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Question Number 21038 by ajfour last updated on 10/Sep/17

Commented by ajfour last updated on 10/Sep/17

$$Findangularvelocityandangular$$$$accelerationofrodifendAmoves$$$$totherightwithconstantvelocity$$$$v.$$

Answered by mrW1 last updated on 11/Sep/17

$$\tan \theta =\frac{\mathrm{h}}{\mathrm{x}}$$$$\frac{1}{{\cos }^2\theta }\times \frac{\mathrm{d}\theta }{\mathrm{dt}}=-\frac{\mathrm{h}}{{\mathrm{x}}^2}\times \frac{\mathrm{dx}}{\mathrm{dt}}=-{\left(\frac{\mathrm{h}}{\mathrm{x}}\right)}^2\times \frac{1}{\mathrm{h}}\times \frac{\mathrm{dx}}{\mathrm{dt}}$$$$\frac{1}{{\cos }^2\theta }\times \frac{\mathrm{d}\theta }{\mathrm{dt}}=-{\tan }^2\theta \times \frac{1}{\mathrm{h}}\times \frac{\mathrm{dx}}{\mathrm{dt}}$$$$\frac{\mathrm{d}\theta }{\mathrm{dt}}=-{\sin }^2\theta \times \frac{1}{\mathrm{h}}\times \frac{\mathrm{dx}}{\mathrm{dt}}$$$$\Rightarrow \mathrm{angular}\mathrm{velocity}\omega ={\sin }^2\theta \times \frac{\mathrm{v}}{\mathrm{h}}$$$$\frac{\mathrm{d}\omega }{\mathrm{dt}}=2\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}\times \frac{\mathrm{v}}{\mathrm{h}}+{\sin }^2\theta \times \frac{\mathrm{dv}}{\mathrm{dt}}$$$$\frac{\mathrm{dv}}{\mathrm{dt}}=0$$$$\frac{\mathrm{d}\theta }{\mathrm{dt}}=\omega ={\sin }^2\theta \times \frac{\mathrm{v}}{\mathrm{h}}$$$$\Rightarrow \mathrm{angular}\mathrm{acceleration}\frac{\mathrm{d}\omega }{\mathrm{dt}}=\sin 2\theta {\sin }^2\theta {\left(\frac{\mathrm{v}}{\mathrm{h}}\right)}^2$$

Commented by ajfour last updated on 11/Sep/17

$$thankssir,ihadmissedmarking$$$$\mathit{h}inquestion,thanksfor$$$$assumingsoandsolvingperfectly.$$

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