Question

A boy whirls a stone in a horizontal circle of radius $1.5m$ and at a height of $2m$ above ground level. The string breaks and the stone flies off horizontally and strikes the ground after travelling a horizontal distance of $10m$ What is the magnitude of the centripetal acceleration of the stone while in circular motion.

Answer 1

In order to solve this question, we will first calculate the time taken by stone to cover vertical height to fall and then using this time period we will find horizontal velocity with given information and later use will use general formula of centripetal acceleration to find its magnitude.

Complete step by step answer:
According to the question, we have given that
$h = 2m$ vertical height of stone above ground level.
$g = 9.8m{s^{ - 2}}$ acceleration due to gravity,
Let ‘t’ be the time taken by stone to fall on ground after covering a vertical distance of $h = 2m$ then, we have,
using newton’s equation of motion
$h = ut + \dfrac{1}{2}g{t^2}$ since u is initial velocity and initial vertical velocity of stone is zero so we get,
$t = \sqrt {\dfrac{{2h}}{g}}$ on putting the value of parameters we get,
$t = \sqrt {\dfrac{4}{{9.8}}}$
$\Rightarrow t = 0.64s$
Now, as we have given that horizontal distance covered by stone to fall on ground is $S = 10m$ let v be the velocity of stone horizontally and $t = 0.64s$ is the time to fall on ground, then using velocity formula we get,
$v = \dfrac{S}{t}$ on putting the value of parameters we get,
$v = \dfrac{{10}}{{0.64}}$
$v = 15.62m{s^{ - 1}}$
Now, we have the horizontal velocity of stone as $v = 15.62m{s^{ - 1}}$ and radius of circular path is given as $R = 1.5m$ and centripetal acceleration a is calculated as
$a = \dfrac{{{v^2}}}{R}$ on putting the value of parameters we get,
$a = \dfrac{{{{(15.62)}^2}}}{{1.5}}$
$a = \dfrac{{243.98}}{{1.5}}$
$\therefore a = 162.6m{s^{ - 2}}$
Hence, the value of centripetal acceleration of the stone is $a = 162.6m{s^{ - 2}}$

Note: It should be remembered that, the motion of stone was circular but in horizontal plane above the ground so, stone don’t have any vertical velocity and the velocity with which it was performing circular motion was the same velocity with which it covers horizontal distance after string breaks.