A pendulum consists of a \(320-\mathrm{g}\) solid ball \(15.0 \mathrm{cm}\) in diameter, suspended by an essentially massless string \(80.0 \mathrm{cm}\) long. Calculate the period of this pendulum, treating it first as a simple pendulum and then as a physical pendulum. What's the error in the simple-pendulum approximation? (Hint: Remember the parallelaxis theorem.)

Understanding the period of a pendulum is key when studying oscillatory motion. The period, denoted by \( T \), is the time it takes for one complete cycle of swing, back and forth, of the pendulum. A simple pendulum, which approximates the ball as a point mass and the string as massless and inextensible, has its period determined by the formula:
\[ T = 2 \times \.pi \times \.sqrt{\frac{l}{g}} \]
where \( l \) is the length of the string and \( g \) is the acceleration due to gravity, typically \( 9.80 \, m/s^2 \).

However, in reality, objects have dimensions and mass distribution, which must be factored into calculations for accuracy. The formula above works best for small angular displacements. We consider this in the exercise by also calculating the period treating the ball not as a point mass, but as a physical pendulum with a distributed mass.

A physical pendulum consists of a real object that has size and shape, such as the solid ball in this exercise. The period of a physical pendulum depends on the object's moment of inertia and its distance from the pivot point. The formula used differs slightly from that of a simple pendulum. It is given by:
\[ T = 2 \times \.pi \times \.sqrt{\frac{I}{mgl}} \]
Here, \( I \) represents the moment of inertia around the pivot point, \( m \) is the mass, \( g \) remains the acceleration due to gravity, and \( l \) is the distance from the pivot to the center of mass. In our exercise, we used the moment of inertia for a sphere and adjusted it for the pendulum using the parallel-axis theorem, to get more accurate results compared to the simple pendulum approximation.

The parallel-axis theorem is a powerful tool in classical mechanics. It allows you to find the moment of inertia of an object about any axis, given you know the moment of inertia about a parallel axis that passes through the center of mass and the distance between the two axes.

The theorem states that if \( I_{cm} \) is the moment of inertia of the body about an axis through its center of mass, and \( d \) is the distance between the center of mass and the new axis, then the moment of inertia about the new axis, \( I \), is given by:
\[ I = I_{cm} + md^2 \]
where \( m \) is the mass of the object. In the step by step solution, we applied this theorem to find the moment of inertia for the solid ball pendulum about the pivot point. This correction is crucial for calculating the period of a physical pendulum because it accounts for the distribution of the ball's mass relative to the pivot point.