Oscillations

A physical pendulum is released from rest at an angle $\theta$ from the vertical. Neglecting friction, what principle can be used to find its maximum angular velocity?

A uniform rod of mass $m$ and length $L$ is pivoted at one end. If a small mass $\frac{m}{2}$ is attached at the midpoint of the rod, how does this affect the period of the pendulum, assuming the moment of inertia of the rod about its end is $\frac{1}{3}mL^2$?

A uniform rod of mass $M$ and length $L$ is pivoted at one end. If the moment of inertia about its center of mass is $\frac{1}{12}ML^2$, what is its moment of inertia about the pivot point using the parallel axis theorem?

The angular frequency of a physical pendulum is given by $\omega = \sqrt{\frac{mgd}{I}}$. How is the period $T$ related to this angular frequency $\omega$?

Under what condition does the motion of a physical pendulum approximate simple harmonic motion (SHM)?

Which of the following is a key difference between a simple pendulum and a physical pendulum?

In the formula for the period of a physical pendulum, $T = 2\pi \sqrt{\frac{I}{mgd}}$, what does 'd' represent?

A physical pendulum has a moment of inertia $I = 0.5 \text{ kg} \cdot \text{m}^2$, a mass $m = 0.25 \text{ kg}$, and the distance from the pivot to the center of mass is $d = 0.2 \text{ m}$. What is the period of the pendulum's oscillation?

A thin rod of length $L$ and mass $M$ is pivoted at a distance $L/4$ from one end. What is the moment of inertia of the rod about this pivot point?

A physical pendulum with moment of inertia $I$ is released from an initial angular displacement of $\theta_0$. If its potential energy at the release point is $U$, and its kinetic energy at the lowest point is $K$, and the period is defined as $T = 2\pi \sqrt{\frac{I}{mgd}}$, how can you relate these parameters to find the period?