Question

A particle of mass m is fixed to one end of a light spring having force constant $k$ and unstretched length is $l$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is

• A. $\frac{ml{\omega }^2}{k-\omega m}$
• B. $\frac{ml{\omega }^2}{k+m{\omega }^2}$
• C. $\frac{ml{\omega }^2}{k-m{\omega }^2}$
• D. $\frac{ml{\omega }^2}{k+\omega m}$

Answer 1

The correct option is C $\frac{ml{\omega }^2}{k-m{\omega }^2}$

Step-1: Draw rough diagram for the given situation.

Let the elongation of the spring be $\Delta l$.

Step-2:Find elongation of the spring.

Formula used:$F_{\text{radial}}=mr{\omega }^2$.

The spring force will provide required centripital force for the rotation of the block
$k\Delta l=m(l+\Delta l){\omega }^2$

$\Delta l=\frac{ml{\omega }^2}{k-m{\omega }^2}$
Final answer is (b)