Why is angular momentum conserved?

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Abhishek11235
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So,this is problem from David Morin's Classical Mechanics(Screenshot 1). I solved the problem. Then I went to see the solution in manual hoping for out of box thinking. As in screenshot 2 is solution by Morin. My question is why he conserves angular momentum about the point (R-h) below C.M? There is torque due to its own weight. Actually I want to understand meaning of the statement he has written in bracket.

If texts is small here is what he has written:
"torque from gravity will be relevant during the subsequent rising-up motion. But during the instantaneous collision, L does not change."
Textbook page showing problem 8.22 'Rolling over a bump' and the answer equation 8.63: V0 ≥ sqrt(10gh/7) · (1 − 5h/7R)^(−1).
Printed textbook page, section 8.22 Rolling over a bump, with derivation of angular momentum conservation about the step corner P, equations 8.138 to 8.140 for the ball's energy after collision.
 
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  • Textbook page showing problem 8.22 'Rolling over a bump' and the answer equation 8.63: V0 ≥ sqrt(10gh/7) · (1 − 5h/7R)^(−1).
    Screenshot_2019-02-15-23-46-47.jpeg
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  • Printed textbook page, section 8.22 Rolling over a bump, with derivation of angular momentum conservation about the step corner P, equations 8.138 to 8.140 for the ball's energy after collision.
    Screenshot_2019-02-15-23-52-23.jpeg
    49.4 KB · Views: 523
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You are looking at a collision that is being modeled as instantaneous. As long as this is a good approximation is appropriate, the time of the collision is too short for the torque to transfer any relevant angular momentum to the system.
 
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The argument is that although there is a torque due to gravity while the ball is rotating about the corner, the time required for the ball to go over the corner is short enough so that the angular momentum does not change appreciably.
 
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