A stone tied to the end of a string of \(1 \text{ m}\) long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?
(2005)
\(\pi^2 \text{ m/s}^2\) and direction along the radius towards the centre
\(\pi^2 \text{ m/s}^2\) and direction along the radius away from the centre
\(\pi^2 \text{ m/s}^2\) and direction along the tangent to the circle
\(\pi^2 \text{ m/s}^2\) and direction along the radius towards the centre
Solution:
Concept: Centripetal acceleration in uniform circular motion.
Formula: \(a = \omega^2 r\), where \(\omega = 2\pi f\) and \(f\) is frequency.
Solution: Frequency \(f = 22 \text{ rev}/44 \text{ s} = 0.5 \text{ Hz}\). Angular velocity \(omega = 2\pi(0.5) = \pi \text{ rad/s}\). Radius \(r = 1 \text{ m}\). So, \(a = (\pi)^2 (1) = \pi^2 \text{ m/s}^2\). Direction is always towards the centre.
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