Question

IF THE RADIUS OF THE CIRCULAR PATH OF PARTICLE GOING AROUND THE CIRCLE IS DOUBLED WITHOUT CHANGING ITS FREQUENCY OF ROTATION THEN WHAT HAPPENDS TO ITS CENTRIPETAL FORCE

Answer 1

F = mv^2/r
when f becomes double
F' = F/2

Answer 2

Explanation :

Centripetal force is given by :

$F_c=mr\omega^2$

Where

m is the mass of an object

r is the radius of the circular path

$\omega$ is the angular velocity and it is equal to $\omega=\dfrac{2\pi}{T}$

The centripetal force is directly proportional to the mass of an object and the radius of the circular path.

So, if the radius of the circle is doubled, its centripetal force will also be doubled.

$F_c'=2F_c$

$F_c$ is the force when the radius is r

$F_c'$ is the force when the radius is 2r

Hence, this is the required solution.