00:01
Hello students, let's solve this question.
00:02
Here, movement of inertia of pulley is i equals to half multiplied by m multiplied by r1 square plus r2 square.
00:11
Here, by putting the values of m, r1 and r2, we get i equals to half multiplied by 0 .350 multiplied by 1 .0200 square plus 0 .0300 square.
00:29
Now after solving this we get the value of i equals to 0 .002275.
00:36
This is the moment of inertia of pulling.
00:39
Now initial kinetic energy of hanging block is k1i equals to half m1 v i square and this will be equal to half putting values of m1 and v i we get half multiplied by 0 .365 multiplied by 0 .820.
00:59
And by solving this, we get the initial kinetic energy of hanging block equals to 0 .12713 joules.
01:08
Now initial potential energy of hanging block, that is, u1i is equals to m1 g delta x.
01:18
And by putting values of g and delta x, we get u1i equals to 0 .365 multiplied by 9 .81 multiplied by 0 .700 and this will be equals to 2 .50645 jules.
01:34
This is the initial potential energy for hanging blow.
01:37
Now initial kinetic energy for sliding blow.
01:42
K2i equals to half m2 multiplied by v i square.
01:48
Now putting values of m2 and v i we get k2 i equals to half multiplied by 0 .845 multiplied by 2.
01:56
By 0 .820 and this will be equal to 0 .284089 joules.
02:02
This is the initial kinetic energy of sliding blow.
02:06
Now initial potential energy of sliding blow, u2i, u2i equals to 0.
02:12
Initial rotational kinetic energy of the pulley, that is k r i equals to half i omega square and this will be equal to half multiplied by i multiplied by v i divided by r2 square.
02:27
As we know, omega is equal to v .i divided by r2...