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A uniform rod has mass $m = 1 k g$ and length $l = 13 m$ . One end of the rod is pulled with a constant velocity of $v_{0} = 1 m / s$ 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 $x^{2} = 20 y .$ If the angular velocity of the rod (in minutes/s) when the end point ‘B’ is at (10, 5) is $ω$ . Find $\frac{ω}{10}$

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answer is 20.23.

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Detailed Solution

$= v_{0} . \frac{12}{13} x^{2} = 20 y$

$\begin{array}{l} \frac{d y}{d x} = \frac{x}{10} = \tan θ \\ θ = 45^{\circ} \end{array}$

$v_{0} \frac{12}{13} = V \cos θ = \frac{V}{\sqrt{2}} V = \frac{12 \sqrt{2}}{13}$

$ω_{r e l} = \frac{V \sin θ + V_{0} \sin α}{13}$

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