A thin vertical uniform wooden rod is pivoted | vedclass

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Question

(B) Let $L$ be the total length of the rod,$A$ be its cross-sectional area,and $\rho_w$ and $\rho_r$ be the densities of water and the rod respectivel

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $L$ be the total length of the rod,$A$ be its cross-sectional area,and $ho_w$ and $ho_r$ be the densities of water and the rod respectively. Given $ho_w = \frac{9}{5} ho_r$. Let $x$ be the submerged length of the rod.The weight of the rod acts at its center of mass,which is at a distance $L/2$ from the pivot. The buoyant force $F_B$ acts at the center of the submerged part,which is at a distance $(L - x/2)$ from the pivot.For small angular displacement $	heta$,the torque due to gravity is $	au_g = (ho_r A L g) \frac{L}{2} \sin 	heta$.The torque due to the buoyant force is $	au_B = F_B (L - x/2) \sin 	heta$,where $F_B = ho_w (A x) g$.Equilibrium is unstable when the net restoring torque becomes zero or negative. The condition for stability is that the restoring torque (gravity) must be greater than or equal to the destabilizing torque (buoyancy).$(ho_r A L g) \frac{L}{2} \sin 	heta = (ho_w A x g) (L - x/2) \sin 	heta$$ho_r L^2 / 2 = ho_w x (L - x/2)$Substitute $ho_w = \frac{9}{5} ho_r$:$ho_r L^2 / 2 = \frac{9}{5} ho_r x (L - x/2)$$L^2 / 2 = \frac{9}{5} (Lx - x^2/2)$$5L^2 = 18Lx - 9x^2$$9x^2 - 18Lx + 5L^2 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{18L \pm \sqrt{324L^2 - 180L^2}}{18} = \frac{18L \pm \sqrt{144L^2}}{18} = \frac{18L \pm 12L}{18}$$x = \frac{30L}{18} = \frac{5}{3}L$ (not possible as $x Thus,$x = L/3$,which means $L/x = 3$.

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