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Momentum and torque about different points 
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#1
Jun1114, 08:52 AM

PF Gold
P: 101

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? 


#2
Jun1114, 09:04 AM

P: 110

There is no paradox. Torque about origin is nonzero and about the center is zero.



#3
Jun1114, 09:19 AM

PF Gold
P: 101

will there be any difference if the angular momentum is taken about a point that is not on the rotation axis?



#4
Jun1114, 09:25 AM

P: 110

Momentum and torque about different points
(Notice the use of 'can' over 'will' there can be two different points which can yield the same magnitude and direction of those quantities). 


#5
Jun1114, 09:33 AM

PF Gold
P: 101

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.



#6
Jun1114, 09:36 AM

Mentor
P: 3,967




#7
Jun1114, 10:30 AM

PF Gold
P: 101

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?



#8
Jun1114, 11:51 AM

P: 127

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


#9
Jun1114, 12:25 PM

Thanks
P: 1,948

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.



#10
Jun1114, 12:42 PM

P: 110




#11
Jun1114, 09:24 PM

PF Gold
P: 101

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. 


#12
Jun1114, 09:55 PM

P: 999




#13
Jun1114, 11:46 PM

PF Gold
P: 101

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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