Question

A stone is tied to a rope of length $2$ m and is whirled above his head horizontally by a man at a height of $9\text{m}$ from the ground. The stone covers a horizontal distance of $20\text{m}$ when the rope breaks. What is the magnitude of the centripetal acceleration when the stone was in the circular motion?

• A. $111.11{\text{m/s}}^2$
• B. $123.5{\text{m/s}}^2$
• C. $99.25{\text{m/s}}^2$
• D. $61.72{\text{m/s}}^2$

Answer 1

The correct option is A $111.11{\text{m/s}}^2$

Let $v$ be speed of the stone before rope breaks
When the rope gets cut and the stone flies off, it experiences an acceleration only in the Y direction, Hence, $a_y=g$.
The rope is being whirled horizontally, so $u_y=0$.

Time of flight-
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 9}{10}}=\sqrt{1.8}\text{s}$
$\therefore \text{Time taken for the time to reach ground}=\sqrt{1.8}\text{s}$.

Initial horizontal velocity of stone be $v_x=v$
$\Rightarrow vt=\text{Horizontal distance travelled by stone}$

$\Rightarrow v=\frac{\text{Horizontal distance travelled by stone}}{t}v=\frac{20}{\sqrt{1.8}}\text{m/s}$

We know centripetal acceleration$a_c=\frac{v^2}{r}=\frac{(\frac{20}{\sqrt{1.8}}{)}^2}{2}=111.11{\text{m/s}}^2$