Question

A ladder is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is $3\text{m/s}$ and the ladder makes an angle ${30}^{\circ }$ with horizontal. Then, the speed of the ladder's centre of mass must be

• A. $6\text{m/s}$
• B. $\sqrt{3}\text{m/s}$
• C. $\frac{3}{\sqrt{2}}\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is D $3\text{m/s}$

Let $l$ be the length of ladder.


$u$ = speed of point B along horizontal floor
$v$ = speed of point A along vertical wall.
Here, length of ladder, $AB=l$ = constant.
$\therefore$ velocity component of $v$ along AB = component of $u$ along AB
$\Rightarrow v\cos {60}^{\circ }=u\cos {30}^{\circ }$
$\Rightarrow \frac{v}{2}=\frac{u\sqrt{3}}{2}$
$\Rightarrow v=\sqrt{3}u$

Now,

$\frac{dx}{dt}=u$ and $-\frac{dy}{dt}=v=\sqrt{3}u$

Since C is the mid point of ladder AB, so position vector
$\overrightarrow{r_c}=\frac{x}{2}\hat{i}+\frac{y}{2}\hat{j}$
[taking origin at bottom of vertical wall]
and $\overrightarrow{v_c}=\frac{d\overrightarrow{r_c}}{dt}=\frac{1}{2}[\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}]$
$\overrightarrow{v_c}=\frac{1}{2}(u\hat{i}-\sqrt{3}u\hat{j})$
$\overrightarrow{v_c}=\frac{1}{2}(3\hat{i}-3\sqrt{3}\hat{j})(\because u=3\text{m/s})$
Therefore, Magnitude of velocity of COM
$|\overrightarrow{v_c}|=\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{9}{4}+\frac{27}{4}}=3\text{m/s}$