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.
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 9.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
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