Question

An object rotates describing a horizontal circumference when subjected to a centripetal force F. What centripetal force will act on the object if the radius of the circle is doubled and the kinetic energy of the object is halved? A) F/4 B) F/2 C) F D) 4F please explain this to me : (

Answer 1

Option A: F/4

Step-by-step explanation:

Centripetal force,
`F = (mv^(2) )/(r)`

Where v = speed

r = radius

Since Kinetic Energy,
`E_(k) = 0.5 mv^(2)`

Writing centripetal force in terms of kinetic energy,
`F = (2 E_k)/(r)`

If the initial radius of the circle, r₁ =r

The doubled radius, r₂ = 2r

If the initial kinetic energy,
`KE_1 = E_(k)`

The halved kinetic energy,
`KE_2 = 0.5E_(k)`

Therefore, the new Centripetal force becomes:

`F_(2) = (2(0.5 E_k))/(2r) \\F_(2) = (0.5 E_k)/(r)\\F_(2) =(1)/(4) * (2 E_k)/(r)\\Since, F = (2 E_k)/(r) \\F_(2) =(1)/(4) * F\\F_(2) =(F)/(4)`