# Why does torque increase with radius or distance from the centre?

Why is it that making a perpendicular force farther away from the axis of rotation increases the rate of change of angular momentum (hence, torque)?

• We have extremely good physical reasons for this to be the case, but you have to first decide what kind of answer it is that you are looking for. After all, "Why not?" is also a valid answer to this question. Commented Dec 5, 2023 at 5:57
• Voting to reopen. A perfectly clear question with some good answers below. Commented Dec 5, 2023 at 16:13

As with everything in physics, it's about modelling the real world, based on experiments.

We all experienced that, e.g. for turning a screw, the same force applied to a longer lever has a better effect. Regarding rotation, experiments showed that applying half the force at twice the lever length gives the same results.

So it made sense to introduce something called "torque", multiplying the force with the lever length, to describe the "turning power" for that screw. And the same "torque" quantity is also useful for describing the rotational acceleration of a body if some force is applied outside its center of mass, and for many other physical effects.

If experiments had not shown this linear relationship, then probably physics had introduced different quantities instead of torque.

To sum it up: Physics found that "torque", being the product of force and lever length, is a useful abstraction when describing rotations.

If your question should be understood as "Why does a perpendicular force farther away from the axis of rotation have a greater effect?", then physics just answers: "It's what our experiments showed us".

Well, torque does not necessarily increase with radius since you didn't hold the force constant since $$\tau=\vec r \times \vec F$$
The reason why we are considering torque is because of the angular momentum: if there is no external torque in a resting frame of reference for a system, then such system's total angular momentum is conserved.

Angular momentum is defined as $$\mathbf L = \mathbf r\times\mathbf p$$ This can be changed in 2 ways: by changing $$\mathbf r$$ and by changing $$\mathbf p$$ (or a combination of both): $$\frac{\mathrm d\mathbf L}{\mathrm dt} = \left(\frac{\mathrm d\mathbf r}{\mathrm dt}\times\mathbf p\right) + \left(\mathbf r\times\frac{\mathrm d\mathbf p}{\mathrm dt}\right)$$ In either case, the effect of a change of one variable is proportional to the value of the other, since they are multiplied. Torque is defined such that it affects angular momentum by changing only $$\mathbf p$$ over time $$\tau = \mathbf r\times\frac{\mathrm d\mathbf p}{\mathrm dt}$$ From Newton's 2nd law we know that any change in momentum is due to a force, $$\mathbf F = \frac{\mathrm d\mathbf p}{\mathrm dt}$$, so we can assert that all torques are derived from forces: $$\mathbf \tau = \mathbf r\times\mathbf F$$

All this to say: Torques change angular momentum by change tangential momentum over time, and the rate of this change is strength of the associated force. Since angular momentum is the product of tangential momentum and distance of the point from the center, the angular momentum changing effect of a given change in momentum is multiplied (literally) by the distance from the center. Hence the amount of torque produced by a given perpendicular force is directly proportional to the distance of the force from the center.

As a sort of separate answer I think its also insightful to think of torque in terms of work:

The work done by a force over a small displacement is $$\mathrm dW = F\mathrm d x$$ (in 1 dimension). We can rearrange this as $$F = \frac{\mathrm dW}{\mathrm dx}$$ to get an alternate interpretation of force as "the rate of work done per unit displacement".

Now considering torque as the rotational analogue of force, the formula for the work done by a torque over a small rotation $$\mathrm dW = \tau\mathrm d\theta$$ can be rearranged as $$\tau = \frac{\mathrm dW}{\mathrm d\theta}$$ to give torque a similar interpretation as "the rate of work done per unit rotation".

If a perpendicular force is acting a distant $$r$$ from the axis of rotation the displacement associated with a rotation by angle $$\mathrm d\theta$$ over which the force does work is equal to $$\mathrm dx = r\mathrm d \theta$$, and so the work done by the force due to this rotation is $$\mathrm dW = Fr\mathrm d\theta$$. Now examining how the perpendicular force acts as a torque, we see that $$\tau = \frac{\mathrm dW}{\mathrm d\theta} = rF$$ Intuitively, increasing the distance of the force from the center increases the amount of work it does for a given amount of rotation, making it a stronger torque.