Question

A bead of mass $m$ can slide without friction along a vertical ring of radius $R$ . One end of a spring of force constant $k=\frac{3mg}{R}$ is connected to the bead and the other end is fixed at the centre of the ring. Initially, the bead is at the point $A$ and due to a small push it starts sliding down the ring. If the bead momentarily loses contact with the ring at the instant when the spring makes an angle of $60^{\circ }$ with the vertical, then the natural length of the spring is

• A. $\frac{5R}{9}$
• B. $\frac{3R}{4}$
• C. $\frac{5R}{6}$
• D. $\frac{4R}{7}$

Answer 1

Answer: (C)

Let's say the bead loses contact with the ring at point $B$ , then $N_2=0$ and
$mgcos60^{\circ }+kx=\frac{m{v}^2}{R}$
$\frac{mg}{2}+kx=\frac{m{v}^2}{R}$
Applying the conservation of mechanical energy principle between $A$ & $B$
$\frac{m{v}^2}{2}=mgR\left(1-cos60^{\circ }\right)=\frac{mgR}{2}$
$\Rightarrow \frac{m{v}^2}{R}=mg$
$\frac{mg}{2}+kx=mg$
$\Rightarrow x=\frac{mg}{2k}=\frac{R}{6}$
The natural length of the spring $l=R-x=R-\frac{R}{6}=\frac{5R}{6}$