Proof of Newton's second law of rotation for rigid body

Question

I'm reading over my lectures notes on rigid body dynamics and I'm trying to prove as many results as possible. However I can't seem to prove that $\tau = \frac{dL}{dt}$ for a rigid body, where $\tau = r \times F$ is the torque and $L = m\cdot r\times v$ is the angular momentum.

I could already prove that that's the case for a point-like object ( this can be done deriving the definition of $L$ and using the fact that $\frac{dr}{dt} \times v = 0$). However none of my (many) attempts was a success when considering a rigid body. What's more I couldn't find any proof of it on internet depsite a thorough search.

Here is what I tried :

For a rigid body, we have the following : $L = \int dL = \int r\times v \cdot dm$

Thus : $\frac{dL}{dt} = \int \frac{d(dL)}{dt} = \int r\times dF$ where $dF$ is the force applied to the tiny mass $dm$.

However I can't find any way of calculating the force $dF$ knowing the total force $F$, its point of application and all the caracteristics of the rigid body. To be honest I'm not much convinced by this approach but it's the only one that got me to the beginning of somewhere (although I imagine there must be some elegant way of proving this).

Thanks for any hint or full answer !

By the way I'm french, hope it might explain any english mistake : )

Answer 1

For a particle in the rigid body we have

$$\tau = \frac{dL}{dt}$$ And $\mathbf{\tau} =\mathbf{ r\times F}$

This gives us $$\frac{dL_i}{dt}=\mathbf{ r_{i}\times F_{net,i}}$$ Where $$F_{net,i}$$ is the net force on the $i^{th}$ particle which includes internal forces from other particles and the external forces.

Summing the equation up for all particles we get

Now consider two particles $i$ and $j$ which exert internal forces on each other which act along the line joining them.

The net torque due to their forces will be

$$r_{i}\times F_{j,i}+r_{j}\times F_{i,j}$$

From Newton's third law, $$F_{j,i}=-F_{j,i}$$

Hence out expression becomes $$(r_{i}-r_{j})\times F_{j,i}$$

But $r_{i}-r_{j}$ is a vector along the line joining them and since the forces also act along that line, the torque would be zero due to internal forces.

Hence summing our equation over all particles we get

$$\sum_{i} \frac{dL_i}{dt}=\sum_{i} r_{i}\times F_{ext,i}$$

Where $F_{ext,i}$ is the net external force on the $i^{th}$ particle since the internal forces cancell out.

But that is the net external torque on the $i^{th}$ particle. Hence we get the required result since the sum of external torques on each particle gives us the net torque in the entire body.

$$\frac{dL_{total}}{dt}=\tau_{net,ext}$$

Answer 2

Use the equation for a single particle and apply summation over all of them. Then use newton's third law to cancell out internal forces.