Question

The rider in circus rides along a circular track in a vertical plane. The minimum velocity at the highest point of the track will be

• A. $\sqrt{gR}$
• B. $\sqrt{2gR}$
• C. $\sqrt{3gR}$
• D. $\sqrt{5gR}$

Answer 1

The correct option is A $\sqrt{gR}$

the particle of mass m is attached to an inextensible string of length R

which provides the necessary centripetal force along with gravity.

Let the initial velocity be u

at the lowest point of the vertical circle. After time dt, it moves to some other point transversing angle θ. The height at which it is now located is

h=R(1−cosθ)

. Neglecting all non-conservative force, from work-kinetic energy theorem, we get the velocity at the point after dt is
$v^2=u^2-2gh$

. The necessary centripetal force is
$T-mgcos\left(\theta \right)=m{v}^2/r$

The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π

. There must be a centripetal force to make this happen. To find the minimum centripetal force, we make T=0 so that our new centripetal force is
$mg=m{v}^2/r⟹v=\sqrt{\left(rg\right)}$