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Consider the image below where we have two point masses $m_1$ and $m_2$ with different masses which are rotating around a fixed axis with angular velocity $\omega$. If the origin is placed on the axis between the masses (left image), then angular momentum vectors $L = r\times p$ are parallel with the axis and does not change upon rotation. Hence torque is zero. If we move the origin away from the line between the masses, the angular momentum of each particle does not lie along the rotation axis and the total momentum will neither lie along the rotation axis. Therefor there must be a torque present to change the angular momentum during rotation. So in one case there is a torque and in the other there isn't???

Of course the two systems are physically the same and the same forces must be acting so there must be an error in the reasoning. Where is it?

enter image description here

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    $\begingroup$ The torque $\tau = \frac{\mathrm{d} \mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d} \mathbf{r}}{\mathrm{d}t} \times \mathbf{p} + \mathbf{r} \times \frac{\mathrm{d} \mathbf{p}}{\mathrm{d}t} = \mathbf{v} \times \mathbf{p} + \mathbf{r}\times\mathbf{F} = \mathbf{r}\times\mathbf{F}$. Clearly this is not independent of your choice of axis $\endgroup$
    – By Symmetry
    Commented Dec 21, 2015 at 10:03
  • $\begingroup$ Clearly, but that does not make it less confusing: depending on what I choose as the origin there will or will not be a torque (which will or will not wear down the bearings on the rotation axle - that does not make sense) $\endgroup$
    – DiggingDeep
    Commented Dec 21, 2015 at 11:33
  • $\begingroup$ You are calculating $L$ in one co-ordinate system (@2nd figure) and calculating torque due to this for a different co-ordinate system (1st figure), I don't think this is right $\endgroup$
    – Courage
    Commented Dec 21, 2015 at 11:36
  • $\begingroup$ No, I am not calculating the torque, I am just saying that the torque is equal to change in momentum $$\tau =\frac{dL/dt}$$. In the first case it is zero. In the second case the angular momentum vector is moving, hence there must be a torque (in that co-ordinate system) $\endgroup$
    – DiggingDeep
    Commented Dec 21, 2015 at 11:59
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    $\begingroup$ Could you expand on exactly you think torque being dependant on coordinates is a problem? You say something about torque wearing down bearings, but if you actually think about what happens to the bearings in more detail it is the force on the bearing, not the torque that will directly determine the wear. It may, however, be natural to view that force as being determined by the torque about some particular axis, which make the problem look simpler. $\endgroup$
    – By Symmetry
    Commented Dec 21, 2015 at 14:10

4 Answers 4

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You are correct and there is nothing wrong with it. In many systems there is a "special" choice for the axis of rotation where you can make either $L$ or $\tau$ equal to zero, by choosing $r⃗$ to be parallel to the velocity or the force respectively. Another example: a mass moving along a straight line experiencing a force along the line. If you choose your origin to be on the line, both $L$ and $\tau$ will be zero, which will not be true if you choose an axis not along the line.

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Look at the comment by "By Symmetry"

The calculation is correct but the conclusion is wrong.

We are looking at an isolated system : if the two particles "are rotating around a fixed axis with angular velocity $\omega$ it means there are no other forces acting on them besides the forces they exert on each other.

Now for each one of these particles, the torque does indeed depend on the point you compute it, since it is indeed $r\times F$.

But consider the system formed by the two particles together. The force on one particle due to the other one is exactly the opposite of the force on the second one by the first one. The sum of these forces is zero, and therefore the total torque on the system of the two particles is the same whatever point you compute it ! So if it is zero somewhere, it is zero everywhere.

That it is more complicated to compute the torque on each particle does not change the absolute fact that it will perforce be the same everywhere !

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What you seem to be overlooking is the force imposed by the fixed axle, which acts in the direction of the lighter particle. When the origin is taken as the point at which the two particles are pivoting around the axle- as in your left hand figure- the force does not result in a torque. However, when the origin is taken elsewhere- as in your right hand figure- the force does result in a torque.

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In order for the first system to rotate about $O$, a fixed point located along the line connecting $m_1$ and $m_2$ either

  • $O$ must be at the center of mass of the system, or
  • there is something (a force) constraining $O$ in the plane of rotation, acting in the plane

When you move the point of calculation for $L$ (let's call it $O'$) to a point along the axis of rotation (collinear with $\vec{\omega}$) at some distance $z$ from the plane of rotation. Either

  • the direction of the angular momentum does not change if line $O'\to O$ passes through the center of mass, or
  • the constraining force produces a torque about $O'$ which causes the total angular momentum about $O'$ to have a constant direction. Without that torque, the angular momentum of the two masses would precess.

Another way to consider the system is that $\vec{L}=\mathscr{I}\omega$, which means that the angular momentum must be in the same direction as the angular velocity. If the angular velocity never changes direction, neither can the angular momentum.

Bottom Line: If there is a constraint in the rotation plane, there will be a torque about $O'$. If there is no constraint, then $O$ is at the center of mass.

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