Question

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

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

Answer 1

Answer: (A)

When stone is tied to a $80cm$ 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 $\omega =\frac{2\pi }{T}$
$T=\frac{\text{ number of revolutions }}{\text{ time taken }}=\frac{25}{14}=1.78s$
and $\omega =\frac{2\pi }{1.78}=3.52rad/s$
Hence, $\alpha =(3.52{)}^2\times 80=991.23\approx 990cm/s^2$