question_answer

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A ladder PQ of length 5 m is inclined to a vertical wall is slipping over horizontal floor with velocity of \[2\,m{{s}^{-1}}.\]

Velocity of centre of mass of ladder, when Q is 3 m from the wall is

A) \[2\,m{{s}^{-\,1}}\]

B) \[1.25\,m{{s}^{-\,1}}\]

C) \[175\,m{{s}^{-\,1}}\]

D) \[2.75\,m{{s}^{-\,1}}\]

Correct Answer: B

Solution :

At any instant, \[y=\sqrt{{{L}^{2}}-{{x}^{2}}}\]

\[\Rightarrow \]\[\frac{dy}{dt}=\frac{-x}{\sqrt{{{L}^{2}}-{{x}^{2}}}}\cdot \frac{dx}{dt}=\frac{-\,3}{\sqrt{{{5}^{2}}-{{3}^{2}}}}\times 2\]

or         \[\frac{dy}{dt}=-\frac{3}{2}m{{s}^{-1}}\]

At any instant, position of centre of mass is at \[\left( \frac{x}{2},\frac{y}{2} \right)\]

\[{{({{v}_{CM}})}_{x}}={{v}_{x}}=\frac{d}{dt}\left( \frac{x}{2} \right)\]or \[{{v}_{x}}=\frac{1dx}{2dt}=1m{{s}^{-\,1}}\]

Similarly, \[{{v}_{y}}=\frac{1}{2}\left( \frac{3}{2} \right)=\frac{3}{4}m{{s}^{-\,1}}\]

So, velocity of centre of mass of ladder is

\[v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}=1.25m{{s}^{-1}}\]