A uniform rod of mass m = 5.0kg and length t = 90cm rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse J = 3.0 N -S in a horizontal direction perpendicular to the rod. Find the force with which one half of the rod will act on the other in the process of motion in newtons.

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answer is 9.

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Detailed Solution

$\begin{array}{l} j \times \frac{1}{2} = \frac{{mℓ}^{2}}{12} \times ω \\ ω = \frac{6 j}{μℓ} \\ F = \int_{0}^{1 / 2} {dmω}^{2} x \\ F = \int_{0}^{1 / 2} \frac{m}{ℓ} {dxω}^{2} x \\ = 9 N \end{array}$

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