Question

A ladder of length $L$ 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 $v$ and the ladder makes an angle $\theta ={30}^{\circ }$ with horizontal. Then, the speed of the ladder's centre of mass must be-

• A. $\frac{\sqrt{3}}{2}v$
• B. $\frac{v}{2}$
• C. $v$
• D. $2v$

Answer 1

The correct option is C $v$

Let $v^{\prime }$ be the velocity of end $\text{A}$.



Using, the properties of rigid body,

Distance $\text{AB}=L$ = constant

And velocity along the length of the rod remains the same,

$v^{\prime }\cos {60}^{\circ }=v\cos {30}^{\circ }$

$v^{\prime }=\frac{v\cos {30}^{\circ }}{\cos {60}^{\circ }}$

$\therefore v^{\prime }=\sqrt{3}v$

Hence, velocity of the ends of the rod $\text{A}$ and $\text{B}$ are $v$ and $\sqrt{3}v$ respectively.

$\therefore \overrightarrow{v_A}=\frac{dx}{dt}=v\hat{i}$

$\therefore \overrightarrow{v_B}=\frac{dy}{dt}=-\sqrt{3}v\hat{j}$

Now, coordinates of the COM of the rod is,

${\overrightarrow{r}}_{com}=\frac{x}{2}\hat{i}+\frac{y}{2}\hat{j}$

On differentiating the above equation we get,

$\therefore {\overrightarrow{v}}_{com}=\frac{{\overrightarrow{dr}}_{com}}{dt}=\frac{1}{2}[\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}]$

${\overrightarrow{v}}_{com}=\frac{1}{2}[v\hat{i}-\sqrt{3}v\hat{j}]$

$\therefore |v_{com}|=\left[\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right]v=v$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(C\right)$ is the correct answer.