Question

A spring mass system (mass $m,$ spring constant $k$ and natural length $l$ ) rests 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 $\omega (k>>m{\omega }^2)$ , the relative change in the length of the spring is best given by the option

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

Answer 1

The correct option is D $\frac{m{\omega }^2}{k}$

Step-1 :Draw $FBD$ of mass $m$ in frame of disc.

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

Step-2 :Find relative change in length of spring.

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

The spring force will provide required centripetal force for the rotation of the block.

$\Rightarrow k\Delta l=m{\omega }^2(l+\Delta l)$

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

As given, $(k>>m{\omega }^2)$

$\frac{\Delta l}{l}=\text{Relative change}=\frac{m{\omega }^2}{k}$
Final answer is (d)