MCQ
A stone ties to the end of a string $1\,m$ long is whirled in a horizontal circle with a constant speed. If the stone makes $22$ revolution in $44$ seconds, what is the magnitude and direction of acceleration of the stone
  • A
    $\frac{{{\pi ^2}}}{4}\;m{s^{ - 2}}$ and direction along the radius towards the centre
  • B
    ${\pi ^2}\;m{s^{ - 2}}$ and direction along the radius away from the centre
  • ${\pi ^2}\;m{s^{ - 2}}$ and direction along the radius towards the centre
  • D
    ${\pi ^2}\;m{s^{ - 2}}$ and direction along the tangent to the circle

Answer

Correct option: C.
${\pi ^2}\;m{s^{ - 2}}$ and direction along the radius towards the centre
c
(c) $a  = \frac{{{v^2}}}{r} = {\omega ^2}r$ $ = 4{\pi ^2}{n^2}r = 4{\pi ^2}{\left( {\frac{{22}}{{44}}} \right)^2} \times 1$ $ = {\pi ^2}\,m/{s^2}$

and its direction is always along the radius and towards the centre.

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