In any situation, while applying torque upon any rigid body, can the reference point/axis be any arbitrary point/axis?

I mean, is it necessary for the reference axis to be taken only through the centre of mass?

Also should the reference axis be stationary? That is, can we take a moving or accelerating axis as reference axis? If yes, then do we need to apply some pseudo forces somewhere to evaluate free body diagram?

Related question:
Why we always apply torque choosing reference axis about O or P? Why not about some other point?enter image description here

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    $\begingroup$ -1. Not clear what you are asking. Please provide more details to explain your difficulty. $\endgroup$ – sammy gerbil Apr 24 '17 at 11:31
  • $\begingroup$ I've edited my question to be more clear to answer. @JohnRennie $\endgroup$ – alto santa Apr 28 '17 at 19:22
  • $\begingroup$ I had to dig deep to pin point exactly where i needed help. It was difficult for me 'coz i was very unclear on this topic- i'm just a beginner. Thanks SE for waiting for me to improve my question. Thanks @sammygerbil for guidance. $\endgroup$ – alto santa Apr 28 '17 at 19:24
  • $\begingroup$ Do you have a particular problem in mind? If so, it would be helpful for you to post the problem and ask your questions about that problem. $\endgroup$ – sammy gerbil Apr 28 '17 at 19:28
  • $\begingroup$ While solving many problems[1] , i was told to take reference axis about only centre of mass. All the solutions did the same. i wondered why there is such restriction on choosing reference axis. so basically my question about isn't related to one particular problem but it is related to little fundametals in solving a problem. That is about choosing reference axis(like accelerating axis ) and where to put in pseudo forces (or torque) if such axis is chosen. @sammygerbil. $\endgroup$ – alto santa Apr 28 '17 at 19:38

The restrictions on the useful set of refernce points that we can use for rotation dynamics comes from the following algebra.

Let $$ {\bf L}_{\bf R}= \sum ({\bf r}_i-{\bf R}(t))\times m_i\dot {\bf r}_i $$ be the total angular momentum of the collection of mass $m_i$ particles about the point ${\bf R}(t)$. Then $$ \frac{d}{dt} {\bf L}_{\bf R}= \sum ({\bf r}_i-{\bf R})\times m_i\ddot {\bf r}_i+\sum (\dot {\bf r}_i-\dot {\bf R})\times m_i\dot {\bf r}\\ = \sum ({\bf r}_i-{\bf R})\times {\bf F}_i -\dot {\bf R}\times \sum m_i \dot {\bf r}_i\\ = {\bf T} - \dot {\bf R}\times M{\bf V} $$ where ${\bf T}$ is total torque about ${\bf R}$, $M=\sum m_i$ and ${\bf V}$ is the velocity of the center of mass. We see that if we want the torque about ${\bf R}$ to equal the rate of change of angular momentum about ${\bf R}$ we must have $\dot{\bf R}\times {\bf V}=0$. This will be true if any of the following conditions hold:

  1. ${\bf R}$ is the center of mass, so $\dot {\bf R}= {\bf V}$.
  2. $\dot {\bf R}=0$, so ${\bf R}$ is a point that is (possibly momentarily) at rest. This case is useful for rolling problems where the point in contact with the ground is momentarily stationary.
  3. ${\bf R}$ is moving parallel to of the center of mass.

The last condition is also useful for rolling problems, as we can take ${\bf R}(t)$ to be the time dependent sequence of points where the wheel touches the ground, rather than the fixed point that just happens to be the place where the wheel touches the ground at some particular time.

  • $\begingroup$ Your solution exactly answers my question. I was asking the 'why' on these 'restrictions'. And you nailed it with your accurate answer. Thanks a lot! @mikestone $\endgroup$ – alto santa Aug 25 '18 at 20:32

It is not exactly necessary to take the reference axis to pass through CM. This is done to lower the mathematical complications. But if we take some other point all the data has to referred with respect to that point.

  • the torque is to be taken $wrt$ that reference axis.

  • also important to consider the moment of inertia about that axis using parallel axis theorem if needed.

  • angular momentum is also to be taken $wrt$ that point Hence it is advisable to choose an easy axis for mathematical calculations.

  • all other data on rotation is to be taken $wrt$ that point

Hence it is advisable to chose an easy axis for mathematical calculations.


It is not clear exactly what you are asking. Maybe you have a specific sitation in mind and you need to add more details to the question.

In general here are some rules about free rigid body motion. In the absense of any external force and torque a rigid body maintains constant linear and angular momentum

  • The CM moves with constant velocity (or zero).
  • And the body rotates about the CM with varying rotational velocity such that the tensor-vector product $ {\rm I}\, {\boldsymbol \omega}$ is constant. If the rotation axis is along a principal inertia axis then the rotational velocity vector ${\boldsymbol \omega}$ remain constant.
  • The combined motion is instaneneous rotation about an axis on or away from the CM combined with a parallel translation. This is called screw motion.

Even when external loads are applied, the general case is still a screw motion (at every instant).

  • The axis of the screw is through the CM if a pure torque is applied.
  • A non-zero force away from the CM (or force thorugh CM combined with a torque) causes the screw axis to be away from the center of mass.
  • Finally a force though the center of mass with zero net torque causes the screw axis to be at infinity (pure translation).
  • The direction of the screw axis is always the direction of ${\boldsymbol \omega}$, and the magnitude equal to $\omega = \| {\boldsymbol \omega} \|$.
  • The parallel motion of along the screw axis has speed $u = \frac{{\boldsymbol \omega} \cdot {\bf v_{CM}}}{\| {\boldsymbol \omega} \|} $
  • $\begingroup$ I have now updated my question to be more clear to answer. @ja72 still thanks for answering to the previous question. $\endgroup$ – alto santa Apr 28 '17 at 19:28

The first question ... u need not take reference point as the CM always. Generally people use CM as reference because it is easy to evaluate answer when it is a symmetric figure. There are many cases where reference point is taken outside the body also.You can see this case while solving ladder problems when friction comes into picture.
Second question... yep the reference axis should stationary


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