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Question: A stone tied to the string of \(1m\) long is whirled in a horizontal circle with a constant speed.if...

A stone tied to the string of 1m1m long is whirled in a horizontal circle with a constant speed.if the stone makes 2222 revolutions in 4444 seconds what would be the magnitude and direction of the stone.

A.π2ms−2{{\pi }^{2}}m{{s}^{-2}} and direction along the tangent to the circle

B.π2ms−2{{\pi }^{2}}m{{s}^{-2}} and direction along the radius towards the centre

C.2π2ms−22{{\pi }^{2}}m{{s}^{-2}}and direction along the radius towards the centre

D.π2ms−2{{\pi }^{2}}m{{s}^{-2}} and direction along the radius away from the centre

Explanation

Solution

The concept of centripetal acceleration is to be applied here, Motion of a particle in a horizontal circle is an example of uniform circular motion. Centripetal acceleration is towards the centre that gives rise to the force which actually is responsible for particles to move in a circular path.

Complete step by step answer:
A particle while moving in a circular path has two types of acceleration
Centripetal and tangential acceleration.
Since a horizontal circle is an example of uniform circular motion it does not have any tangential force or a tangential acceleration along the tangent and hence only has centripetal acceleration that promotes its motion in a horizontal circle.
Centripetal acceleration arises due to centripetal force which is acting along centre and hence direction of centripetal acceleration is towards the centre or radius of the circle.
Centripetal acceleration in terms of angular velocity ww, radius rr is defined as w2×r{{w}^{2}}\times r where w=2πfw=2\pi f (f is frequency).
frequency=no of revolutionstime in sec⁡ondfrequency=\dfrac{no\,of\,revolutions}{time\,in\,\sec ond}
r=1mr=1m
No. of revolutions = 22
Time = 44 seconds
frequency=2244hertz=12hertzfrequency=\dfrac{22}{44}hertz=\dfrac{1}{2}hertz
The magnitude of centripetal acceleration=w2r{{w}^{2}}r
⇒(2×π×frequency)2×1m\Rightarrow {{\left( 2\times \pi \times frequency \right)}^{2}}\times 1m
⇒(2×π×12)2×1m\Rightarrow {{\left( 2\times \pi \times \dfrac{1}{2} \right)}^{2}}\times 1m
∴π2m/s2\therefore {{\pi }^{2}}m/{{s}^{2}} is the magnitude and direction is towards the center.

Hence the correct option is B.

Note:
Motion of a particle in a vertical circle is an example of non-uniform vertical motion. In this motion the net acceleration of the particle is the vector sum of tangential velocity and centripetal velocity.