Question

A stone tied to the end of string of $80\, cm$ long, is whirled in a horizontal circle with a constant speed. If the stone makes $25$ revolutions in $14\, s$. Then, magnitude of acceleration of the same will be:

• A. $990\, cm / s ^2$
• B. $680\, cm / s ^2$
• C. $750\, cm / s ^2$
• D. $650\, cm / s ^2$

Answer 1

Answer: (A)

When stone is tied to a $80 \,cm$ long string and whirled in a circle the centripetal force is provided by the tension in the string created by drawing the string inwards.
If angular velocity is $\omega$ and radius of circular path is $r$ then
acceleration $\alpha=\omega^2 r$
where $\omega=\frac{2 \pi}{T}$
$T=\frac{\text { number of revolutions }}{\text { time taken }}=\frac{25}{14}=1.78 \,s$
and $ \omega=\frac{2 \pi}{1.78}=3.52 \,rad / s$
Hence, $\alpha=(3.52)^2 \times 80=991.23 \approx 990 \,cm / s ^2$