00:01
Hello everyone we are going to understand this question here given in the question length of rod is equal to l length of rod that is l so center of mass of rod be at l by 2 center of mass of rod at l by 2 now now when rod is released from the rest from vertical position then we have to find the speed of center of of rod at any point of time.
00:56
So, initial center of mass of rod be at h by 2.
01:03
So when it released, when it was making angle theta, when rod was making angle theta, then height dropped by the center of mass, height dropped by, height dropped by rod, height dropped by center of mass of rod, that is equal to, ly2 minus ly2 into cost theta so here h is equal to l by 2 minus ly2 cost theta so we can write h is equal to l y2 into 1 minus cost theta now applying the conservation of energy now using conservation of energy using conservation of energy so we can we can write m into g into h is equal to half i omega square.
03:04
So we can write omega square is equal to 2mg h upon i.
03:17
Now substituting the value in this 2mg into value of h is l y2 into 1 minus cost theta upon value of i 1 by 3 ml squared.
03:43
Now doing further calculation we will get omega is equal to square root of here m and m will get cancelled and 2 and 2 will get cancelled so we will get 3 into g into 1 minus g x theta upon l this is the angular speed of the center of mass so so the required velocity of the center of mass of rod b.
04:27
So the required speed...