Question

A particle is projected with velocity $\sqrt{3gL}$ at point A (lowest point of the circle) in the vertical plane. Find the maximum height above horizontal level of point A if the string slacks at the point B as shown. Given $L$ is radius of circle.

• A. $\frac{4L}{3}$
• B. L
• C. $\frac{4L}{9}$
• D. $\frac{L}{3}$

Answer 1

The correct option is A $\frac{4L}{3}$

At point B, as string slacks, tension becomes zero.

So balancing forces in the direction of string at B,

$\frac{m{v}_B^2}{L}=mgcos\theta ⟹\frac{m{v}_B^2}{2}=\frac{mgLcos\theta }{2}$

So Kinetic energy at point B is $\frac{mgLcos\theta }{2}$

Potential energy at point B with respect to point A is $mg(L+Lcos\theta )=mgL(1+cos\theta )$

Total energy at point B is, $E_B=mgL(1+\frac{3cos\theta }{2})$

Total energy at point A is, $E_A=\frac{m{V}_A^2}{2}=\frac{3mgL}{2}$

By conservation of mechanical energy, under gravitational force,

$E_A=E_B⟹mgL(1+\frac{3cos\theta }{2})=\frac{3mgL}{2}$

$⟹cos\theta =\frac{1}{3}$

So height of point B with respect to A is,

$L(1+cos\theta )=\frac{4L}{3}$