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Question Number 170005 by ajfour last updated on 13/May/22

Commented by ajfour last updated on 14/May/22

$$Amovesincirclewith$$$$speedu,whileBdownthe$$$$verticaldiameterwithspeed$$$$u\cos \theta ,where\theta isshown,$$$$whattomean.Giveninitial$$$$positions,iftheycollidein$$$$P(0,a),findb/a.$$

Commented by mr W last updated on 14/May/22

$$\frac{b-a}{u\cos \theta }=\frac{\pi a}{2u}$$$$\frac{(b-a)\sqrt{b^2+a^2}}{b}=\frac{\pi a}{2}$$$$let\lambda =\frac{b}{a}$$$$(\lambda -1)\sqrt{1+{\lambda }^2}=\frac{\pi \lambda }{2}$$$$\Rightarrow \lambda \approx 2.4547$$

Commented by ajfour last updated on 14/May/22

$$\theta isvariable,along$$$$instantaeousBA;sir.$$

Answered by ajfour last updated on 14/May/22

$$letattimet,$$$$A(p,q)\mathrm{\forall }{p}^2+q^2=a^2$$$$B(0,y)$$$$\frac{ds}{dt}=u$$$$\tan \theta =\frac{p}{y-q}$$$$u\cos \theta =-\frac{dy}{dt}=-\frac{udy}{ds}$$$$\frac{1}{\sqrt{1+{\left(\frac{a\cos \varphi }{y-a\sin \varphi }\right)}^2}}=-\frac{dy}{ad\varphi }$$$$\Rightarrow a^2{\left(\frac{d\varphi }{dy}\right)}^2=1+{\left(\frac{a\cos \varphi }{y-a\sin \varphi }\right)}^2$$$$.......$$$$$$

Answered by mr W last updated on 14/May/22

Commented by mr W last updated on 14/May/22

$$attimet:$$$$Pat\phi =\frac{ut}{a}=\omega twith\omega =\frac{u}{a}$$$$Q(0,h)$$$$\tan \theta =\frac{a\cos \phi }{h-a\sin \phi }=\frac{a\cos \omega t}{h-a\sin \omega t}$$$$\frac{dh}{dt}=-u\cos \theta =-\frac{u}{\sqrt{1+{\left(\frac{a\cos \omega t}{h-a\sin \omega t}\right)}^2}}$$$$let\xi =\frac{h}{a}$$$$\frac{d\xi }{dt}+\frac{\omega }{\sqrt{1+{\left(\frac{\cos \omega t}{\xi -\sin \omega t}\right)}^2}}=0$$$$or$$$$\frac{d\xi }{d\phi }+\frac{1}{\sqrt{1+{\left(\frac{\cos \phi }{\xi -\sin \phi }\right)}^2}}=0$$$$\xi (\phi =0)=\frac{b}{a}=\lambda$$$$\xi (\phi =\frac{\pi }{2})=1$$$$....$$

Commented by ajfour last updated on 14/May/22

$$thankssir,difficultd.e.!$$

Commented by mr W last updated on 14/May/22

$$yes!i{don}^{\prime }tknowhowtosolve.$$

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