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A disc of mass $m$ is attached to a spring of stiffness $k$ as shown in the figure. The disc rolls without slipping on a horizontal surface. The natural frequency of vibration of the system is

1. $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$
2. $\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$
3. $\frac{1}{2\pi}\sqrt{\frac{2k}{3m}}$
4. $\frac{1}{2\pi}\sqrt{\frac{3k}{2m}}$

Correct answer: $\frac{1}{2\pi}\sqrt{\frac{2k}{3m}}$

Solution

For rolling without slipping, the disc has both translational and rotational kinetic energy. For a solid disc, $I=\frac{1}{2}mR^2$, so the effective mass becomes $m+I/R^2=m+\frac{1}{2}m=\frac{3m}{2}$. Hence $\omega_n=\sqrt{k/(3m/2)}=\sqrt{2k/3m}$ and $f_n=\omega_n/(2\pi)$.

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