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

Answer 1

Answer: (D)

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