Question

The minimum speed for a particle at the lowest point of a vertical circle of
radius R to describe the circle is 'v'. If the radius of the circle is reduced
to one-fourth of its value, the corresponding minimum speed will be​

Answer 1

Answer:

v₂ =  $\frac{1}{2}\times\sqrt{5gR}$ or v₂ = $\frac{v}{2}$

Explanation:

The minimum speed (v) of a particle at the lowest point is given by the formula

v = $\sqrt{5gR}$

where,

g is the acceleration due to the gravity

R is the radius

Now,

for the first case Radius = R

and for the cases 2 Radius = R/4

let the speed for the case 2 be v₂

thus,

v₂ = $\sqrt{5g(\frac{R}{4})}$

or

v₂ =  $\frac{1}{2}\times\sqrt{5gR}$

or we can say

v₂ = $\frac{v}{2}$

hence, the value of the speed for radius R/4 becomes half the value of speed for radius = R