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Momentum and torque about different points

by kelvin490
Tags: momentum, points, torque
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kelvin490
#1
Jun11-14, 08:52 AM
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Suppose there is a point mass performing circular motion about an axis shown below:



If the angular momentum is taken about the origin O, then the angular momentum L is not parallel to the axis of rotation. As the mass is moving in the circular orbit a torque is needed because L is keep on changing direction.

However, if the angular momentum is taken about the center of circular motion, then L is always parallel to the axis of rotation and no torque is needed to maintain the circular motion.

How to solve this paradox?
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Adithyan
#2
Jun11-14, 09:04 AM
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There is no paradox. Torque about origin is non-zero and about the center is zero.
kelvin490
#3
Jun11-14, 09:19 AM
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will there be any difference if the angular momentum is taken about a point that is not on the rotation axis?

Adithyan
#4
Jun11-14, 09:25 AM
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Momentum and torque about different points

Quote Quote by kelvin490 View Post
will there be any difference if the angular momentum is taken about a point that is not on the rotation axis?
Quantities like angular momentum, Torque, Angular velocity etc always depend on the axis/ point about which they are measured (since they are cross products with radius). Different points of consideration can yield different magnitudes/direction.

(Notice the use of 'can' over 'will'- there can be two different points which can yield the same magnitude and direction of those quantities).
kelvin490
#5
Jun11-14, 09:33 AM
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But if we take arbitrary points to calculate angular momentum, Torque, Angular velocity, will these quantities varies accordingly so that we always obtain the same resultant motion? Because I think there could only be one results for one set of input parameters.
Nugatory
#6
Jun11-14, 09:36 AM
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Quote Quote by kelvin490 View Post
But if we take arbitrary points to calculate angular momentum, Torque, Angular velocity, will these quantities varies accordingly so that we always obtain the same resultant motion? Because I think there could only be one results for one set of input parameters.
Try a few examples and see if you can make it come out differently...
kelvin490
#7
Jun11-14, 10:30 AM
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If the centripetal force that maintains this circular motion is suddenly disappeared, the mass will move linearly. The mass will rotate about its center of mass so that the angular momentum about the previous rotation axis is conserved. What is the mechanism that initiate this rotation?
mattt
#8
Jun11-14, 11:51 AM
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Quote Quote by kelvin490 View Post
If the centripetal force that maintains this circular motion is suddenly disappeared, the mass will move linearly. The mass will rotate about its center of mass
You said it was a point mass. Anyway, the torque of the centripetal force with respect to the center of the circle is zero, so the angular momentum vector (with respect to the center of the circle) remains constant, during the period of time that the centripetal force is acting, and also after it.

so that the angular momentum about the previous rotation axis is conserved. What is the mechanism that initiate this rotation?
Do you mean the rotation (spinning) of a rigid body?

If a rigid body is spinning (let us say at constant angular velocity) and its center of mass is moving with uniform circular motion, then the total angular momentum vector of your system (with respect to the point O = center of the circle) is:

[tex]\vec{L}_O=\sum_i \vec{r}_i\times\vec{p}_i = \sum_i (\vec{r}_{O-CM}+\vec{r'}_i)\times\vec{p}_i=\sum_i (\vec{r}_{O-CM}+\vec{r'}_i)\times(m_i \frac{d \vec{r}_{O-CM}}{dt} + m_i \frac{d \vec{r'}_i}{dt}) = [/tex]


[tex]=\vec{r}_{O-CM}\times M\vec{v}_{O-CM} + \vec{r}_{O-CM}\times(\sum_i\vec{p'}_i) + \sum_i(m_i \vec{r'}_i)\times\frac{d \vec{r}_{O-CM}}{dt} + \sum_i \vec{r'}_i\times\vec{p'}_i =[/tex]


[tex]=\vec{r}_{O-CM}\times M\vec{v}_{O-CM}+ \sum_i \vec{r'}_i\times\vec{p'}_i [/tex]


(given that [tex]\sum_i\vec{p'}_i = \vec{0}[/tex] and [tex]\sum_i(m_i \vec{r'}_i)=\vec{0}[/tex] )

In this situation, [tex]\vec{L}_O[/tex] does not vary with time, because the net exterior force is centripetal and its torque with respect to the point O is zero (we assume the internal forces are "newtonian" and "central" and the exterior force is applied at the center of mass).

If the total exterior force (centripetal in this example) suddenly dissapear, the torque of the total exterior force with respect to the point O is again zero (because now total exterior force is zero), so [tex]\vec{L}_O[/tex] remains constant.

If you mean why the rigid body is spinning, or when did it start spinning.....well, at some point in the past a force couple or pure moment was applied to the body (for example), so it remains spinning at the same constant angular velocity (wrt its center of mass) regardless of the centripetal force acting on its center of mass, until we apply another moment to change its spin.
dauto
#9
Jun11-14, 12:25 PM
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No, the mass won't rotate. The angular momentum will be preset without the mass rotation because the angular momentum is given by r cross p which won't be zero. A mass moving in a straight line has angular momentum when measure around a point that is not in the line which is the case here.
Adithyan
#10
Jun11-14, 12:42 PM
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Quote Quote by dauto View Post
No, the mass won't rotate. The angular momentum will be preset without the mass rotation because the angular momentum is given by r cross p which won't be zero. A mass moving in a straight line has angular momentum when measure around a point that is not in the line which is the case here.
I agree with him. Even if we assume the mass is not a point mass, the mass had a resultant angular momentum 'L' which is similar to a condition where the mass has been replaced by a point mass rotating with 'L'. So as soon as the centripetal force disappears, different particle will develop different angular momenta whose resultant is still "L".
kelvin490
#11
Jun11-14, 09:24 PM
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Sorry for the confusion about whether the mass is a point mass. Let's assume the mass is a solid sphere.

When the mass is performing circular motion about center O. The sphere is also spinning about its center of mass because it's always the same point that facing the center O. If the centripetal force disappears the sphere should keep on spinning while its center of mass moves linearly with constant velocity. How to show the angular momentum is conserved? Should it spin faster or slower?

What if the mass is a point mass? It seems that a point should not be able to spin because it is one dimension and you cannot distinguish different face of the point. In kinetic theory, monoatomic ideal gas is modeled as points and thus no rotational KE is possible.
jbriggs444
#12
Jun11-14, 09:55 PM
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Quote Quote by kelvin490 View Post
When the mass is performing circular motion about center O. The sphere is also spinning about its center of mass because it's always the same point that facing the center O. If the centripetal force disappears the sphere should keep on spinning while its center of mass moves linearly with constant velocity. How to show the angular momentum is conserved? Should it spin faster or slower?
Re-read post 9. Just because it is no longer moving in a circle does not mean that it stops having angular momentum.
kelvin490
#13
Jun11-14, 11:46 PM
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Thanks all. It's clear now. Since the perpendicular distance between the original center of circular motion and the vector V is always the same, mrxv is kept constant while the mass moves linearly after the centripetal force is removed.

For a point mass, to conserve angular momentum it doesn't spin (and it cannot) and simply move forward.

For a sphere that always has the same point facing the center O during circular motion, after the centripetal force is removed it spins with same self spinning angular speed as before. It also has linear motion which its center of mass moves like the case of point mass.


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