Question

The end $B$ of uniform rod $AB$ which makes angle $\theta$ with the floor is being pulled with a velocity $v_0$ as shown. Taking the length of rod as $l$, calculate the following at the instant when $\theta ={37}^{\circ }$.
(a) The velocity of end $A$
(b) The angular velocity of rod
(c) Velocity of $CM$ of the rod.

Answer 1

(i)$v_A=v_B\tan \theta =v+\tan 37°=\frac{3}{4}$

(ii) Angular speed $\omega$

$v_A=\omega \times x=\omega \times \frac{l}{2}\cos 37°=\omega \times l\frac{2}{5}=\omega \frac{v_A}{\left(\frac{2l}{5}\right)}=\frac{3}{4v}\frac{v}{\frac{2l}{5}}=\frac{15v}{8l}$

(iii)Let downward velocity of com is vd

$\frac{\frac{1}{2}}{v_d\cos \theta }=\frac{l}{v_A}=\cos \theta$

$vd=\frac{1}{2}vA=\frac{3}{8}v$

Horizontal velocity $=v_h=\frac{v}{2}$

$(\frac{v}{2}\hat{i}-\frac{3}{8}\hat{j})$