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

Answer:

Option A:  F/4

Explanation:

Centripetal force, $F = \frac{mv^{2} }{r}$

Where v = speed

r = radius

Since Kinetic Energy, $E_{k} = 0.5 mv^{2}$

Writing centripetal force in terms of kinetic energy, $F = \frac{2 E_k}{r}$

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:

$F_{2} = \frac{2(0.5 E_k)}{2r} \\F_{2} = \frac{0.5 E_k}{r}\\F_{2} =\frac{1}{4} * \frac{2 E_k}{r}\\Since, F = \frac{2 E_k}{r} \\F_{2} =\frac{1}{4} * F\\F_{2} =\frac{F}{4}$