Question

A small lead ball, attached to a 1.25-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.5 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?

Answer 1

Step-by-step explanation:

Radius of the circular path, r = 1.25 m

Angular velocity of the ball,
`\omega=3\ rev/s=18.84\ rad/s`

It is placed 1.5 meters above the ground. Using the conservation of energy to find the height of the ball.

`mgh=(1)/(2)mv^2`

`h=(v^2)/(2g)`

Since,
`v=r* \omega`

`h=((r\omega)^2)/(2g)`

`h=((1.25* 18.84)^2)/(2* 9.8)`

h = 28.29 meter

So, the ball will rise to a maximum height of 1.5 m + 28.29 m = 29.796 meters. Hence, this is the required solution.