Question

A smooth ring A of mass m can slide on a fixed horizontal rod XY. A string tied to the ring passes over a fixed pulley B and carries a block C of mass $M(=2m)$ as shown in $Fig.6.286$. T an instant the string between the ring and pulley makes an angle $\theta$ with the rod.
a. Show that if the ring slides with speed $v$, the block descends with speed $v\cos \theta$.
b. With what acceleration will the ring start moving if the system is released from rest with $\theta ={30}^0?$

Answer 1

$AD=x$

$BC=y$

$l=length$

$AB=l-y$

$DB=d$

In$△ABD,{(l-y)}^2=x^2+d^2$

$2(l-y)(-\frac{dy}{dt})=2x\left(\frac{dx}{dt}\right)+0$

$\frac{dy}{dx}=\left(\frac{x}{l-y}\right)(-\frac{dx}{dt})$

$\frac{dy}{dt}=speedofblockc$

$\frac{x}{l-y}=cos\theta$

$speedofblockc=vcos\theta$

$(l-y)(-\frac{d^{2}y}{d{t}^2}){\left(\frac{dy}{dt}\right)}^2=x\left(\frac{d^{2}x}{d{t}^2}\right)+{\left(\frac{dx}{dt}\right)}^2$

$(-\frac{d^{2}y}{d{t}^2})=\frac{x}{l-y}\left(\frac{d^{2}x}{d{t}^2}\right)$

$a_C=a,a_A=2a$

$Tcos{30}^o=m\left(2a\right)$

$T=\frac{4ma}{\sqrt{3}}$

$g-a=\frac{2a}{\sqrt{3}}$

$g=\frac{2a}{\sqrt{3}}+a$

$\frac{2a+\sqrt{3a}}{\sqrt{3}}=g$

$a(2+\sqrt{3})=g\sqrt{3}$

$a=\frac{g\sqrt{3}}{(2+\sqrt{3})}$