I've heard it from two teachers and saw a task with a solution based on this assumption:

If the net torque is zero when measured from one frame of reference, it is equal to zero in all other frames of reference

This actually seems clear, especially if you put it to simple words like:

If a body doesn't rotate from one point of view, it won't start rotating if you look at it from another point of view

Still, I wanted to test it and made such an image: torques

I calculated the net torque measured from point A (I'll skip the designations such as $m_1, r_1, m_2$ etc. and get to the value):

$$1m\cdot1N + 5m\cdot1N - 4m\cdot1.5N = 0$$

Now, let's calculate the net torque from B:

$$-2m\cdot1N + 2m\cdot1N - 1m\cdot1.5N = -1.5Nm$$

From this it seems, that the body doesn't rotate around an axis in A, but does around the axis in B.

  • $\begingroup$ To avoid this kinds of question, engineers often decompose groups of forces into couples (which are translationally neutral but impart rotation) and purely translational forces. In doing so they would necessarily have asked if there was a reaction force at the pivot. $\endgroup$ Jan 15, 2014 at 0:54
  • 1
    $\begingroup$ Putting in numbers makes it much more difficult to understand than using symbols. $\endgroup$
    – Bernhard
    Jan 15, 2014 at 7:01

3 Answers 3


The sum of the torques is equal only when the sum of the forces is zero. In general the torque transfer from point A to point B is

$$\vec{M_B} = \vec{M_A} + \vec{r}_{AB} \times \vec{F} $$

you can see in your case the sum of the forces is $0.5N$, and the distance between A and B is $3m$ so the difference is $1.5 Nm$ which is what you get.

Change the downward force to $2N$ and try it again, only to see the sum of torques is the same regardless of point of consideration.

BTW: In dynamics, the sum of torque is always taken about the center of mass, in order to yield the correct angular acceleration, because the sum of forces is generally not zero.

  • $\begingroup$ Thank you, that's actually precisely the answer I was looking for, maybe I just didn't put it in the right words. This equation is also quite easy to proof and to memorize. It seems to me, that the teacher I'd asked maybe misinterpreted my words, yet the problem which caused me to think about this question still seems inacurrate to me. If you'd be so kind, would you look at this? The problem. The answer to a), b) and c) are "Yes". Is it somehow omitted, that the forces also balance ($F_{net}=0$) or am I still missing something? $\endgroup$
    – neverneve
    Jan 16, 2014 at 16:18
  • $\begingroup$ There is always a point where torques balance. This is the line of action of the force. Wherever this line is, the instance center of motion, is going to be on the other side of the center of mass. The line and center are called Pole & Polar $\endgroup$ Jan 16, 2014 at 17:16
  • $\begingroup$ If you know the vector sum of torques at a point P as $\vec{M}_P$ and the sum of vector forces $\vec{F}$ then the line of action of the forces is located at $$\vec{r} = \vec{r}_P - \frac{\vec{M}_P\times\vec{F}}{|\vec{F}|^2} $$ For the problem, it is ommited that $\vec{F}=0$ which causes the special case discussed above. $\endgroup$ Jan 16, 2014 at 17:20
  • $\begingroup$ I didn't get the BTW comment. What do you mean by "correct" angular acceleration? if I take torques around some different point in such a case, won't that angular acceleration be correct when measured from that point? $\endgroup$
    – Sal_99
    Aug 12, 2020 at 15:07
  • $\begingroup$ I you sum the torques about a point not the center of mass and use it to calculate angular acceleration, then you will get an incorrect answer. See why torque about the center of mass is important. $\endgroup$ Aug 12, 2020 at 22:58

I'm going to interpret your "frame of reference" as meaning "reference point." This is usually the main concern when dealing with rotating objects. If by frame of reference you really mean different inertial frames, then ignore me.

Short answer: "Rotating" doesn't necessarily mean non-zero angular momentum.

Long answer: See below.

When calculating torque, one must choose a reference point, as you've done. In general, you will get different values of torque when choosing different reference points. Your example is right on.

Here's how to resolve the rotating vs not rotating that you're concerned about. Torque is useful because it relates to angular momentum: $$\vec{\tau}_\text{net}=\frac{d \vec{L}}{dt}=\frac{d}{dt}\int\vec{r}\times\vec{v}\rho\ dV$$ As you can see from the rightmost expression, angular momentum is also reference point-dependent. It's value is different depending on where you choose your reference point to be. It is very easy to conflate a zero (non-zero) value of angular momentum with something spinning (not spinning). That is, just because something has a zero value of angular momentum, that doesn't mean it's not spinning. I suppose this association is so strong because one usually measures the angular momentum (and torque), for simple systems, about the rotation axis. But this doesn't have to be done.

Here's a quick example of what I mean. A spool of mass $m$ is wrapped many times with a string of negligible mass. If you pull the string as shown in the leftmost picture, the spool will translate downward and will also rotate about the center of mass. In this example, you can choose to measure angular momentum (and torque) about the center of mass of the spool. (center picture). By doing so you'll find a non-zero rate of change of angular momentum and a non-zero torque. If instead you measure it along the line of the string (rightmost picuter), you'll find zero torque and zero constant angular momentum, even though it is rotating.

enter image description here

  • $\begingroup$ So what does this mean , the body is rotating . What does it signify Having 0 Angular Momentum about the the line of string $\endgroup$
    – Shashaank
    Dec 24, 2016 at 16:17
  • $\begingroup$ @Shashaank I think it means that body is not rotating about line of string $\endgroup$ Oct 7, 2020 at 10:28

You are confusing the words "frame of reference", "point of viewing" and "point of application":

If a body doesn't rotate from one point of view, it won't start rotating if you look at it from another point of view

This is valid only if both points of viewing are in same state, i.e. at rest, or both moving with same velocity or with same acceleration.

Next, when you are doing the experiment, what you have done is calculated torque about 2 different points i.e point A and point B. In this case, you are not seeing a constant torque from 2 point A and B, what you have done is calculated moment of the different forces about different points which is why they are different in magnitude.

To what you heard your teachers say

If the net torque is zero when measured from one frame of reference, it is equal to zero in all other frames of reference

This is inaccurate, suppose a horizontal wheel, if you just stand and see this wheel, it's angular momentum is not changing hence no torque, but if you start running about it with more and more speed, it would seem that the wheel's angular momentum is changing hence in this frame you get a torque!

  • $\begingroup$ "it would seem that the wheel's angular momentum is changing hence in this frame you get a torque!" Which frame are you talking about? $\endgroup$ Oct 7, 2020 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.