Question

A stone tied to the end of a string $1\text{m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes $22$ revolutions in $44$ seconds, what is the magnitude and direction of acceleration of the stone?

• A. ${\pi }^2{\text{ms}}^{-2}$ and direction along the radius towards the centre.
• B. ${\pi }^2{\text{ms}}^{-2}$ and direction along the radius away from the centre.
• C. ${\pi }^2{\text{ms}}^{-2}$ and direction along the tangent to the circle.
• D. $\frac{{\pi }^2}{4}{\text{ms}}^{-2}$ and direction along the radius towards the centre.

Answer 1

The correct option is A ${\pi }^2{\text{ms}}^{-2}$ and direction along the radius towards the centre.

We know that the angular velocity $\omega =2\pi f$, where $f$ is the frequency.
It is also given that the stone makes $22$ revolutions in $44$ seconds.
Hence, $f=\frac{22}{44}=0.5\text{Hz}$
So, $\omega =2\pi f=\pi$.

We also know that,
$a_c={\omega }^{2}R={\pi }^2\left(1\right)={\pi }^2{\text{m/s}}^2$


As it is moving with constant speed, tangential acceleration
$a_t=\frac{dv}{dt}=0$
Therefore, net acceleration
$a_{net}=a_r={\pi }^{2}m{s}^{-2}$
and direction along the radius towards the centre.

Hence option A is the correct answer.