# How can you take torque about an accelerating point that isn't the center of mass?

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. 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. 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)

• 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? – knzhou May 5 at 19:28
• apcentral.collegeboard.org/pdf/… – Imajinary May 5 at 19:32
• 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. – knzhou May 5 at 19:37
• 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. – Imajinary May 5 at 19:40
• 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. – knzhou May 5 at 19:42

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).