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 ?

Answer 1

Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency v =$\frac{\text{Number of revolutions }} {\text{Time taken}} = \frac{14}{25}$ Hz
Angular frequency, ω = 2πν

$= 2 x\frac{ 22}{27} x\frac{ 14}{ 25} = \frac{88 }{ 25} rad\,\, s^{-1}$

Centripetal acceleration, ac = ω2 r

= $(\frac{88}{25})^2$ x 0.8= 9.991 m/s2

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