A uniform rod has mass $m = 2 k g$ and length $l = 13 m$ . One end of the rod is pulled with a constant velocity of $v_{0} = 34 m / s$ along a frictionless horizontal floor in the negative x direction. The other end is moving along a parabolic fixed curve. The equation of the parabola is $x^{2} = 20 y$ . Find the angular velocity of the rod (in rad/s) when the end point ‘B’ is at (10, 5)

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$\tan ϕ = \frac{d y}{d x} = \frac{2 x}{20} = \frac{2 \times 10}{20} = 1 ϕ = 45 ° {\vec{v}}_{B} = {\vec{v}}_{A} + \vec{ω} \times \vec{A B} \Rightarrow \frac{v_{B}}{\sqrt{2}} (- \overset{\land}{i} - \overset{\land}{j}) = 34 (- \overset{\land}{i}) + ω \times 13 {\sin θ \overset{\land}{i} + \cos θ (- \overset{\land}{j})} \Rightarrow (\frac{- v_{B}}{\sqrt{2}}) \overset{\land}{i} + (\frac{- v_{B}}{\sqrt{2}}) \overset{\land}{j} = \overset{\land}{i} {- 34 + 13 ω \times \frac{5}{13}} + (- 13 ω \times \frac{12}{13}) \overset{\land}{j} \Rightarrow \frac{v_{B}}{\sqrt{2}} = 34 - 5 ω & \frac{v_{B}}{\sqrt{2}} = 12 ω \Rightarrow 34 - 5 ω = 12 ω \Rightarrow ω = 2 r a d / s$