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.


1 Answer 1


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.


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