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

Answer:

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:

$a_{c} = \frac{v^{2}}{R}$        

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:

$h = ut + \frac{1}{2}gt^{2}$                

$h = 0.t + \frac{1}{2}gt^{2}$                

$t = \sqrt{\frac{2h}{g}}$                

$t = \sqrt{\frac{2\times 1.8}{9.8}}$

t = 0.606 s

Now, for the horizontal component of velocity:

x = vt

$v =\frac{x}{t}$

$v =\frac{9.9}{0.606} = 16.34\ s$

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

$a_{c} = \frac{16.34^{2}}{1.8} = 148.33\ m/s^{2}$