All Problems

Dynamics of a solid body

Problem 1.273

A uniform rod of mass m=5.0 kgm=5.0 \mathrm{~kg} and length l=90 cml=90 \mathrm{~cm} rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse J=3.0 NsJ=3.0 \mathrm{~N} \cdot \mathrm{s} in a horizontal direction perpendicular to the rod. As a result, the rod obtains the momentum p=3.0 Nsp=3.0 \mathrm{~N} \cdot \mathrm{s}. Find the force with which one half of the rod will act on the other in the process of motion.

Reveal Answer
F=9/222/ml=9 NF=9 / 2^{22} / m l=9 \mathrm{~N}