Shear stress in Interstellar docking scene
The movie Interstellar has a scene where a huge spaceship (64 meter diameter ring) called Endurance is spinning out of control after an explosion. The protagonist slows its angular velocity down by docking his smaller craft (Ranger) to Endurance and using Ranger's retro thrusters for about 30 seconds until their (Ranger+Endurance) angular velocity is reduced to zero.
The first time I watched the scene, I couldn't stop thinking the docking mechanism would have to be massively strong to resist the shear stress caused by the torque applied to Endurance.
I did some back of the envelope calculations to estimate the shear stress it would have to withstand.
$$ \omega = \frac{2\pi}{T} $$
$$ \alpha = -\frac{\omega}{\Delta{t}} $$
$$ I = MR^2 $$
$$ \mathcal{T} = I\alpha = -\frac{2{\pi}MR^{2}}{T\Delta{t}} $$
Where:
$$ \omega \text{ is the angular velocity of the Endurance.}\ \alpha \text{ is the angular acceleration required to stop the spin.}\ I \text{ is the moment of inertia of the Endurance approximating it to a ring.}\ \mathcal{T}\text{ is the torque required to slow the Endurance down.} $$
My estimates follow: $$ T \text{ is visually about 6 rpm, although TARS says 60 rpm. So } \text{0.1 }s^{-1}.\ \Delta{t} \text{ is about } 30 \text{ s}.\ M \text{ is unknown. 394
I will use 445,000 kg to match the ISS mass.}\ R \text{ is 32 m, Endurance being 64-meter wide.} $$
The required torque would be $\mathcal{T} = -954\text{ MNm}$.
That is huge! And it should come from retro thrusters!
So I estimated the shear stress considering the docking mechanism to be a solid round shaft. It is a best case scenario, since it would have to be hollow to allow flow of people.
$$ \tau = \frac{\mathcal{T}r}{I_p} $$
Where:
$$ \tau \text{ is the shear stress at distance }r\text { from the center}\ I_p \text{ is the shaft polar moment of inertia.} $$
For a solid round shaft $I_p = \pi{r}^4/2$. So if we consider a $\text{2 m}$ diameter solid shaft, the maximum shear stress would be:
$$ \tau_{\text{max}} = \frac{2\mathcal{T}}{\pi{r}^3} $$
And that would give $\tau_{\text{max}} = 607 \text{ MPa}$.
This is above what most types of steel would withstand.
Do my calculations make sense? Would that be possible in the real world anyhow? 15