Before I ask this question, let me describe the set-up. You have a simple pendulum attached to the ceiling. At the bottom of the pendulum there is a sphere. The sphere rotates counterclockwise when going to the right, and then rotates clockwise while going to the left.

Rotating ball on a pendulum

The question was to find what direction the net torque is on the ball with respect to to point of attachment between the sphere and the ball when the pendulum is at its max rightward displacement. Now, this is what confused me. When I learned torque, I learned that measuring torque is only meaningful (ex. you can apply $τ = Iα$ and $τ = \mathrm dL/\mathrm dt$) when you measure torque about the center of mass or a point that is fixed (not moving or moving at constant velocity).

The point of attachment is essentially in free-fall at the max rightward displacement (and is hence accelerating) and is not the center-of-mass of the ball, so from what I learned, this is an "invalid" point of taking torque (ex. you can't make assumptions on the motion of the ball from taking torque about this point). Instead of taking torque about the point of attachment, I took torque about the center point when the ball was at its max rightward displacement. From this perspective, the ball rotates from a counterclockwise direction to a clockwise direction. Therefore, from the point of the center-of-mass of the ball, the torque must be clockwise.

That was my answer, but as I said the question asked what was the torque relative to the point of attachment. The answer was

At the sphere’s maximum rightward displacement, the gravitational force (taken to act at the center of the sphere) exerts a clockwise torque about the point of attachment to the string.

I understand everything except how they took torque about the point of attachment. From my understanding, taking torque about these "non-valid" origins can lead to wrong conclusions. For example, look at this diagram of a block in space. If you take torque about point a, one will find torque to be zero and hence conclude that the object is not spinning. However, taking torque about the center of mass reveals there is a net torque and hence a change in rotation. Block in space

So, my question is, am I wrong in thinking that taking the torque about the point-of-attachment is invalid?

Credits of first picture and problem goes to AP College Board (this is a public problem you can find online)

Credits of second picture to another physics exchange user (neverneve)

  • 1
    $\begingroup$ I generally agree with you, but what the CollegeBoard did might make more sense in context. Could you give a link to the AP problem? $\endgroup$
    – knzhou
    Commented May 5, 2020 at 19:28
  • $\begingroup$ apcentral.collegeboard.org/pdf/… $\endgroup$
    – Imajinary
    Commented May 5, 2020 at 19:32
  • $\begingroup$ I assume what they mean is to consider torques about the fixed point that coincides with the attachment point at a certain moment, not about the attachment point in general. You're right that the second would be more subtle. $\endgroup$
    – knzhou
    Commented May 5, 2020 at 19:37
  • $\begingroup$ Wait, that actually makes a lot of sense. I think that was my whole confusion. So, if I understand correctly, if you take torque about that fixed origin, then the point of attachment provides no torque. But, gravity still provides a clockwise torque relative to that origin. Hence, at that instant, the net torque (and hence direction of acceleration) is clockwise. $\endgroup$
    – Imajinary
    Commented May 5, 2020 at 19:40
  • 1
    $\begingroup$ Yes, exactly. Then you might ask, why did the AP exam writers write the question ambiguously? I assume it's because they figured most people wouldn't even realize the subtlety is there (given how simply torque is treated in AP Physics 1), so trying to point out the subtlety would produce more confusion than it would fix. $\endgroup$
    – knzhou
    Commented May 5, 2020 at 19:42

3 Answers 3


You have two choices if you want to consider a point that is not the CoM.

  • consider the point fixed in an inertial frame.

Under this method, you are doing an instantaneous calculation. The object will move away from that point, but the point itself is not accelerating. The main problem is that if the CoM of the object is not in line with the point, its acceleration will change the angular momentum without necessarily changing the angular velocity.

If you consider a rod in freefall, torque about the right end would appear to be positive from the force of gravity. But this torque instead of rotating the rod is accelerating it downward. The angular momentum of $L = mvd$ is increasing as $v$ increases. Non-zero torque, increasing angular momentum (but not rotation).

  • consider the point moving with the object.

Since the point considered is no longer at rest in an inertial frame, fictitious forces arise. These forces are in opposite direction of the acceleration of the frame, applied through the CoM. In this case when we examine the rod from the right end, a torque appears from the pull of gravity. But a counter-torque appears from the non-inertial frame. These two torques cancel out exactly, and the rod does not rotate (as expected).


Here is the full concept. You can find torque about any point.

Problem is you cannot apply $\tau = I \alpha$ about any point.

To apply $\tau = I \alpha$, we need to make sure that the point is inertial. That leaves us with three options (not two)

  1. Choose fixed axis. It is an inertial point. So we don't face any problem.
  2. Choose centre of mass. If it is not inertial, torque of pseudo force has to be considered. Fortunately, pseudo force acts at CoM. So its torque around CoM is always zero.
  3. Choose any arbitrary point. But make sure that you consider torque of pseudo force while while applying $\tau = I \alpha$.

Now try the question again!


At the instant of maximum displacement, the point of attachment is instantaneously at rest. That point in space can be taken as a point of reference for the rate of change of the angular momentum of the sphere caused by gravity. This would include only the angular momentum associated with the motion of the center of mass.


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