Looking for classes? Ksquare Career Institute, Bengaluru →
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
$\pi^2$ m s$^{-2}$ and direction along the tangent to the circle
B
$\pi^2$ m s$^{-2}$ and direction along the radius towards the centre
C
$\dfrac{\pi^2}{4}$ m s$^{-2}$ and direction along the radius towards the centre
D
$\pi^2$ m s$^{-2}$ and direction along the radius away from the centre
Detailed Solution
Frequency: $n = \dfrac{22}{44} = \dfrac{1}{2}$ revolution per second
Angular speed: $\omega = 2\pi n = 2\pi \times \dfrac{1}{2} = \pi$ rad/s
Centripetal acceleration: $a_r = \omega^2 r = (2\pi n)^2 r$
$a_r = (\pi)^2 \times 1 = \pi^2$ m s$^{-2}$
The speed is constant, so the tangential acceleration is zero: $a_T = 0$.
So the acceleration is $\pi^2$ m s$^{-2}$, directed along the radius towards the centre.
Angular speed: $\omega = 2\pi n = 2\pi \times \dfrac{1}{2} = \pi$ rad/s
Centripetal acceleration: $a_r = \omega^2 r = (2\pi n)^2 r$
$a_r = (\pi)^2 \times 1 = \pi^2$ m s$^{-2}$
The speed is constant, so the tangential acceleration is zero: $a_T = 0$.
So the acceleration is $\pi^2$ m s$^{-2}$, directed along the radius towards the centre.