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

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(A) Let the coordinates of the ends of the ladder be $(x, 0)$ and $(0, y)$. Then $x^2 + y^2 = L^2$. Differentiating with respect to time $t$: $2x(dx/d

Vedclass · Accepted Answer

$2v^2/L^2$

Vedclass · Answer

(A) Let the coordinates of the ends of the ladder be $(x, 0)$ and $(0, y)$. Then $x^2 + y^2 = L^2$. Differentiating with respect to time $t$: $2x(dx/dt) + 2y(dy/dt) = 0$. Given $dx/dt = -v$ (speed of end on floor),so $x(-v) + y(dy/dt) = 0 \Rightarrow dy/dt = xv/y = v \cot \alpha$. The angular velocity $\omega$ of the ladder is given by $\omega = |d\alpha/dt|$. Since $x = L \cos \alpha$,$dx/dt = -L \sin \alpha (d\alpha/dt) = -v$. Thus,$\omega = v / (L \sin \alpha)$. The angular acceleration $\alpha_{ang} = d\omega/dt = (d\omega/d\alpha) \cdot (d\alpha/dt)$. $d\omega/d\alpha = -(v / L \sin^2 \alpha) \cos \alpha$. Since $d\alpha/dt = -\omega = -v / (L \sin \alpha)$,we have $\alpha_{ang} = [-(v \cos \alpha) / (L \sin^2 \alpha)] \cdot [-v / (L \sin \alpha)] = (v^2 \cos \alpha) / (L^2 \sin^3 \alpha)$. At $\alpha = 45^o$: $\alpha_{ang} = (v^2 \cos 45^o) / (L^2 \sin^3 45^o) = (v^2 \cdot (1/\sqrt{2})) / (L^2 \cdot (1/\sqrt{2})^3) = (v^2 / \sqrt{2}) / (L^2 / 2\sqrt{2}) = 2v^2 / L^2$.

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