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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

$${\omega }_{max}=\frac{\sqrt{u^2+v^2}}{s}$$$$s=r\mid \sin (\varphi -\theta )\mid$$$$=\mid \sin \varphi \left(r\cos \theta \right)-\cos \varphi \left(r\sin \theta \right)\mid$$$$=\mid \frac{va}{\sqrt{u^2+v^2}}-\frac{ub}{\sqrt{u^2+v^2}}\mid$$$$\Rightarrow {\omega }_{max}=\frac{u^2+v^2}{\mid va-ub\mid }.$$

Commented by Tinkutara last updated on 03/Mar/18

Thanks Sir! ����

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