Question

The end $B$ of the rod $AB$ which makes angle $\theta$ with the floor is being pulled with a constant velocity $v_0$ as shown. The length of the rod is $l$

• A. At $\theta ={37}^{\circ }$ velocity of end A is $\frac{4}{3}{v}_0$ downwards
• B. At $\theta ={37}^{\circ }$ angular velocity of rod is $\frac{5v_0}{3l}$
• C. Angular velocity of rod is constant
• D. Velocity of end $A$ is constant

Answer 1

Answer: (A), (B)

The correct options are
A At $\theta ={37}^{\circ }$ velocity of end A is $\frac{4}{3}{v}_0$ downwards
B At $\theta ={37}^{\circ }$ angular velocity of rod is $\frac{5v_0}{3l}$

Velocity component along $AB=0$
$\therefore v_0\cos \theta =v\sin \theta$ or $v=v_0\cot \theta =f_1\left(\theta \right)$
at $\theta ={37}^{\circ },v=\frac{4}{3}{v}_0$
$\omega =\frac{Relativevelocity\perp toAB}{l}$
$=\frac{v_0\sin \theta +v\cos \theta }{l}=f_2\left(\theta \right)$
$=\frac{\left(v_0\right)\left(3/5\right)+\left(4/3v_0\right)\left(4/5\right)}{l}$
$=\frac{5v_0}{3I}$