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Question Number 90819 by ajfour last updated on 26/Apr/20

Commented by ajfour last updated on 26/Apr/20

$$massofrodisM,and\alpha =0,but$$$$rodrotateswithconstantangular$$$$veloity{\omega }_0,thenfindrelative$$$$speedofsmallmassmwith$$$$respecttorod,asitreaches$$$$theextremeend(a=l)fromthe$$$$hingedend(a=0).$$

Commented by ajfour last updated on 26/Apr/20

$$thisonefromarenowned$$$$physicsbook,haveanswer.$$

Answered by ajfour last updated on 26/Apr/20

Answered by mr W last updated on 26/Apr/20

$$whensleeveisatpositionx:$$$$angularspeedofrod=\omega$$$$speedofsleevealongrod=u$$$$acc.ofsleevealongrod=a=u\frac{du}{dx}$$$$I{\omega }_0=(I+{mx}^2)\omega$$$$\Rightarrow \omega =\frac{\frac{{Ml}^2}{3}}{\frac{{Ml}^2}{3}+{mx}^2}{\omega }_0=\frac{{Ml}^2}{{Ml}^2+3{mx}^2}{\omega }_0$$$$ma-m{\omega }^{2}x=0$$$$\Rightarrow u\frac{du}{dx}-{\omega }^{2}x=0$$$$\Rightarrow u\frac{du}{dx}-{\left(\frac{{Ml}^2{\omega }_0}{{Ml}^2+3{mx}^2}\right)}^{2}x=0$$$$\Rightarrow {\int }_0^{u_{max}}udu={\int }_0^l{\left(\frac{{Ml}^2{\omega }_0}{{Ml}^2+3{mx}^2}\right)}^{2}xdx$$$$\Rightarrow \frac{u_{mac}^2}{2}={\left(\frac{{Ml}^2{\omega }_0}{3m}\right)}^2{\int }_0^l\frac{xdx}{{(\frac{{Ml}^2}{3m}+x^2)}^2}$$$$\Rightarrow u_{max}^2={\left(\frac{{Ml}^2{\omega }_0}{3m}\right)}^2{\int }_0^l\frac{d(\frac{{Ml}^2}{3m}+x^2)}{{(\frac{{Ml}^2}{3m}+x^2)}^2}$$$$\Rightarrow u_{max}^2={\left(\frac{{Ml}^2{\omega }_0}{3m}\right)}^2{\left[\frac{1}{\frac{{Ml}^2}{3m}+x^2}\right]}_l^0$$$$\Rightarrow u_{max}^2={\left(\frac{Ml{\omega }_0}{3m}\right)}^2[\frac{1}{\frac{M}{3m}}-\frac{1}{\frac{M}{3m}+1}]$$$$\Rightarrow u_{max}^2={\left(\frac{Ml{\omega }_0}{3m}\right)}^2\left[\frac{1}{\frac{M}{3m}(\frac{M}{3m}+1)}\right]$$$$\Rightarrow u_{max}=l{\omega }_0\sqrt{\frac{M}{M+3m}}$$

Commented by ajfour last updated on 26/Apr/20

$$ExcellentSir!Verybasicway.$$$$thanksSir!$$

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