Question

One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of ${37}^{\circ }$ with the horizontal as shown in Fig. 6.39. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. $(\sin {37}^{\circ }=3/5)$

Answer 1

If l is the stretched length of the spring, then from figure
$\frac{d}{l}=\cos {37}^{\circ }=\frac{4}{5},i.e.,l=\frac{5}{4}d$
So, the stretch $y=l-d=\frac{5}{4}d-d=\frac{d}{4}$
and $h=l\sin {37}^{\circ }=\frac{5}{4}d\times \frac{3}{5}=\frac{3}{4}d$
Now, taking point B as reference level and applying law of conservation of mechanical energy between A and B,
$E_A=E_B$
or $mgh+\frac{1}{2}k{y}^2=\frac{1}{2}m{v}^2$ [as for B, h=0 and y=0]
or $\frac{3}{4}mgd+\frac{1}{2}k{\left(\frac{d}{4}\right)}^2=\frac{1}{2}m{v}^2$ [as for A, $h=\frac{3}{4}dandy=\frac{1}{4}d$]
or $v=d\sqrt{\frac{3g}{2d.}+\frac{h}{16m}}$