Question

A boy whirls a stone in a horizontal circle $1.8m$ above the ground by means of a string of radius $1.2m$, while whirling the stone string was horizontal, it breaks and stone files off horizontally, striking the ground $9.1m$ away. The centripetal acceleration during the circular motion was nearly: ($g=9.8m/s^2$)

• A. $198m/s^2$
• B. $19.8m/s^2$
• C. $162.7m/s^2$
• D. $16.27m/s^2$

Answer 1

The correct option is C $162.7m/s^2$

Centripetal acceleration is given by,

$a=\frac{v^2}{r}$

Considering the vertical component of the motion only we can say,

$s=\frac{1}{2}g{t}^2$

$\therefore t=\sqrt{\frac{2s}{g}}$

$\therefore t=\sqrt{\frac{2\times 1.8}{9.8}}=0.61sec$

The horizontal component of velocity shares the same time of flight and is constant so,

$v=\frac{9.1}{0.61}=15.01m/s$

So now we can get the centripetal acceleration,

$a=\frac{v^2}{r}=\frac{{15.01}^2}{1.2}=162.7m/s^2$