A stone tied to the end of a string of 1 $\mathrm{m}$ long is whirled in a horizontal circle with a constant…

A stone tied to the end of a string of 1 $\mathrm{m}$ 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?
  1. $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre
  2. $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius away from the centre
  3. $\pi^2 \mathrm{~ms}^{-2}$ and direction along the tangent to the circle
  4. $\pi^2 / 4 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre

Solution

\(a=\frac{v^2}{r}=\omega^2 r=4 \pi^2 n^2 r=4 \pi^2\left(\frac{22}{44}\right)^2 \times 1=\pi^2 \mathrm{~m} / \mathrm{s}^2\)
and its direction is always along the radius and towards the centre.

Asked in: NEET 2005

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