Answer to Question #133292 in Mechanics | Relativity for Siddh

Question

Two identical beads are attached to free ends of two identical springs of spring constant k=((2+(3^(1/2)))mg)/((3^(1/2))R). Initially both springs make angle of 60° at fixed point N where one end of both springs are fixed. Normal length of springs is 2R. Where R is radius of smooth ring over which beads are sliding. Ring is placed in vertical plane and beads are at symmetry with respect to vertical line passing through diameter of ring. What is the normal force and relative acceleration between the beads at initial moment?

Answer 1

Explanations & Calculations

• Refer to the sketch
• At the just start of the motion there exists an angular acceleration hence a tangential acceleration but not a angular velocity. It achieves some angular velocity just after the start.
• To calculate the normal force (P) apply Newtons second formula on any of the bead towards the center.

$\qquad\qquad \begin{aligned} \small F &= \small ma\\ \small kx\cos30^0+mg\cos60^0 -P &= \small mr\omega^2\\ \small P&= \small \frac{\sqrt3}{2}kx+\frac{mg}{2}\cdots(1) \end{aligned}$ : no $\small \omega$ at the very beginning

• x is the net extension of the spring that is equal to

$\qquad\qquad \begin{aligned} \small x&= \small 2R\cos30-R \\ &= \small (\sqrt3-1)R \end{aligned}$

• Therefore, spring force is

$\qquad\qquad \begin{aligned} \small kx&= \small \frac{(2+\sqrt3)mg}{\sqrt3R}\times (\sqrt3-1)R\\ &= \small \frac{(\sqrt3+1)mg}{\sqrt3} \end{aligned}$

• Therefore, by (1)

$\qquad\qquad \begin{aligned} \small P &= \small \frac{\sqrt3}{2}\times\frac{(\sqrt3+1)mg}{\sqrt3}+\frac{mg}{2}\\ &= \small \frac{(\sqrt3+2)mg}{2} \end{aligned}$

• Tangential force on each bead is

$\qquad\qquad \begin{aligned}\small f_t&= \small mg\sin60+kx\sin30\\ &= \small \frac{\sqrt3 mg}{2}+\frac{(\sqrt3+1)mg}{\sqrt3}\times \frac{1}{2}\\ &= \small \frac{\sqrt3(\sqrt3+4)mg}{6} \end{aligned}$

• Then the tangential acceleration is

$\qquad\qquad \begin{aligned} \small f_t&=\small ma_t\\ \small a_t &= \small \frac{\sqrt3(\sqrt3+4)g}{6} \end{aligned}$

• Acceleration of the right side bead (relative to the left side one )

$\qquad\qquad \begin{aligned} \small \rightarrow a&= \small 2a_t\cos30\\ &= \small \frac{(\sqrt3+4)g}{2} \end{aligned}$