00:02
In this problem, a meter stick is pivoted at this end, let's say this end is o.
00:09
Okay, and then it is released to rotate freely about this horizontal axis.
00:16
And let's say the final position when the stick is vertical is this.
00:22
Now, since the stick is uniformly, the mass of this stick is uniformly distributed.
00:27
The center of mass of the stick is at half of its distance.
00:32
So this is the center of mass of the stick.
00:35
And the final position of the center of mass is at this position.
00:40
Let's consider this position has the zero gravitational potential energy.
00:47
So from this we can say the gravitational potential energy is zero at this location.
00:52
And the distance of center of mass from the excess rotation is l .2, that is a half of the length of the stick.
00:59
So for part a, the change in gravitational energy is equivalent to the final gravitational potential energy minus the initial gravitational potential energy and that will be the final is zero and the initial is m g into l upon 2 okay so from this we get the change in gravitational potential energy as minus of 0 .18 kilograms into 9 .8 meters per second square into half meters or the change in gravitational potential energy is equivalent to 0 .882 jules...