Question

The ends $A$ and $B$ of a rod of length $l$ have velocities of magnitudes $\left|{\overrightarrow{v}}_A\right|=v$ and $\left|{\overrightarrow{v}}_B\right|=2v$, respectively. If the inclination of ${\overrightarrow{v}}_A$ relative to the rid is $\alpha$, find the:
(a) inclination $\beta$ of ${\overrightarrow{v}}_B$ relative to the rod.
(b) angular velocity of the rod

Answer 1

We know in a rigid body the velocities of two points along the line joining two should remain same.

i.e $v\cos \theta =2v\cos \beta$

$\cos \beta =\frac{1}{2}\cos \alpha or\beta ={\cos }^{-1}\left(\frac{1}{2}\cos \alpha \right)$

(b) For angular velocity of the rod we can write

$\omega =\frac{\left|{\left({\overrightarrow{v}}_{AB}\right)}_{\perp }\right|}{l}=\frac{\left|{({\overrightarrow{v}}_A-{\overrightarrow{v}}_B)}_{\perp }\right|}{l}$

$\left|{\overrightarrow{v}}_{AB}\right|=|{\overrightarrow{v}}_A-{\overrightarrow{v}}_B|=v\sin \alpha -(-2v\sin \beta )=v(\sin \alpha +2\sin \beta )$

$\Rightarrow \omega =\frac{((\sin \alpha +2\sin \beta )v}{l}$ in clockwise direction

We have calculated $\cos \beta =\frac{\cos \alpha }{2}$

Thus,

$\sin \beta =\sqrt{1-{\cos }^2\beta }\Rightarrow \sin \beta =\sqrt{\frac{3+{\sin }^2\alpha }{2}}$

Hence $\omega =\frac{\sin \alpha +\sqrt{3+\sin \alpha }}{l}$ (in clockwise direction)