A solid rectangular box, of dimensions \(100 \mathrm{~mm} \times 60 \mathrm{~mm} \times 20 \mathrm{~mm}\), is spinning freely with angular velocity 240 r.p.m. Determine the frequency of small oscillations of the axis, if the axis of rotation is (a) the longest, and (b) the shortest, axis.

For a solid rectangular box with dimensions \(a\), \(b\), and \(c\), the moments of inertia are given by: $$I_a = \frac{1}{12}mb(a^2 + c^2)$$ $$I_b = \frac{1}{12}mc(b^2 + a^2)$$ $$I_c = \frac{1}{12}ma(b^2 + c^2)$$ Using the given dimensions, let \(a = 100~mm\), \(b = 60~mm\), and \(c = 20~mm\). We don't have the value of mass \(m\), but since it will just cancel out later, we can proceed without knowing its value.

The angular velocity is given in r.p.m (revolutions per minute), and we need to convert it to radians per second. Use the conversion factor \(1~\text{rev/min}=2\pi~\text{rad/min}\) and \(1~\text{min} = 60~\text{s}\). $$\omega = 240~\text{r.p.m} \times \frac{2\pi ~\text{rad}}{1 ~\text{rev}} \times \frac{1 ~\text{min}}{60 ~\text{s}} = 16\pi ~\text{rad/s}$$

Assuming no external torques, the angular momentum \(L\) is conserved. The equation for the angular momentum is \(L = I\omega\). As the moment of inertia changes because of the different axes of rotation but the angular momentum remains constant, the angular velocity will also change. The frequencies of small oscillations will be the difference between the new and initial angular velocities. Determine the initial angular momentum \(L_0\) of the box for both cases and find the new angular velocity \(\omega'\) and frequency \(f\).

For the longest axis, \(a = 100~mm\), the initial moment of inertia \(I_a\) and initial angular momentum \(L_0\) are: $$I_a = \frac{1}{12}m(100^2 + 20^2)$$ $$L_0 = I_a\omega = \frac{1}{12}m(100^2 + 20^2)(16\pi)$$ Now, find the new moment of inertia \(I_b'\) for the axis \(b = 60~mm\). $$I_b' = \frac{1}{12}m(60^2 + 100^2)$$ Using the conservation of angular momentum, find the new angular velocity \(\omega'\). $$L_0 = I_b'\omega'$$ $$\Rightarrow \omega' = \frac{L_0}{I_b'} = \frac{\frac{1}{12}m(100^2 + 20^2)(16\pi)}{\frac{1}{12}m(60^2 + 100^2)} = \frac{2}{3}(16\pi) = \frac{32}{3}\pi ~\text{rad/s}$$ Finally, calculate the frequency of small oscillations (in Hz) for case (a): $$f = \frac{\omega' - \omega}{2\pi} = \frac{\frac{32}{3}\pi - 16\pi}{2\pi} = \frac{16}{3} ~\text{Hz}$$

For the shortest axis, \(c = 20~mm\), the initial moment of inertia \(I_c\) and initial angular momentum \(L_0\) are: $$I_c = \frac{1}{12}m(60^2 + 20^2)$$ $$L_0 = I_c\omega = \frac{1}{12}m(60^2 + 20^2)(16\pi)$$ Now, find the new moment of inertia \(I_b'\) for the axis \(b = 60~mm\). $$I_b' = \frac{1}{12}m(60^2 + 100^2)$$ Using the conservation of angular momentum, find the new angular velocity \(\omega'\). $$L_0 = I_b'\omega'$$ $$\Rightarrow \omega' = \frac{L_0}{I_b'} = \frac{\frac{1}{12}m(60^2 + 20^2)(16\pi)}{\frac{1}{12}m(60^2 + 100^2)} = \frac{40}{3}\pi ~\text{rad/s}$$ Finally, calculate the frequency of small oscillations (in Hz) for case (b): $$f = \frac{\omega' - \omega}{2\pi} = \frac{\frac{40}{3}\pi - 16\pi}{2\pi} = \frac{8}{3} ~\text{Hz}$$ So, the frequency of small oscillations of the axis, if the axis of rotation is (a) the longest, is \(16/3 ~\text{Hz}\), and (b) the shortest, axis is \(8/3 ~\text{Hz}\).