A smooth ring of mass m and radius R = 1 m is pulled at P with a constant acceleration a = 4 m s-2 on a horizontal surface such that the plane of the ring lies on the surface. Find the angular acceleration of the ring at the given position. (in rad/s2)
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 2.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The pseudo force acting at the CM of the ring is F→ps=−ma i^ which produces a torque about P given as τ→ps=−(ma)(R)k^This produces an angular acceleration α→=τ→psIP=−maR k^2mR2 ∵IP=mR2+mR2=2mR2=−a2Rk^α=42×1=2 rad/s2