Question 49 of 183

A bead of mass ' $m$ ' slides without friction on the wall of a vertical circular hoop of radius ' $R$ ' as shown in figure. The bead moves under the combined action of gravity and a massless spring ( k ) attached to the bottom of the hoop. The equilibrium length of the spring is ' $R$ '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' R ', would be (spring constant is ' k ', g is accleration due to gravity)

A$\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$B$2 \sqrt{\mathrm{gR}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$C$\sqrt{2 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$D$\sqrt{2 \mathrm{Rg}+\frac{4 \mathrm{kR}^2}{\mathrm{~m}}}$

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Answer: (A) $\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$

$\begin{aligned} \text { Work done by gravity } & =m g\left(2 R-R \cos 60^{\circ}\right) \\ & =\frac{3 m g R}{2}\end{aligned}$
$\begin{aligned} \text { Work done by spring } & =-\frac{1}{2} k\left(0^2-R^2\right) \\ & =\frac{1}{2} k R^2\end{aligned}$
Net work = change in kinetic energy
i.e. $\frac{3 m g R}{2}+\frac{k R^2}{2}=\frac{1}{2} m v^2$
or $\quad v^2=3 g R+\frac{k R^2}{m}$
or $\quad v=\sqrt{3 g R+\frac{k R^2}{m}}$