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A uniform rod of mass $m = 5.0 k g$ length $L = 90 c m$ 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. As a result the rod obtains the momentum $p = 3.0 N . s .$ Find the force with which one half of the rod will act on the other in the process of motion.

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$6 N$

$9 N$

$7 N$

$8 N$

answer is A.

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Detailed Solution

$p \times \frac{L}{2} = (\frac{m L^{2}}{12}) ω \Rightarrow ω = \frac{6 p}{m L}$

Rod will rotate about its c.m., one half exerts centrifugal force on the other half, therefore $F = \frac{m ω^{2}}{2} \times \frac{L}{4}$

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