Question

A spring-mass system (mass $m$, spring constant $k$ and natural length $l$) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring-mass system, rotates about its axis with an angular velocity $(k>>>m^2)$, the relative change in the length of the spring is best given by the option:

• A. $\frac{m{\omega }^2}{3k}$
• B. $\sqrt{\frac{2}{3}}\frac{m{\omega }^2}{k}$
• C. $\frac{m{\omega }^2}{k}$
• D. $\frac{2m{\omega }^2}{k}$

Answer 1

The correct option is C

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

Step 1: Given data and drawing the diagram

Mass $=m$,

Spring constant $=k$

Angular velocity$=\omega$

Natural length $=l_0$

$(k>>>m^2)$

Step 2: Find relative change in the length of the spring

Let $x$ be the extension in length of spring.

The total length of spring be $l_0+x$.

Since, the disc is rotating with angular velocity $\omega$ about its axis, it will experience a outward force called centrifugal force. It is given by $m{\omega }^2(l_0+x)$.

Spring will has a restoring force acting inward which is given by $kx$.

If centrifugal force is greater, mass will go out of disc. If restoring force is more than mass will move towards the axis of disc. Since, mass is in equilibrium, both forces must be equal.

Putting both forces equal,

$$\begin{gathered} m{\omega }^2(l_0+x)=kx \\ (\frac{l_0}{x}+1)=\frac{kx}{m{\omega }^2} \\ x=\left(\frac{l_{0}m{\omega }^2}{k-m{\omega }^2}\right) \\ k>>m{\omega }^2 \\ k-m{\omega }^2=k \end{gathered}$$

So, $(x/l_0)$ is equal to$m{\omega }^2/k$.

Hence, option C is correct.