A ladder of length L is slipping with its end | vedclass

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(D) Let the ladder be $AB$ with length $L$. The end $A$ is on the floor and $B$ is on the wall. The instantaneous axis of rotation $(IAR)$ is at point

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None

Vedclass · Answer

(D) Let the ladder be $AB$ with length $L$. The end $A$ is on the floor and $B$ is on the wall. The instantaneous axis of rotation $(IAR)$ is at point $O$,which is the intersection of the perpendiculars to the velocities of $A$ and $B$.Since $A$ moves horizontally with speed $v$ and $B$ moves vertically,the $IAR$ is at a distance $x = L \sin \alpha$ from the wall and $y = L \cos \alpha$ from the floor.The angular velocity $\omega$ of the ladder is given by $\omega = v / (L \cos \alpha)$.Given $\alpha = 30^o$,$\cos 30^o = \sqrt{3}/2$,so $\omega = v / (L \cdot \sqrt{3}/2) = 2v / (L \sqrt{3})$.The center of mass $C$ is at a distance $r = L/2$ from the $IAR$ $O$.The speed of the center of mass is $v_c = r \omega = (L/2) \cdot (2v / (L \sqrt{3})) = v / \sqrt{3}$.Wait,re-evaluating using the provided diagram: The distance from the $IAR$ $O$ to the center $C$ is $r = \sqrt{(L/4)^2 + (\sqrt{3}L/4)^2} = \sqrt{L^2/16 + 3L^2/16} = \sqrt{4L^2/16} = L/2$.The angular velocity $\omega = v / (L \cos 30^o) = v / (L \sqrt{3}/2) = 2v / (L \sqrt{3})$.Thus,$v_c = (L/2) \cdot (2v / (L \sqrt{3})) = v / \sqrt{3}$.Since $v/\sqrt{3}$ is not among the options,let's re-check the calculation. If the velocity $v$ is at the end making $30^o$ with the floor,the distance to $IAR$ is $L \cos 30^o$. $\omega = v / (L \cos 30^o)$. The distance from $IAR$ to center is $L/2$. $v_c = (L/2) \cdot (v / (L \cos 30^o)) = v / (2 \cos 30^o) = v / \sqrt{3}$. Given the options,the correct answer is $v$ if the geometry was different,but based on the standard ladder problem,it is $v/\sqrt{3}$. Since this is not an option,we select 'None'.

$A$ 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 $\alpha = 30^o$ with the horizontal. Then the speed of the ladder's center must be

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