farcher
Mar 20 at 17:20

$begingroup$
For pure rolling $omega$ and $v$ are constant. In this case, $v$ is decreasing and $omega$ is increasing due to slipping. By the whole system I mean the ground and the slipping disk.
$endgroup$
– Bill Watts
Mar 20 at 17:31

$begingroup$
BillWatts Is the pseudo force concept by @Farcher wrong? Please confirm, i dont know who is right.
$endgroup$
– Lelouche Lamperouge
Mar 21 at 6:45

|
show 4 more comments

$begingroup$
How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too.
$endgroup$
– Vishal Jain
Mar 20 at 8:11

$begingroup$
So the answer by @Farcher is wrong?
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:49

$begingroup$
@BillWatts What is your “whole system”?
$endgroup$
– Farcher
Mar 20 at 17:20

$begingroup$
For pure rolling $omega$ and $v$ are constant. In this case, $v$ is decreasing and $omega$ is increasing due to slipping. By the whole system I mean the ground and the slipping disk.
$endgroup$
– Bill Watts
Mar 20 at 17:31

$begingroup$
BillWatts Is the pseudo force concept by @Farcher wrong? Please confirm, i dont know who is right.
$endgroup$
– Lelouche Lamperouge
Mar 21 at 6:45

How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too.

– Vishal Jain
Mar 20 at 8:11

So the answer by @Farcher is wrong?

– Lelouche Lamperouge
Mar 20 at 14:49

@BillWatts What is your “whole system”?

– Farcher
Mar 20 at 17:20

For pure rolling $omega$ and $v$ are constant. In this case, $v$ is decreasing and $omega$ is increasing due to slipping. By the whole system I mean the ground and the slipping disk.

– Bill Watts
Mar 20 at 17:31

BillWatts Is the pseudo force concept by @Farcher wrong? Please confirm, i dont know who is right.

– Lelouche Lamperouge
Mar 21 at 6:45

|
show 4 more comments

answered Mar 12 at 10:10

Farcher

52k340109

$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
Mar 12 at 10:17

$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
Mar 12 at 13:51

$begingroup$
@Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work.
$endgroup$
– Farcher
Mar 12 at 13:57

$begingroup$
@Farcher please review the 2nd answer. It seems to be opposite to your answer.
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:55

$begingroup$
@LeloucheLamperouge I answer your question But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change. You subsequently added an edit which has been answered by BillWatts.
$endgroup$
– Farcher
Mar 21 at 6:59

|
show 2 more comments

answered Mar 12 at 10:10

Farcher

52k340109

$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
Mar 12 at 10:17

$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
Mar 12 at 13:51

$begingroup$
@Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work.
$endgroup$
– Farcher
Mar 12 at 13:57

$begingroup$
@Farcher please review the 2nd answer. It seems to be opposite to your answer.
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:55

$begingroup$
@LeloucheLamperouge I answer your question But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change. You subsequently added an edit which has been answered by BillWatts.
$endgroup$
– Farcher
Mar 21 at 6:59

|
show 2 more comments

answered Mar 12 at 10:10

Farcher

52k340109

answered Mar 12 at 10:10

Farcher

52k340109

answered Mar 12 at 10:10

Farcher

52k340109

answered Mar 12 at 10:10

Farcher

52k340109

answered Mar 12 at 10:10

Farcher

52k340109

$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
Mar 12 at 10:17

$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
Mar 12 at 13:51

$begingroup$
@Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work.
$endgroup$
– Farcher
Mar 12 at 13:57

$begingroup$
@Farcher please review the 2nd answer. It seems to be opposite to your answer.
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:55

$begingroup$
@LeloucheLamperouge I answer your question But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change. You subsequently added an edit which has been answered by BillWatts.
$endgroup$
– Farcher
Mar 21 at 6:59

|
show 2 more comments

$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
Mar 12 at 10:17

$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
Mar 12 at 13:51

$begingroup$
@Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work.
$endgroup$
– Farcher
Mar 12 at 13:57

$begingroup$
@Farcher please review the 2nd answer. It seems to be opposite to your answer.
$endgroup$
– Lelouche Lamperouge
Mar 20 at 14:55

$begingroup$
@LeloucheLamperouge I answer your question But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change. You subsequently added an edit which has been answered by BillWatts.
$endgroup$
– Farcher
Mar 21 at 6:59

ohh that makes sense. thanks a lot :)

– Lelouche Lamperouge
Mar 12 at 10:17

Where this pseudo-force comes from?

– Eli
Mar 12 at 13:51

@Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work.

– Farcher
Mar 12 at 13:57

@Farcher please review the 2nd answer. It seems to be opposite to your answer.

– Lelouche Lamperouge
Mar 20 at 14:55

@LeloucheLamperouge I answer your question But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change. You subsequently added an edit which has been answered by BillWatts.

– Farcher
Mar 21 at 6:59

|
show 2 more comments

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