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 $\theta ={30}^o$ with horizontal. Then, the speed of the ladder's centre of mass must be

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

Answer 1

The correct option is C $v$

Let the length of the ladder be $L$.

Since the ladder is constrained to move and the x and y-axis, let the position co-ordinate of the ends of the ladder be (x,0), (y,0).

$AB=\sqrt{x^2+y^2}=L$; where $L$ is constant.

$\frac{1}{2}\left(x^2+y^2{)}^{\frac{-1}{2}}2\right(x\frac{dx}{dt}+y\frac{dy}{dt})=0$

$\Rightarrow (x\frac{dx}{dt}+y\frac{dy}{dt})=0$ ......(1)

Let the velocity of the upper end along the y-axis is $v_y$.

The velocity of the lower end along the x-axis is$v_x=v$ as given.

Component of $v_y$ along AB $=$ Component of $v$ along AB.

$v_y\cos {60}^0=v\sin {60}^0$

$\Rightarrow v_y\times \frac{1}{2}=v\times \frac{\sqrt{3}}{2}$

$\Rightarrow v_y=\sqrt{3}v$

The end points of the ladder have co-ordinates $(x,0),(0,y)$.

Position co-ordinate of the cm of the ladder $=\frac{1}{2}(x+y)$

Since $v_y$ is along the -ve y-axis, we take it to be negative.

$v_{cm}=\frac{d\overrightarrow{r_{cm}}}{dt}=\frac{1}{2}(\hat{i}\frac{dx}{dt}+\hat{j}\frac{dy}{dt})$

$\Rightarrow v_{cm}=\frac{1}{2}(\hat{i}{v}_x+\hat{j}{v}_y)$

$\Rightarrow v_{cm}=\frac{1}{2}(\hat{i}v-\hat{j}\sqrt{3}v)$

$\therefore$ Magnitude of the velocity of cm: $\left|v_{cm}\right|=\frac{1}{2}\sqrt{(}(v^2+(\sqrt{3}v{)}^2)=v$