Question

A ring of mass $m=1 \,kg$ can slide over a smooth vertical rod. A light string attached to the ring passing over a smooth fixed pulley at a distance of $L=0.7\, m$ from the rod is shown in the figure. At the other end of the string, mass $M=5\, kg$ is attached, lying over a smooth fixed inclined plane of inclination angle $37^{\circ}$. The ring is held in level with the pulley and released. Determine the velocity of ring (in $m / s$ ) when the string makes an angle $\left(\alpha=37^{\circ}\right)$ with the horizontal. $\left[\sin 37^{\circ}=0.6\right.$ ]

Answer 1

Let $x$ is the vertical distance covered by the ring. Then
$x=L \tan 37^{\circ}=0.7 \times \frac{3}{4}$
$\Delta l=L \sec 37^{\circ}-L=L\left(\sec 37^{\circ}-1\right)$
$\Rightarrow \frac{L}{4}=\Delta l$

$\Delta l=$ distance moved by block $M$
Now, from constraint relation
$v_{M}=v_{r} \cos 37^{\circ}=\frac{4}{5} v_{r}$ ...(ii)
$v_{r}=$ velocity of ring,
$v_{M}=$ velocity of the block at this instant
From work-energy theorem, we get $W_{\text {gravity }}=\Delta KE$
$-m g x+M g \Delta l \sin 37^{\circ}+\frac{1}{2} m v_{r}^{2}+\frac{1}{2} M v_{M}^{2}=0$ ...(iii)
On solving equations (i) and (ii), we get
$v{r}=0\, m / s$