00:01
In this problem, we have been given that there is a ball which is attached to the end of a string, and the ball rotates in a vertical circle.
00:11
So let's say this is the ball here, and we have been given the length of the string, which is the radius of this vertical circle.
00:18
So let's denote that as r.
00:20
So this is the length of the string.
00:22
That's 50 centimeters.
00:23
So in terms of meter, it will be 0 .5 meters.
00:26
And the mass of the ball is 200 grams.
00:29
That's 0 .2 kilograms.
00:31
So we need to determine the velocity of the ball at the top of the circle as well as the tension at the topmost point.
00:41
So this is the topmost point represented by t.
00:44
We need to figure out the velocity as well as the tension.
00:48
So let's represent the topmost point as a here.
00:50
And it's given to us that at the bottom or the lowest point, the speed is 10 meter per second.
00:57
So let's denote vb.
00:59
As the speed at the lowest point that's 10 meter per second.
01:03
So here we use the idea of conservation of energy to get the velocity at the topmost point.
01:09
So at b we can see that the total energy is only because of the kinetic energy considering that this is near the surface of earth, there will be no gravitational potential energy.
01:19
So it will be half m vb square as the speed at lowest point b and the total energy at e it should be equal to the total energy at b according to conservation of energy.
01:31
So total energy at a will be half m into v square plus the gravitational potential energy which will be because of the vertical height equal to twice the length of the string.
01:42
So it will be m g into 2r.
01:45
Here we can cut m from both the sides and we can observe that vb square by 2 minus 2gr will be equal to v square by 2.
01:57
Putting the values here we're going to get 10 square by 2 minus 2 times, let's take g as 9 .8, r is given as 0 .5, and we multiply this with 2, and that's equal to v square...