# Why is the sum of torques for each particle equal to the external torque?

Let's assume we have a rigid body. The internal forces all have equal and opposite counterparts so the they will produce a net zero torque. We can therefore ignore internal forces when calculating the total torque. Let's say we apply a tangential force $$F$$ at a distance $$r$$ from the axis and this is the only force. Why is sum of each particle's torque $$\sum_{i}\tau_{i} = Fr$$?

I have an argument for this based on conservation of energy, but is there a simpler explanation? Here's my approach:

Suppose the object has zero kinetic energy. Apply that force $$F$$ over a small angle $$\Delta\theta$$. Then the displacement is $$r\Delta\theta$$ and so the work done is $$Fr\Delta\theta$$. The total kinetic energy of the object is now $$Fr\Delta\theta$$. Likewise it can be shown $$\sum_{i}\frac{1}{2}m_iv_i^2 = \frac{1}{2}Iw^2$$. So the two must equal, $$Fr\Delta\theta = K = \frac{1}{2}Iw^2$$. Now $$\Delta\theta = \frac{1}{2}\alpha t^2$$ from which we can calculate how much time has passed during the angular displacement $$\Delta\theta$$, multiplying this time by $$\alpha$$ gives us the final angular speed of $$w=\sqrt{2\Delta\theta\alpha}$$. Plugging this into $$\frac{1}{2}Iw^2$$ and canceling $$\Delta\theta$$ from both sides of prior kinetic energy equation gives $$Fr=I\alpha$$. Notice that $$\sum_{i}\tau_{i} = \sum_{i}m_ia_ir_i = \sum_{i}m_i(\alpha r_i)r_i = \alpha\sum_im_ir_i^2 = I\alpha$$. So in fact $$Fr = \sum_{i}\tau_{i}$$.

This energy approach seems overly complicated, perhaps I'm missing a simpler way of looking at it.

I think I found a simpler way. First you prove $$(\sum_i{\tau_i})\Delta\theta$$ is the change in total kinetic energy for a small $$\Delta\theta$$. You know the change in a single particle's kinetic energy for a small angle is $$\tau_i\Delta\theta$$ so the total change in K.E. is just the sum of all these. Then you know $$Fr\Delta\theta$$ is also the change in kinetic energy of the object. So $$(\sum_i{\tau_i})\Delta\theta = Fr\Delta\theta$$ and the $$\Delta\theta$$'s cancel.