00:01
Hello friend as shown in the figure the two uniform rods a .v and b c having the mass rod av having the mass 2 .4 kg and length of av is 0 .36 meter.
00:22
Mass of vc rod is given 4 kg and length of vc is 0 .60 meter.
00:40
As shown in the figure we will see having the velocity.
00:49
Meter per second towards right we have to calculate the velocity of pin b when a v rod rotates by 90 degree calculating moment of inertia of the each rod moment of inertia of the rod av about a 1 by 3 mass of a v and to length of ab square mass is given 2 .4 kg length is given 0 .36.
02:08
You will solve it.
02:12
Then moment of inertia above a point of the rod ab is 0 .103 km meter square.
02:28
Moment of the rod bc about its center of mass 1 by 12 mass of bc into length of bc square.
02:45
Mass of bc rod is 4 kg length is 0 .6 mc so it would be 0 .1 to 0 .00 kg meter squared.
03:06
Now at position 1, potential energy will be mass of av, g into length of ab by 2.
03:29
Similarly, mass of vc, g into height of vc, center of mass.
03:45
I can show you in the diagram by blue color.
03:48
So this is hvc and this is hav calculating the value.
04:12
So gravitational potential energy in this position we will get 2 .4 into.
04:20
We are just substituting the value...