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?
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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.
Farcher
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Thank you for your answer. There is still something I would like to understand more. You said: ''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 do you mean by choosing the point of contact between the step and the cube? – Sam Jefferson Jan 28 '19 at 10:37
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@SamJefferson It is at the point of contact between the cube and step that the impulsive force(s) acts so when finding the torques on the cube due to the forces acting on the cube, the torque due to the force at the step is zero. – Farcher Jan 28 '19 at 11:11
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So we essentially choose the origin of our reference frame to be at the point of contact? – Sam Jefferson Jan 28 '19 at 11:29
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@SamJefferson This enable you to assume that the angular momentum of the cube about the point of contact just before hitting the step is equal to the angular momentum of the cube about the point of contact just after hitting the step. If you chose any other point you would have to include the significant impulse torque due to the step. – Farcher Jan 28 '19 at 12:04
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Ah I see. So then after the lower face of the cube lifts off the ground, the gravity will exert torque for non infinitesimal amount of time, but then we can use conservation of energy to solve problems. Am I saying it right? – Sam Jefferson Jan 29 '19 at 14:08