The real force 'F' acting on a particle of mass 'm' performing circular motion acts along the radius of circle 'r' and is directed towards the centre of circle. The square root of magnitude of such force is (T=periodic time)

Options

A. `(2pi)/Tsqrt(mr)`
B. `(Tmr)/(4pi)`
C. `(2piT)/(sqrt(mr))`
D. `(T^(2)mr)/(4pi)`

Answer

A

Explanation

Froce F acting on a body of mass m performing circular motion of radius r,
`F = (mv^(2))/(r) " "("centripetal force") ……(i)`
where V = velocity fo the particle
The tim period of one complete cycle
`T = ("perimeter of a circular path")/("velocity of body") = (2pir)/(v)`
`rArr " "v = (2pir)/(T) " "......(i)`
From Eqs. (i) and (ii) , we get
`F=(m)/(r)((2pir)/(T))^(2) = mr((2pi)/(T))^(2) or sqrt(F) =(2pi)/(T) sqrt(mr)`
So ,option (a) is correct.