Question

In the figure shown one end of a light spring of natural length $l_0=\sqrt{\frac{5}{8}}m(\approx 0.8m)$ is fixed at a point D and other end is attached to the centre B of a uniform rod AF of length $l_0/\sqrt{3}$ and mass $10kg$. The rod is free to rotate in a vertical plane about a fixed horizontal axis passing through the end A of the rod. The rod is held at rest in horizontal position and the spring is in relaxed state. It is found that, when the rod is released to move it makes an angle of ${60}^o$ with the horizontal when it comes to rest for the first time. Find the
(a) the maximum elongation in the spring. (approximately)
(b) the spring constant. (approximately)

Answer 1

(a) In triangle ABC
$AB=BC$
From geometry in $△ABC$
$\therefore BC=\frac{l_0}{2\sqrt{3}}$
So, $\angle ABC=\angle ACB={60}^o$
Also in triangle $BCD,\angle DBC={150}^o$
Applying cosine law
${DC}^2={DB}^2+{BC}^2-2DB.BC\cos {150}^o$
$=\frac{5}{8}+\frac{1}{12}\times \frac{5}{8}+2\sqrt{\frac{5}{8}}.\frac{1}{2\sqrt{3}}\times \sqrt{\frac{5}{8}}\times \frac{\sqrt{3}}{2}$
$DC\approx 1m$
Expansion in the spring is $x=1-0.8\approx 0.2m$
Vertical displacement of centre of mass of the rod
$BE=AC\sin {60}^o=\frac{l_0}{2\sqrt{3}}\left(\frac{\sqrt{3}}{2}\right)=0.2m$
Using conservation of mechanical energy
$Mg\left(BE\right)=\frac{1}{2}k{x}^2$
$10\times 10\times 0.2=\frac{1}{2}k{\left(0.2\right)}^2\Rightarrow K={10}^{3}N/m$

$\Rightarrow k\approx {10}^{3}N/m$