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

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.

$mgh=\dfrac{1}{2}mv^2$

$h=\dfrac{v^2}{2g}$

Since, $v=r\times \omega$

$h=\dfrac{(r\omega)^2}{2g}$

$h=\dfrac{(1.25\times 18.84)^2}{2\times 9.8}$

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.