Is Angular momentum conserved in Impure rolling
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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.
newtonian-mechanics angular-momentum rotational-dynamics
New contributor
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add a comment |
$begingroup$
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.
newtonian-mechanics angular-momentum rotational-dynamics
New contributor
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1
$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
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– Mick
14 hours ago
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Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
8 hours ago
add a comment |
$begingroup$
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.
newtonian-mechanics angular-momentum rotational-dynamics
New contributor
$endgroup$
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.
newtonian-mechanics angular-momentum rotational-dynamics
newtonian-mechanics angular-momentum rotational-dynamics
New contributor
New contributor
edited 11 hours ago
Qmechanic♦
106k121941217
106k121941217
New contributor
asked 15 hours ago
Lelouche LamperougeLelouche Lamperouge
384
384
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1
$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
14 hours ago
$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
8 hours ago
add a comment |
1
$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
14 hours ago
$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
8 hours ago
1
1
$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
14 hours ago
$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
14 hours ago
$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
8 hours ago
$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
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$begingroup$
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.
$endgroup$
$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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 12 hours ago
FarcherFarcher
50.7k338105
50.7k338105
$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
ohh that makes sense. thanks a lot :)
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
Where this pseudo-force comes from?
$endgroup$
– Eli
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
Lelouche Lamperouge is a new contributor. Be nice, and check out our Code of Conduct.
Lelouche Lamperouge is a new contributor. Be nice, and check out our Code of Conduct.
Lelouche Lamperouge is a new contributor. Be nice, and check out our Code of Conduct.
Lelouche Lamperouge is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Possible duplicate of Is $v$ not always equal to $omega r$ in angular motion?
$endgroup$
– Mick
14 hours ago
$begingroup$
Only if the sum of the external torque is zero the angular momentum is conserved
$endgroup$
– Eli
8 hours ago