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 $44$ seconds, what is the magnitude and direction of acceleration of the stone?

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