Q. A particle of mass $m$ is fixed to one end of a light spring of force constant $k$ and unstretched length $l$. The system is rotated about the other end of the spring with an angular velocity $\omega$, in gravity free space. The increase in length of the spring will be:Physics Question Image

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Solution:

Let increase in lengh of spring is $x$
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At equilibrium,
$k x =m \omega^{2} x+m \omega^{2} l$
$k x-m \omega^{2} x =m \omega^{2} l $
$x =\frac{m \omega^{2} l}{k-m \omega^{2}}$