Question

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

Answer 1

$x=2R\cos 60=2R\frac{1}{2}=R$

$l-x=\sqrt{3}R-R=(\sqrt{3}R-1)R$

Then spring force is

$F_s=k(l-x)=\frac{(\sqrt{3}+1)mg}{R}(\sqrt{3}-1)R=2mg$

The weight is $W=mg$

The projection of spring force on the normal is

$-2mg\cos 60=2mg\frac{1}{2}=-mg$

The projection of weight on the normal is

$mg\sin 60=mg\frac{\sqrt{3}}{2}$

The sum of all forces in the normal direction must be zero:

$N-mg+mg\frac{\sqrt{3}}{2}=0$

$N=(1-\frac{\sqrt{3}}{2})mg$