Question

A bead of mass $m$ can slide without friction on a fixed circular horizontal ring of radius $3R$ having centre at the point $C$. The bead is attached to one of the ends of spring of spring constant $k$. Natural length of spring is $R$ and the other end of the spring is fixed at point $O$ as shown in figure. Bead is released from position $A$, what will be kinetic energy of the bead when it reaches at point $B$?

• A. $\frac{25}{2}k{R}^2$
• B. $\frac{9}{2}k{R}^2$
• C. $8k{R}^2$
• D. $12k{R}^2$

Answer 1

The correct option is C $8k{R}^2$

Given,
mass of bead $=m$
radius of circular horizontal ring $=3R$
natural length of spring $=R$
According to conservation law of energy,
$K{E}_i+P{E}_i=K{E}_f+P{E}_f$
$\Rightarrow 0+\frac{1}{2}k[OA-R{]}^2=K{E}_f+\frac{1}{2}k[OB-R{]}^2$
$\Rightarrow \frac{1}{2}-k[5R-R{]}^2=K{E}_f+\frac{1}{2}k[R-R{]}^2$
$\Rightarrow K{E}_f=8k{R}^2$.