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 angular velocity of the rod

• A. $\frac{5v_0}{3l}$
• B. $\frac{3v_0}{5l}$
• C. $\frac{5v_0}{4l}$
• D. $\frac{4v_0}{5l}$

Answer 1

The correct option is A $\frac{5v_0}{3l}$

Let the instantaneous center of rotation (ICR) is as shown,

$v_b=R\times \omega$
where, $v_b=\frac{4}{3}{v}_0\frac{4}{3}{v}_0=lcos\theta \times \omega \frac{4}{3}{v}_0=lcos{37}^o\times \omega \frac{4}{3}{v}_0=l\times \frac{4}{3}\times \omega$
$\omega =\frac{5v_0}{3l}$