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Questions  

A spring mass system ( mass m , spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc together with spring mass system, rotates about it’s axis with an angular velocity ω,k>>mω2 the relative change in the length of the spring is best given by the option

a
2mω2k
b
mω2k
c
23mω2k
d
mω23k

detailed solution

Correct option is B

Spring force=centrifugal force  kx=ml+xω2⇒kx−mxω2=mlω2     ⇒x=mlω2k−mω2      ⇒x=mlω2k      [k>>mω2]       Relative change is  xl=mω2k

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Similar Questions

When a particle of mass m is suspended from a massless spring of natural length , the length of the spring becomes 2 . When the same mass moves in conical pendulum as shown in figure, the length of the spring becomes L. The radius of the circle of conical pendulum is:


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