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

Explanation

Here, ` r= 80 cm =0.8 , v= 14// 25 s^(-1)`. :.` omega = 2 pi = 2xx (22) / 7 xx (14)/(25) = (88)/(25) rad. S^(-10`.
The centripetal acceleration , ` a = omega^2 r = ( (88)/( 25))^2 xx 0.80 = 9. 90 ms^(-2)`
The direction of centripetal acceleration is along the string directed towards the centre of circular path.