00:01
Traveling around a circle on the table as shown.
00:03
The attached cord shown in the diagram is pulled down with a constant speed.
00:07
And we then want to calculate how much work is done to pull the cord down.
00:13
So what we can do is write down the conservation of angular momentum.
00:19
That is the initial angular momentum must equal to the final angular momentum of the ball.
00:26
So initially before the cord is pulled, the angular momentum of the ball is its mass, which is for pounds, its weight over g, 32 .2 feet per square second, multiplied by its speed, 6 feet per second, times the distance to the hole, which is 3 feet.
00:50
The final angular momentum is, again, its mass 32 .2 times the distance that the quad is pulled, and hence the distance r2 which is two feet and this is multiplied by its speed v theta so the theta component of the velocity so if we rearrange this equation we can find v theta and so v theta if we rearrange this is nine feet a second so that's the theta component of the ball's velocity so the actual final velocity v2 will call it the magnitude of v2 at least is the square root of its of the square of the r and theta components so that's nine squared plus two squared under the square root sign and hence we get v2 to be 9 .22 feet per second so the absolute velocity of the ball is 9 .22 feet per second after the cord is pulled now taking the work energy theorem, we know that the initial kinetic energy of the ball plus the work done in pulling the cord, u, must equal to the final kinetic energy of the ball...