Physics - Circular Motion Question with Solution | TestHub

Let $-{\vec{v}}_B$ makes an angle $\theta$ with the line CD (a line perpendicular to AB ). ${\vec{v}}_{AB}or{\vec{v}}_A-{\vec{v}}_B$ is along $AB$ i.e components of ${\vec{v}}_{AB}$ perpendicular to $AB$ or along $CD$ is zero. Hence $v\cos \alpha =2v\cos \theta$ Or $\cos \alpha =2\cos \theta$ $\_\_\_\_$ (1)

In triangle $\mathrm{OAB}\frac{\sin \alpha }{r}=\frac{\sin \left(180^{\circ }-\theta \right)}{2r}$

or $\sin \alpha =\frac{\sin \theta }{2}$______(2)

Squaring and adding (1) and (2) we get $1=4{\cos }^2\theta +\frac{1}{4}{\sin }^2\theta$ or $4=16{\cos }^2\theta +\left(1-{\cos }^2\theta \right)$ or $15{\cos }^2\theta =3$ or $\cos \theta =\frac{1}{\sqrt{5}}$ or $\cos \alpha =\frac{2}{\sqrt{5}}$

[from equation (1)]

or $\cot \alpha =2$