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A thin horizontal uniform rod $A B$ of mass $m$ and length $l$ can rotate freely about a vertical axis passing through its end $A$ . At a certain moment the end $B$ starts experiencing a constant force $F$ which is always perpendicular to the original position of the stationary rod and directed in a horizontal plane. Find the angular velocity of the rod as a function of its rotation angle $ϕ$ counted relative to the initial position.

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$\sqrt{6 F^{2} \sin ϕ / m l}$

$\sqrt{2 F \sin ϕ / m l}$

$(6 F \sin ϕ / m l)$

$\sqrt{6 F \sin ϕ / m l}$

answer is C.

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Detailed Solution

$For any angular position T, t he torque of the force F a b o u t A τ = F \times l \cos θ T h u s \int_{0}^{ϕ} τ d θ = \frac{1}{2} l ω^{2}$

$\int_{0}^{ϕ} F l \cos θ d θ = \frac{1}{2} \frac{m l^{2}}{3} ω^{2} o r F l \sin ϕ = \frac{m l^{2}}{6} ω^{2} ∴ ω = \sqrt{6 F \sin ϕ / m l}$

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