Question Number 31133 by Tinkutara last updated on 02/Mar/18
Commented by ajfour last updated on 03/Mar/18
Commented by ajfour last updated on 03/Mar/18
$$\omega_{{max}} =\frac{\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }}{{s}} \\ $$$${s}={r}\mid\mathrm{sin}\:\left(\phi−\theta\right)\mid \\ $$$$\:\:=\mid\mathrm{sin}\:\phi\left({r}\mathrm{cos}\:\theta\right)−\mathrm{cos}\:\phi\left({r}\mathrm{sin}\:\theta\right)\mid \\ $$$$\:\:=\mid\frac{{va}}{\:\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }}−\frac{{ub}}{\:\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }}\mid \\ $$$$\Rightarrow\:\:\omega_{{max}} =\frac{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }{\mid{va}−{ub}\mid}\:\:. \\ $$
Commented by Tinkutara last updated on 03/Mar/18
Thanks Sir! ����