00:01
In question, we have a uniform disk of mass m equals to 1 kg and radius small r equals to 1 meter.
00:06
Start rolling down from point a without slipping on the rough inner surface of the semicircle containing radius r equals to 4 meters.
00:14
So angular velocity of the disk at point b we have to determine.
00:19
So now from the energy conservation we can write that the kinetic energy initial, this is equals to 1 by 2 mv square.
00:29
This is translational kinetic energy plus 1 by 2 multiplied by i omega square so movement of inertia for the disk this will be m r square so we can write m r square by 2 multiplied by omega square which will be v square by r square so from here we get 3 by 4 modplied by m v square okay so this will be the total kinetic energy of the disk okay and this energy will be equals to the potential energy.
00:59
So this energy, ke, this will be equals to the potential energy.
01:03
So we can write that 3x4 mv square, this is equals to mg multiplied by h.
01:09
H is the height between the two centers.
01:12
And we can calculate this height as this diagram...