A ladder AP of length 5, m is inclined to a v | vedclass

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(A) Let the position of end $A$ be $(x, 0)$ and end $P$ be $(0, y)$. The length of the ladder is $L = 5\, m$,so $x^2 + y^2 = L^2 = 25$.Given $x = 3\,

Vedclass · Accepted Answer

$1.25$

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(A) Let the position of end $A$ be $(x, 0)$ and end $P$ be $(0, y)$. The length of the ladder is $L = 5\, m$,so $x^2 + y^2 = L^2 = 25$.Given $x = 3\, m$,then $y = \sqrt{25 - 3^2} = 4\, m$.Differentiating $x^2 + y^2 = 25$ with respect to time $t$,we get $2x(dx/dt) + 2y(dy/dt) = 0$.Given $dx/dt = -2\, m/s$ (moving away from the wall),so $2(3)(-2) + 2(4)(dy/dt) = 0$,which gives $dy/dt = 1.5\, m/s$.The coordinates of the center of mass are $x_{cm} = x/2$ and $y_{cm} = y/2$.The velocity components of the center of mass are $v_{x,cm} = (1/2)(dx/dt) = (1/2)(-2) = -1\, m/s$ and $v_{y,cm} = (1/2)(dy/dt) = (1/2)(1.5) = 0.75\, m/s$.The magnitude of the velocity of the center of mass is $V_{cm} = \sqrt{v_{x,cm}^2 + v_{y,cm}^2} = \sqrt{(-1)^2 + (0.75)^2} = \sqrt{1 + 0.5625} = \sqrt{1.5625} = 1.25\, m/s$.

$A$ ladder $AP$ of length $5\, m$ is inclined to a vertical wall and is slipping over a horizontal surface with a velocity of $2\, m/s$. When the end $A$ is at a distance of $3\, m$ from the wall,what is the velocity of the center of mass $(C.M.)$ of the ladder at this moment (in $, m/s$)?

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