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Q.

A uniform rod of mass m = 5 kg and length L = 90cm rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse J = 3 N.s in a horizontal direction perpendicular to the rod. As a result, the rod attains the momentum p = 3 N.s . Find the force (in newtons) with which one half of the rod will apply on the other half in the process of motion.

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answer is 9.

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Detailed Solution

Angular impulse = change in angular momentum (about point C)3 ℓ2=(m ℓ212)ω  ⇒ ω=3 x 6mℓ =185×0.9=4 rad/sBy Newton's 2 nd Law       T=(m2)(ω2)(ℓ4) ⇒T       = mω2ℓ8=5×16×0.98=9N
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