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 $\omega$, 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)

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

Let $x=$ the increase in the length of spring. Then the particle moves along a circular path of radius $(l+x)$ and the spring force $=kx=$ centripetal force.
$\therefore m{\omega }^2(l+x)=kx$ or $x=\frac{m{\omega }^{2}l}{k-m{\omega }^2}$