00:01
In this problem we have a hoop with mass of 10 kilograms in pure rolling motion on a horizontal surface.
00:13
At the horizontal surface the initial speed of the rolling motion is 3 .6 meters per second.
00:23
The hoop encounters a inclined plane with angle of inclination 26 degrees.
00:32
It moves along the plane by a distance of 1 .2 meter.
00:40
At this point the speed of the center of mass is v2 and the angular speed is omega 2.
00:48
The initial angular speed is omega 1.
00:52
We have to find out the final rotational kinetic energy for the hoop in pure rolling motion.
01:10
At all point the rotational kinetic energy is equivalent to the translational kinetic energy.
01:18
Okay, same as equation 1.
01:23
And also we can conserve the mechanical energy for the hoop in pure rolling motion.
01:31
So the total mechanical energy at the bottom is equivalent to the total mechanical energy at the top...