Question

A boy whirls a stone in a horizontal circle $2m$ above the ground by means of a string $1.25m$ long. The string breaks and the stone flies off horizontally, striking the ground $10m$ away. What is the magnitude of the centripetal acceleration during circular motion? Take $g=10{ms}^{-2}.$

• A. $100{ms}^{-2}$
• B. $200{ms}^{-2}$
• C. $300{ms}^{-2}$
• D. $400{ms}^{-2}$

Answer 1

The correct option is B $200{ms}^{-2}$

Given that $h=2m$ and $r=1.25m$ and a horizontal distance $s=10m$ When the string breaks , the stone is projected in the horizontal direction , which means that there is no initial velocity .

From $s=ut+\frac{1}{2}g{t}^2,$ we have

$h=\frac{1}{2}g{t}^2$ (1)

the horizontal distance travelled in time t is, $s=vt$ (2)

where $v$ is the velocity of the stone in the horizontal direction which is the same as its velocity in circular motion

eliminating $t$ from eq(1) and (2) we get

$v^2=\frac{g{s}^2}{2h}$

So centripetal acceleration =$a_c=\frac{v^2}{r}=\frac{g{s}^2}{2hr}=\frac{10\ast 10\ast 10}{2\ast 2\ast 1.25}=200m/s^2$