The ends $A$ and $B$ of a rod of length $l$ have velocities of magnitudes $\left|\vec{v}_{A}\right|=v$ and $\left|\vec{v}_{B}\right|=2 v$, respectively. If the inclination of $\vec{v}_{A}$ relative to the rod is $\alpha$, find the angular velocity of the rod.

Question

Aicharya · Accepted Answer

(a) We know in a rigid body the velocities of two points along the line joining two should remain same.
i.e., $\left(v_{A}\right)_{||}=\left(v_{B}\right)_{||}$
Hence $v \cos \alpha=2 v \cos \beta$
$\cos \beta=\frac{1}{2} \cos \alpha$ or $\beta=\cos ^{-1}\left(\frac{1}{2} \cos \alpha\right)$
(b) Angular velocity of the rod, $\omega=\frac{\left|\left(\vec{v}_{A B}\right)_{\perp}\right|}{l}=\frac{\left|\left(\vec{v}_{A}-\vec{v}_{B}\right)_{\perp}\right|}{l}$ $\begin{aligned}\left|\vec{v}_{A B}\right| &=\left|\vec{v}_{A}-\vec{v}_{B}\right|=v \sin \alpha-(-2 v \sin \beta) \\ &=v(\sin \alpha+2 \sin \beta) \end{aligned}$ $\Rightarrow \omega=\frac{(\sin \alpha+2 \sin \beta) v}{l}$ (in clockwise direction)
We have calculated $\cos \beta=\frac{\cos \alpha}{2}$. Thus,
$\sin \beta=\sqrt{1-\cos ^{2} \beta} \Rightarrow \sin \beta=\frac{\sqrt{3+\sin ^{2} \alpha}}{2}$
Hence $\omega=\frac{\left(\sin \alpha+\sqrt{3+\sin ^{2} \alpha}\right) v}{l}$ (in clockwise direction)

The ends $A$ and $B$ of a rod of length $l$ have velocities of magnitudes $\left|\vec{v}_{A}\right|=v$ and…

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