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 as shown in Fig. 8.217.
At the other end of the string mass M = 5 kg is attached, lying over a smooth fixed inclined plane of inclination angle ${37}^0$. The ring is held in level with the pulley and released. Determine the velocity of ring when the string makes an angle ($\alpha ={37}^0$) with the horizontal. [sin ${37}^0$ = 0.6].

Answer 1

m=1kg

M=5kg

let, after reaching the final position

L increased by $△L$, the ring descends by $x$

From diagram , $\tan {37}^{\circ }=x/L$ on $x=\frac{3L}{4}$.

$\cos {37}^{\circ }=\frac{L}{L+\Delta L}\Rightarrow \Delta L=L(\sec {37}^{\circ }-1)$

let, at final position, velocity of ring $=V_{\gamma }$ so, from diagram velocity of $M^{\prime }=V_{\gamma }\cos {53}^{\circ }$

$=\frac{3vr}{5}$

now change in potential Energy = change in kinetic energy

$-Mg\sin {37}^{\circ }x\Delta 1+mgx=\frac{1}{2}m{v}_p^2+\frac{1}{2}$ $M{\left(\frac{3vp}{5}\right)}^2$

$0\pi -5\times 10\times \frac{3}{8}\times \frac{2}{9}+10\times \frac{3L}{4}=\frac{1}{2}(v_{\gamma }^2+5\times \frac{9v_r^2}{25})$

$\text{on}{v}_{\gamma }=0$

Ans- Velocity =0m/s