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 the 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

$\left(a\right)$ We use her the constrain relation.

$v_B=$velocity of end B

$v_0=$velocity of end A

Component of $v_0$ along the rod $=$ component of $v_0$ along the rod

$⟹v_{0}cos{37}^0=v_{B}cos{53}^0$

$⟹v_B=v_0\frac{cos{37}^0}{sin{37}^0}=v_{0}cos{37}^0$

$⟹v_B=\frac{4v_0}{3}$

$\left(b\right)$ Angular velocity of rod

$=\frac{4v_0}{3l}$

$\left(c\right)$ Velocity of centre of mass has two components $v_0$ along x-axis $\&\frac{4v_0}{3}$ along negative y-axis.

So, ${\overrightarrow{V}}_{CM}=v_0\hat{i}-\frac{4v_0}{3}\hat{j}$

$|v_{CM}|=\sqrt{(v_0{)}^2+\frac{16(v_0{)}^2}{3^2}}$

$=\frac{5}{3}{v}_0$