Question

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

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

Answer 1

Answer: (C)

Given,
Mass of bead $=m$
The radius of circular horizontal ring $=3R$
The natural length of spring $=R$
According to the conservation of energy,
$KE_{i}+PE_{i}=KE_{f}+PE_{f}$
$\Rightarrow \, 0+\frac{1}{2}k\left[OA - R\right]^{2}=KE_{f}+\frac{1}{2}k\left[OB - R\right]^{2}$
$\Rightarrow \, 0+\frac{1}{2}k\left[5 R - R\right]^{2}=KE_{f}+\frac{1}{2}k\left[R - R\right]^{2}$
$\Rightarrow \, KE_{f}=8kR^{2}$