Motion in a Plane - Result Question 47
50. 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?
[2005]
(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) $\pi^{2} / 4 m s^{-2}$ and direction along the radius towards the centre
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Answer:
Correct Answer: 50. (a)
Solution:
(a) $a_r=\omega^{2} R & a_t=\frac{d v}{d t}=0$
or, $a_r=(2 \pi n)^{2} R=4 \pi^{2} n^{2} R^{2}=4 \pi^{2}(\frac{22}{44})^{2}(1)^{2}$
$a _{\text{net }}=a_r=\pi^{2} ms^{-2}$ and direction along the radius towards the centre.
2 If the force acting on a particle is always perpendicular to the velocity of the particle, then the path of the particle is a circle. The centripetal force $(F_c)$ is always perpendicular to the velocity of the particle, i.e., $F_c \perp v$
