Question

End $A$ of a rod $AB$ is being pulled on the floor with a constant velocity $v_0$ as shown. Taking the length of the rod as $l$, at an instant when the rod makes an angle ${37}^{\circ }$ with the horizontal, calculate



The velocity of the CM of the rod

• A. $\frac{5}{7}{v}_0$ at $ta{n}^{-1}\frac{4}{3}$ below horizontal
• B. $\frac{5}{7}{v}_0$ at $ta{n}^{-1}\frac{3}{4}$ below horizontal
• C. $\frac{5}{6}{v}_0$ at $ta{n}^{-1}\frac{3}{4}$ below horizontal
• D. $\frac{4}{5}{v}_0$ at $ta{n}^{-1}\frac{3}{4}$ below horizontal

Answer 1

The correct option is D $\frac{4}{5}{v}_0$ at $ta{n}^{-1}\frac{3}{4}$ below horizontal

The center of mass of the rod will be at a distance of $\frac{l}{2}$.

where, $ICR$ = insterntenious center of rotation
Velocities at point A:

Along the length
$V_{cm}=V_{0}cos\theta$ ($\theta ={37}^o$)
$V_{cm}=V_0\times \frac{4}{5}$
$V_{cm}=\frac{4}{5}{V}_0$

Hence, option D is correct option.