00:01
Angular momentum is conserved in the interaction between the child and the merry -go -round because there is no external torque.
00:09
Therefore, l -i is equal to l -f, where i stands for initial and f stands for final.
00:19
Therefore, l -i of the child, plus l -i of the merri -go -round is equal to l -f of the child plus l -f of the child plus l -f of the merri -round.
00:36
And since the child is at rest initially, therefore this goes to zero.
00:45
Now angular momentum of, angular momentum is given by i times omega.
00:54
Therefore, for the merry -go -round, this will be equal to i of the merry -go -round times omega of the merry -round initially, which is omega -not.
01:08
And this is equal to lfc plus lfm and lfc is moment of inertia of the child times omega where omega is the final moment of inertia after it slows down after the merry -go round slows down and also lfm will be equal to im times omega since they both since the child is on the merry -go -round they slow down and they have the same angular velocity...