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. As shown in figure the increase in length of the spring will be
A) $\frac{m{\ \omega}^{2}l}{k}$
B) $\frac{m\ \omega^{2}l}{k - m{\ \omega}^{2}}$
C) $\frac{m\ \omega^{2}l}{k + m{\ \omega}^{2}}$
D) None of these
Explanation
(b) Let $x$ be increase in length of the spring. The particle would move in a circular path of radius $(l + x)$. Centripetal force = force due to the spring
$m(l + x)\omega^{2} = kx$
$\therefore\ x = \frac{m\omega^{2}l}{k - m\omega^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.