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 seconds, what is the magnitude of acceleration of the stone?

• A. $7.81{\text{ms}}^{-2}$
• B. $8.81{\text{ms}}^{-2}$
• C. $9.91{\text{ms}}^{-2}$
• D. $10.81{\text{ms}}^{-2}$

Answer 1

The correct option is C $9.91{\text{ms}}^{-2}$

Radius of the circular path, $r=80\text{cm}=0.80\text{m}$
Number of revolutions made by stone in 25 seconds $=14$
frequency, $v=\frac{14}{25}\text{rev/second}$.
Angular velocity, $\omega =2\pi v=2\pi \left(\frac{14}{25}\right)$
Acceleration experienced by the stone $a=r{\omega }^2$
$=0.8\times {(2\times \frac{22}{7})}^2\times {\left(\frac{14}{25}\right)}^2=9.91{\text{m/s}}^2$
This acceleration of $9.91{\text{m/s}}^2$ is directed towards the centre of the circle.