Question

A uniform rod of length $l$ is spinning with an angular velocity $\omega =2vl$ while its centre of mass moves with a velocity $v$. Find the velocity of the end $A$ of the rod.

Answer 1

Velocity of end $A,{\overrightarrow{v}}_A={\overrightarrow{v}}_{AC}+{\overrightarrow{v}}_C$

Hence the velocity of $A$ is

$v_A=\sqrt{v_{AC}^2+v_C^2+2v_{AC}{v}_C\cos \theta }$

We know $v_C=v,v_{AC}=\frac{1}{2}\omega ;\theta ={90}^o+{30}^o={120}^o$

Hence, $v_A=\sqrt{v^2+\frac{l^2{\omega }^2}{4}+lv\omega \cos {120}^o}$

But $\omega =\frac{2v}{l},v_{AC}=\frac{1}{2}\omega =\frac{1}{2}\left(\frac{2v}{l}\right)=v$

which gives $v_A=v$

Let ${\overrightarrow{v}}_A$ make angle $\varphi$ with the direction of $v_C$. Then

$\varphi ={\tan }^{-1}\frac{v\sin \theta }{v_{AC}\cos \theta +v_C}$

We get $\varphi ={\tan }^{-1}\frac{v\sin {120}^o}{v\cos {120}^o+v}=\frac{\pi }{2}$