Question

A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass \(M\) and length \(L\) that is pivoted freely about one end, with a solid sphere of the same mass, \(M\), and a radius of \(L / 2\) centered about the free end of the rod. a) Obtain an expression for the moment of inertia of the pendulum about its pivot point as a function of \(M\) and \(L\). b) Obtain an expression for the period of the pendulum for small oscillations.

Short answer

Answer: The moment of inertia of the pendulum about the pivot point as a function of \(M\) and \(L\) is \(I_{total} = \frac{17}{15}ML^2\). Question: What is the period of oscillation for small oscillations? Answer: The period of oscillation of the pendulum for small oscillations is \(T = 2\pi\sqrt{\frac{17}{30}\frac{L}{g}}\).

Step-by-step solution

Moment of Inertia of the Rod

To find the moment of inertia of the rod about the pivot point, we use the formula for a thin rod where the axis of rotation is through one end: \(I_{rod} = \frac{1}{3}ML^2\).

Moment of Inertia of the Sphere

To find the moment of inertia of the sphere about the pivot point, we first find the moment of inertia about its center of mass and then use the parallel axis theorem to find the moment of inertia about the pivot point. The formula for the moment of inertia of a solid sphere about its center of mass is \(I_{sphere,cm} = \frac{2}{5}MR^2\), where \(R = \frac{L}{2}\). So, \(I_{sphere,cm} = \frac{2}{5}M\left(\frac{L}{2}\right)^2\). Now, using the parallel axis theorem (which states that \(I = I_{cm} + MD^2\)) to find the moment of inertia of the sphere about the pivot point, with \(D = L\) as the distance between the pivot point and the center of mass of the sphere, we have: \(I_{sphere} = I_{sphere,cm} + ML^2 = \frac{2}{5}M\left(\frac{L}{2}\right)^2 + ML^2\).

Moment of Inertia of the Pendulum

Now we can find the total moment of inertia of the pendulum about the pivot point by adding the moment of inertia of the rod and the moment of inertia of the sphere: \(I_{total} = I_{rod} + I_{sphere} = \frac{1}{3}ML^2 + \left(\frac{2}{5}M\left(\frac{L}{2}\right)^2 + ML^2\right)\). Simplifying the expression, we get: \(I_{total} = \frac{1}{3}ML^2 + \frac{1}{5}ML^2 + ML^2 = \left(\frac{1}{3} + \frac{1}{5} + 1\right)ML^2\). So, the moment of inertia of the pendulum about the pivot point as a function of \(M\) and \(L\) is: \(I_{total} = \frac{17}{15}ML^2\).

The Period of Oscillation of a Physical Pendulum

To find the period of oscillation for small oscillations, we can use the formula for the period of oscillation of a physical pendulum: \(T = 2\pi\sqrt{\frac{I}{Mgh}}\), where \(T\) is the period, \(I\) is the moment of inertia, \(M\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the perpendicular distance from the pivot point to the center of mass of the system. In our case, the center of mass of the system lies at the midpoint of the rod, so \(h = \frac{L}{2}\), and \(I = \frac{17}{15}ML^2\). Plugging these values into the formula, we get: \(T = 2\pi\sqrt{\frac{\frac{17}{15}ML^2}{Mg\frac{L}{2}}}\). Simplifying the expression, we get: \(T = 2\pi\sqrt{\frac{17}{30}\frac{L}{g}}\). Hence, the period of oscillation of the pendulum for small oscillations is: \(T = 2\pi\sqrt{\frac{17}{30}\frac{L}{g}}\).

Key concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics, playing a role akin to mass in linear motion. It quantifies an object's resistance to changes in its rotational motion and depends on the object's mass distribution relative to the rotation axis.

For a given object, the moment of inertia changes with the axis of rotation. For instance, the moment of inertia of a rod about an axis through its end, as reflected in the given exercise, is given by the formula \( I = \frac{1}{3}ML^2 \) , where \(M\) is the mass and \(L\) is the length of the rod. This formula suggests that both the mass and how it's spread out along the rod determine how hard it is to start or stop the rod's spinning.

Understanding this concept is crucial when dealing with physical pendulums like grandfather clock pendulums, where the pendulum's timing – and therefore the clock's accuracy – is directly linked to the moment of inertia.

Small Oscillations

Small oscillations, or small-angle approximations, is a critical simplification tool in the study of pendulum mechanics. When a pendulum swings with only a small maximum angle from its vertical equilibrium position, we can make the approximation that \( \text{sin}(\theta) \) is approximately equal to \( \theta \) (when \( \theta \) is in radians).

This approximation allows us to simplify the complex equations describing pendulum motion, making it considerably easier to solve for the period and frequency of oscillation. For the physical pendulum in the exercise, this small oscillation approximation leads to a simple formula to calculate the pendulum's period, ultimately allowing it to be used as a precise timekeeping tool.

Parallel Axis Theorem

The parallel axis theorem is instrumental in calculating moments of inertia for composite systems or objects not rotating around their center of mass. This theorem states that the moment of inertia \( I \) about any axis parallel to and a distance \( D \) away from another axis through the center of mass \( I_{cm} \) is given by \( I = I_{cm} + MD^2 \) , where \( M \) is the mass of the object.

In our exercise scenario, the parallel axis theorem enables us to find the moment of inertia of a sphere attached to a rod—a composite pendulum—about the pivot point. We first calculate the moment of inertia of the sphere about its center and then add the term \( ML^2 \) to account for the rotation about the pivot. This step is vital to achieve the correct total moment of inertia, thus getting an accurate expression for the pendulum's period for small oscillations.