Question

A stone tied to the end of a string of 1m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44s, what is the magnitude and direction of acceleration of the stone?

• A. \(\frac{\pi^2}{4}\)ms^-2 and direction along the radius towards the centre
• B. 2\(\pi^2\) ms^-2 and direction along the radius away from centre
• C. \(\pi^2\) ms^-2 and direction along the radius towards the centre
• D. 4\(\pi^2\) ms^-2 and direction along the tangent to the circle

Answer 1

The Correct Option is C

Given, String length is 1 m.
In 44 seconds, a stone makes 22 rotations.
Stone thus completes \(\frac{22}{44}\) revolutions in 1 second, making frequency = \(\frac{22}{44}\) sec 1 = \(\frac{1}{2}\) Hz.
We are aware that angular speed = 2 frequency w = 2 \(\frac{1}{2}\)= rad/sec.
Now, acceleration equals a=2r, where r is the radius or string length.
Acceleration = \(\frac{1 }{ 9.8596}\) m/s²

Therefore, the correct option is (C): \(\pi^2\) ms^-2 and direction along the radius towards the centre