Question

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

• 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

Answer 1

Answer: (B)

$\frac{m\omega^{2}l}{k - m\omega^{2}}$

Answer 2

In the given condition elastic force will provides the required centripetal force

$kx = m\omega^{2}r$

$kx = m ⥂ ⥂ \omega^{2}(l + x) \Rightarrow kx = m\omega^{2}l + m\omega^{2}x$ $\Rightarrow$

$x(k - m\omega^{2}) = m\omega^{2}l\therefore x = \frac{m\omega^{2}l}{k - m\omega^{2}}$