Question

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

• A. $990{\text{cm /sec}}^2$
• B. $680{\text{cm /sec}}^2$
• C. $750{\text{cm /sec}}^2$
• D. $650{\text{cm /sec}}^2$

Answer 1

The correct option is A $990{\text{cm /sec}}^2$

We know that the angular velocity $\omega =2\pi f$, where $f$ is the frequency.
It is also given that the stone makes $14$ revolutions in $25$ seconds.
Hence, $f=\frac{14}{25}\text{Hz}$
So. $\omega =2\pi f=2\pi \times \frac{14}{25}$

We also know that
$a_c={\omega }^{2}R={(2\times \frac{22}{7}\times \frac{14}{25})}^2\times 80$
(as $\pi =\frac{22}{7}$ and $R=80\text{cm}$ given)
$\therefore a_c\approx 990{\text{cm/s}}^2$


As it is moving with constant speed, then tangential acceleration will be
$a_t=\frac{dv}{dt}=0$
Therefore,
$a_{net}=a_c=990{\text{cm s}}^{-2}$

Hence option A is the correct answer.