JEE Main20208 Jan 2020Morning ShiftPhysicsLaws of MotionActual
A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l . The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:
Options
- Am l ω 2 k - ω m
- Bm l ω 2 k - m ω 2
- Cm l ω 2 k + m ω 2
- Dm l ω 2 k + m ω
Correct answer
B. m l ω 2 k - m ω 2
Step-by-step solution
As we know, the spring force will give the necessary centripetal force for rotation. So, m ω 2 l + x = k x l x + 1 = k m ω 2 Thus, the stretch in the spring x = l m ω 2 k - m ω 2