Question

A stone tied to the end of a string of $1\, m$ 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?

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

Answer 1

Answer: (A), (B)

$a=r \omega^{2} ; \omega=2 \pi v$
$22$ revolution $=44 \sec$.
$ 1 $ revolution $=44 / 22=2 \sec$.
$u =1 / 2 \ ,Hz$
$a=r \omega^{2}=1 \times \frac{4 \pi^{2}}{4}=\pi^{2} m / s ^{2}$.
Towards the centre, the centripetal acceleration $=-\omega^{2} R$
and away from the centre, the centrifugal acceleration is $+\omega^{2} R$
$\therefore$ (a) and (b) are correct as the directions are given