1.

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

Answer»

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


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