Question

A boy whirls a stone in a horizontal circle of radius 1.8 m and at height 1.8 m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 9.9 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?

Answer 1

Acceleration is
`148.33\ m/s^(2)`

Solution:

As per the question:

Radius of the circle, R = 1.8 m

Height above the ground, h = 1.8 m

Horizontal distance, x = 9.9 m

Now,

The magnitude of the centripetal acceleration can be calculated as:

`a_(c) = (v^(2))/(R)`

where

v = velocity

R = radius

`a_(c)` = centripetal acceleration

Now, if we consider the vertical component of motion only, then considering the initial velocity, 'u' = 0, from kinematic eqn:

`h = ut + (1)/(2)gt^(2)`

`h = 0.t + (1)/(2)gt^(2)`

`t = \sqrt{(2h)/(g)}`

`t = \sqrt{(2* 1.8)/(9.8)}`

t = 0.606 s

Now, for the horizontal component of velocity:

x = vt

`v =(x)/(t)`

`v =(9.9)/(0.606) = 16.34\ s`

Now, we know that the centripetal acceleration is given by:

`a_(c) = (16.34^(2))/(1.8) = 148.33\ m/s^(2)`