
As we know, the spring force will give the necessary centripetal force for rotation.
So,
mω2(l+x)=kx
(xl+1)=mω2k
Thus, the stretch in the spring
x=k−mω2lmω2
JEE Main 2020 — Physics Mechanics
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:
Held on 8 Jan 2020 · Verified 6 Jul 2026.
k−ωmmlω2
k−mω2mlω2
k+mω2mlω2
k+mωmlω2
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