Question

A frictionless track ends in a circular loop of radius R. A body slides down the track from point A which is at height h = 5 cm. Maximum value of R, for a body to complete the loop successfully is

• A. 2 cm
• B. $\frac{10}{3}$ cm
• C. $\frac{15}{4}$ cm
• D. $\frac{18}{3}$ cm

Answer 1

The correct option is A 2 cm

To just complete the circle, at the highest point, tension is zero and the gravitational force provides the necessary centripetal force.

Hence, $\frac{m{v}_{top}^2}{R}=mg\dots \dots .\left(i\right)$

Applying conservation of energy at the top and bottom points,
$\frac{1}{2}m{v}_{bottom}^2=mg\times 2R+\frac{1}{2}m{v}_{top}^2\dots \dots .\left(ii\right)$
Substituting $m{v}_{top}^2$ from (i) in (ii) and solving,
$v_{bottom}=\sqrt{5gR}$

But from given velocity at bottom$=\sqrt{2gh}$

So, $\sqrt{2gh}=\sqrt{5gR}$
$2h=5R\Rightarrow R=2cm$