Question

A bead of mass $m$ is attached to one end of a spring of natural length 𝑅 and spring constant $K=\left(\right(\sqrt{3}+1\left)\text{mg}\right)/R$ . The other end of spring is fixed at a point $A$ on a smooth vertical ring of radius $R$ as shown in the figure. The normal reaction at $B$ just after it is released to move is

• A. $mg/2$
• B. $3\sqrt{3}mg$
• C. $\sqrt{3}mg$
• D. $3\sqrt{3}mg/2$

Answer 1

The correct option is D $3\sqrt{3}mg/2$

Force on bead having mass $m$ placed on point B and their corresponding angles are shown in figure.


Where,
$N=$ Normal reaction force on bead.
$mg=$ Gravitational force on bead.
$k\Delta x=$ spring force on particle.
Apply sine rule in isosceles triangle AOB,
$\frac{l_1}{sin{120}^0}=\frac{R}{sin{30}^0}$
$l_1=\sqrt{3}R$ ($l_1$= length of spring when bead is at B)
Hence, elongation in spring is
$\Delta x=l_1-R=\sqrt{3}R-R=(\sqrt{3}-1)R$
Hence, spring force = $k\Delta x=\frac{(\sqrt{3}+1)mg}{R}(\sqrt{3}-1)R=2mg$

Now, draw the F.B.D of bead.


From F.B.D, we can easily take the component of spring force and gravitational force in the the direction of normal force.
$N=mgcos{30}^0+2mgcos{30}^0$
$N=\frac{3\sqrt{3}mg}{2}$