A stone tied to the end of a string 80 cm long is whirled in a

Question.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

solution:

Length of the string, $I=80 \mathrm{~cm}=0.8 \mathrm{~m}$

Number of revolutions $=14$

Frequency, $v=\frac{\text { Number of revolutions }}{\text { Time taken }}=\frac{14}{25} \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi \mathrm{V}$

$=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \mathrm{rad} \mathrm{s}^{-1}$

Centripetal acceleration, $a_{\mathrm{c}}=\omega^{2} r$

$=\left(\frac{88}{25}\right)^{2} \times 0.8$

$=9.91 \mathrm{~m} / \mathrm{s}^{2}$

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.