Question

A bead of mass m is attached to one end of a spring of natural length 3 R and
spring constant k = (31)+mgR . 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 ?

Answer 1

Answer on Question #64345-Physics-Mechanics-Relativity

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?

Solution

$$x = 2 R \cos 6 0 = 2 R \frac {1}{2} = R.$$ $$l - x = \sqrt {3} R - R = (\sqrt {3} - 1) R$$

The spring force is

$$F _ {s} = k (l - x) = \frac {(\sqrt {3} + 1) m g}{R} (\sqrt {3} - 1) R = 2 m g.$$

The weight is

$$W = m g.$$

The projection of spring force on the normal is

$$- 2 m g \cos 6 0 = 2 m g \frac {1}{2} = - m g.$$

The projection of weight on the normal is

$$m g \sin 6 0 = m g \frac {\sqrt {3}}{2}.$$

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

$$N - m g + m g \frac {\sqrt {3}}{2} = 0$$ $$N = \left(1 - \frac {\sqrt {3}}{2}\right) m g.$$

Answer: $\left(1 - \frac{\sqrt{3}}{2}\right)mg$

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