A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l …

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:
  1. mlω2k-ωm
  2. mlω2k-mω2
  3. mlω2k+mω2
  4. mlω2k+mω

Solution

As we know, the spring force will give the necessary centripetal force for rotation.

So, 

mω2l+x=kx
lx+1=kmω2

 Thus, the stretch in the spring
x=lmω2k-mω2

Asked in: JEE Main 2020 (08 Jan Shift 1)

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