Cube tumbling over

Question

When a moving cube hits a low step (the step is perpendicular to its velocity) and tumbles over, is its angular momentum conserved? It is implied that it is so, but certainly gravity produces a torque right?

Answer 1

There are two forces acting on the cube when it hits the step.
The attractive force on the cube due to the Earth (the weight of the cube) and the impulsive force on the cube due to the step.
Each of these forces can produce a torque but if one chooses the point of contact between the step and the cube and there is no torque due to the impulsive force.
What you are left with is the torque due to the weight of the cube $\tau_{\rm weight}(t)$ where $t$ is the time.
The impulsive torque is $\displaystyle \int_{\rm collision} \tau (t) \,dt$ and this will give you the change in angular momentum.

However if the time during which the collision occurs is small this impulse torque due to the weight of the cube is much smaller than the angular momentum of the cube just before it hits the step and so it will not change the angular momentum of the cube about the point of contact by any significant amount.

This is a similar argument to that used when there are mid-air collisions/explosions and linear momentum is assumed to be conserved even though the bodies are attracted by the Earth as explained in this answer.