The figure shows a uniform rod (length L = 1.0 m,mass = 2.0 kg)suspended from a pivot a distance d = 0.25 m above its center of mass.The angular frequency (in rad/s)for small oscillations is approximately

Question

Quizplus · Accepted Answer

The angular frequency for small oscillations of a uniform rod suspended from its center of mass is given by: $\omega=\sqrt{\frac{g}{\frac{L}{12}}}$, where $g$ is the acceleration due to gravity and $L$ is the length of the rod.

However, in this case, the rod is suspended from a distance $d$ above its center of mass. This changes the moment of inertia of the rod, causing the angular frequency to be larger. The new equation for the angular frequency is: $\omega=\sqrt{\frac{3g}{2\frac{L^2}{12}+2d^2}}=\sqrt{\frac{3g}{2\frac{1^2}{12}+2(0.25)^2}}\approx4.1$ rad/s.

Therefore, the best choice would be D.

The Figure Shows a Uniform Rod (Length L = 1