A stone tied to the end of a string of 2 m long is whirled in a horizontal circle with a constant sp

Question

A stone tied to the end of a string of 2 m long is whirled in a horizontal circle with a constant speed. If the stone makes 30 revolutions in 60 seconds, what is the magnitude and direction of acceleration of the stone?

Professor PC Thomas Classes · Answer

(a) : $a=r \omega^{2} ; \omega=2 \pi 
u$ \ 30 revolutions $=60 \mathrm{~s} \Rightarrow 
u = 30 / 60 = 0.5 \mathrm{~Hz}$ \ $\omega = 2 \pi (0.5) = \pi \mathrm{~rad/s}$ \ $a=r \omega^{2}=2 	imes \pi^{2}=2 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}$. \ It is the centripetal acceleration, which always acts towards the centre of the circle.