Q:One end of a masses spring of force constant 100N/m and natural length 0.5m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.

Since the the mass of 0.5Kg is made to rotate with an angular velocity ω=2rad/s, so      mω2r=F=-K0.5+rwhere r is the elongation of the spring.Again   0.5×22×r=-1000.5+ror            2r=-50-100ror                102r=-50or               r=-50102=-049msince  elongation can not be negative so minus sign is ignored.Therefore the elongation of spring is 0.49m.

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