Question

A small lead ball, attached to a 1.70-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 2.0 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

Answer:

H = 54.37

Explanation:

given,

lead ball attached at = 1.70 m

rate of revolution = 3 revolution/sec

height above the ground = 2 m

$velocity = \dfrac{distance}{time}$

circumference of the circle = 2 π r

                                             = 2 x π x 1.7

                                             = 10.68 m

$velocity = \dfrac{3 \times 10.68}{1}$

v = 32.04 m/s

using conservation of energy

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

$\dfrac{1}{2}v^2 + gh = gH$

$\dfrac{1}{2}32.04^2 + 9.8\times 2 = 9.8\times H$

$532.88 = 9.8\times H$

H = 54.37

the maximum height reached by the ball is equal to H = 54.37