So hypothetically, you have a disc or ring. And on opposing edges of that disc/ring, there are two thrusters that given a set amount of Newtons of force (say 120 N) both in the counterclockwise direction.

Assume the axis is along the z dimension, so the direction of torque along that axis of rotation would be in the positive z direction if torque is counterclockwise.

My question is this: I know that if you have a tangential force on a wheel or something at an instantaneous moment, it can cause torque, but what if you maintain a force around the wheel, always in the same place relative to the wheel, and always pointing in a direction tangential to the rotation? Would this still cause torque, over time?

So if you had these two thrusters on either side of the disc, both facing the same rotational direction, would this mean net torque could be summed as tau = 240, because you would just add these two?

Or would you have to calculate the cross product of tangential force and radius for each instant of rotation the wheel spins in, since it's no longer an instantaneous force?


1 Answer 1


The situation you describe would indeed constitute a constant torque, and would cause the disk To increase in rotational speed (revolutions per minute) without limit, as long as the situation persists. It is exactly akin to having a linear constant unbalanced force (like a thruster) on an object, which would cause an unending acceleration as long as the force is applied.

The formula for a torque $\vec\tau$ add you note is

$\vec\tau = \vec r \times \vec F$

i.e. the cross product of the radial distance to the rotation axis and the applied force. Another way to evaluate the cross product is:

$\vec\tau = |\vec r ||\vec F|\sin \theta $

This makes it easy to see that because the magnitudes of $\vec r$ and $\vec F$ are always the same, and $\theta$ is always 90° if the force is tangent to the circle, the torque is constant at all times.


Your Answer

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

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