Problem 37 - University Physics with Modern Physics

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.

Step-by-Step Solution

Answer

Answer: The moment of inertia of the pendulum is \(I_{total} = \frac{8}{15}ML^2\). The period of the pendulum for small oscillations is \(T = 2\pi\sqrt{\frac{8L}{15g}}\).

Step 1: Find the moment of inertia of the rod

The moment of inertia for a uniform thin rod about one end is given by the equation \(I_{rod} = \frac{1}{3}ML^2\), where \(M\) is the mass of the rod and \(L\) is the length of the rod.

Step 2: Find the moment of inertia of the sphere

The moment of inertia for a solid sphere about a diameter is \(I_{sphere} = \frac{2}{5}MR^2\), where \(M\) is the mass of the sphere and \(R\) is the radius of the sphere. For this physical pendulum, the radius of the sphere is \(L / 2\). Therefore, \(I_{sphere} = \frac{2}{5}M(\frac{L}{2})^2\).

Step 3: Calculate the parallel-axis theorem for the sphere

As the sphere is pivoted about the endpoint of the rod, we will use the parallel-axis theorem to find the total moment of inertia about the pivot. The parallel-axis theorem states that \(I = I_{cm} + Md^2\), where \(I_{cm}\) is the moment of inertia of the sphere about its center of mass, \(M\) is the mass of the sphere, and \(d\) is the perpendicular distance between the axes (\(L\) in this case because the sphere is centered about the free end of the rod). So, \(I_{sphere} = \frac{2}{5}M(\frac{L}{2})^2 + M L^2\).

Step 4: Calculate the total moment of inertia of the pendulum

Now, we can find the total moment of inertia for the pendulum by adding the moment of inertia of the rod and the sphere together. \(I_{total} = I_{rod} + I_{sphere} = \frac{1}{3}ML^2 + \frac{2}{5}M(\frac{L}{2})^2 + M L^2\).

Step 5: Simplify the equation for the total moment of inertia

The total moment of inertia equation should be simplified. \(I_{total} = \frac{1}{3}ML^2 + \frac{1}{5}ML^2 + ML^2 = (\frac{1}{3} + \frac{1}{5} + 1)ML^2 = \frac{8}{15}ML^2\).

Step 6: Find the period for small oscillations

For small oscillations of a physical pendulum, the period \(T\) can be found using the equation \(T = 2\pi\sqrt{\frac{I_{total}}{MgR}}\). In this case, \(I_{total}\) is the moment of inertia we found in the previous step, \(M\) is the mass of the pendulum (both rod and sphere), \(g\) is the gravitational acceleration, and \(R\) is the perpendicular distance from the pivot point to the center of mass of the pendulum. As the sphere is at the end of the rod and the mass of the rod is evenly distributed along it, the center of mass of the pendulum is at the midpoint of the rod or \(\frac{L}{2}\) away from the pivot. Hence, \(T = 2\pi\sqrt{\frac{\frac{8}{15}ML^2}{MgL/2}}\).

Step 7: Simplify the period equation

Finally, we can simplify the equation for the period. \(T = 2\pi\sqrt{\frac{\frac{8}{15}ML^2}{\frac{1}{2}MgL}} = 2\pi\sqrt{\frac{8L}{15g}}\). The moment of inertia of the pendulum is \(I_{total} = \frac{8}{15}ML^2\). The period of the pendulum for small oscillations is \(T = 2\pi\sqrt{\frac{8L}{15g}}\).

Key Concepts

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. Imagine it like mass in linear motion, but for rotating objects.

• For a uniform thin rod pivoted at one end, we use the formula: \( I_{rod} = \frac{1}{3}ML^2 \), where \( M \) is mass, and \( L \) is the length of the rod.
• For a solid sphere, the formula for its moment of inertia about an axis through its diameter is \( I_{sphere} = \frac{2}{5}MR^2 \). Now, the radius \( R \) is \( \frac{L}{2} \), giving \( I_{sphere} = \frac{2}{5}M(\frac{L}{2})^2 \).

By using these formulas, we can compute the individual moments of inertia and add them, considering where the sphere is located in relation to the rod.

Parallel-Axis Theorem

The parallel-axis theorem allows us to find the moment of inertia of a body about any axis, given that we know its moment of inertia about a parallel axis through the center of mass. The theorem is expressed as:\[ I = I_{cm} + Md^2 \]Here, \( I_{cm} \) is the moment about the center of mass, \( M \) is the mass of the body, and \( d \) is the distance between the center of mass axis and the new axis. In our pendulum, the solid sphere is at the end of the rod. Thus, we apply the theorem to the moment of inertia of the sphere:\[I_{sphere} = \frac{2}{5}M(\frac{L}{2})^2 + ML^2 \]. This additional \( ML^2 \) considers the rod's endpoint rotation axis.

Period of Oscillation

The period of oscillation defines the time a pendulum takes to complete one full swing back and forth. It is especially insightful for timekeeping mechanisms. For small oscillations, the formula:\[ T = 2\pi\sqrt{\frac{I_{total}}{MgR}} \]is used. Here, \( I_{total} \) is the pendulum's moment of inertia about the pivot, \( M \) is its mass, \( g \) is gravitational acceleration, and \( R \) is the distance from the pivot to the center of mass. This formula shows how the period is affected by factors like mass distribution and gravitation.

Small Oscillations

Small oscillations refer to the assumption that the pendulum usweing through a slight angle. This assumption allows for specific simplifying approximations in calculations. It's a common assumption in physics because:

• It simplifies the mathematical model, allowing us to treat the motion akin to simple harmonic motion.
• Leads to the sinusoidal approximation \( \theta \approx \sin(\theta) \), where \( \theta \) is the angle of displacement.

For a clock pendulum, perfectly modeling small oscillations ensures accurate timekeeping without the complications of large swings.

Gravitational Acceleration

Gravitational acceleration, typically denoted as \( g \), is the rate of acceleration due to Earth's gravity, approximately \( 9.81 \text{ m/s}^2 \). Its presence in formulas like the period of oscillation formula emphasizes its influence on pendulum dynamics.

• Influences how quickly and how strongly pendulums accelerate towards equilibrium.
• A constant in many pendulum problems, providing consistent character to mechanical systems like clocks.

Without \( g \), the pendulum would lack a restoring force, necessary for the characteristic oscillations crucial in timing devices.