Question

A holiday ornament in the shape of a hollow sphere with mass $M=$ 0.015 kg and radius $R=$ 0.050 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. ($Hint$: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

Short answer

The period of the pendulum is approximately 0.451 seconds.

Step-by-step solution

Understanding the Physical Pendulum

A physical pendulum is an object that can oscillate about an axis. The period of a physical pendulum depends on its moment of inertia about the pivot point and the distance from the pivot to its center of mass. The formula for the period is given by: $T=2\pi \sqrt{\frac{I}{mgd}}$, where $I$ is the moment of inertia, $m$ is the mass, $g$ is the acceleration due to gravity (9.81 m/s²), and $d$ is the distance from the pivot to the center of mass.

Moment of Inertia via the Parallel-Axis Theorem

The moment of inertia for a solid sphere about its center is $I_{cm}=\frac{2}{5}M{R}^2$. Using the parallel-axis theorem, we find the moment of inertia about the pivot point at the surface of the sphere: $I=I_{cm}+M{d}^2$. Since the sphere is hollow, we assume it's thin, contributing mostly at the radius, thus $d=R$. Therefore, $I=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$.

Calculate the Moment of Inertia

Substitute the values into the moment of inertia formula: $I=\frac{7}{5}(0.015)\times (0.050{)}^2$. This calculation gives us $I\approx 1.05\times {10}^{-5}{\text{ kg m}}^2$.

Calculate the Period of the Pendulum

Substitute $I=1.05\times {10}^{-5}$, $m=0.015\text{ kg}$, $g=9.81{\text{ m/s}}^2$, and $d=0.050\text{ m}$ into the period equation: $T=2\pi \sqrt{\frac{1.05\times {10}^{-5}}{0.015\times 9.81\times 0.050}}$. This simplifies to $T\approx 0.451\text{ seconds}$.

Key concepts

Moment of Inertia

In physics, the moment of inertia is a measure of how difficult it is to change the rotation of an object. It's often thought as a rotational analogue to mass in linear motion. For different shapes and axis configurations, the moment of inertia can vary greatly.

For a sphere, this quantity is central in the calculations of a physical pendulum. The moment of inertia depends on both the mass distribution of the object and the axis about which it rotates.

• The base formula for a solid sphere rotating about its own center is: $I_{cm}=\frac{2}{5}M{R}^2$.
• For our physical pendulum, because the rotation isn't around the center, we need to adjust using the parallel-axis theorem.

This adapted formula enables correct calculations for rotations different from the base axis.

Parallel-Axis Theorem

The parallel-axis theorem is a powerful tool in physics that allows for calculating the moment of inertia of an object when it is rotating about an axis that is parallel but not through the center of mass.

The formula is: $$I=I_{cm}+M{d}^2$$ where:

• $I_{cm}$ is the moment of inertia through the center of mass,
• $M$ is the mass of the object, and
• $d$ is the distance from the new axis to the axis through the center of mass.

In the given problem, the sphere is hung and rotates around an axis through the hanging loop, which is at its surface; hence the distance $d$ equals $R$. Applying the parallel-axis theorem simplifies the calculation, providing the precise moment of inertia needed for the subsequent steps.

Calculating Period

To determine how long it takes for the pendulum to complete one full swing back and forth, we calculate its period. This involves using key parameters derived from its physical properties.
The formula for a physical pendulum is: $$T=2\pi \sqrt{\frac{I}{mgd}}$$ where:

• $T$ is the period,
• $I$ is the moment of inertia,
• $m$ is mass,
• $g$ is the acceleration due to gravity, and
• $d$ is the distance from the pivot to the center of mass.

This equation perfectly encapsulates how both the distribution of mass and the path it swings through impact the pendulum's movement. With the values substituted from the physical attributes of the sphere, we derive the period as approximately 0.451 seconds—a swift oscillation due to its small size and mass.

Physics Problem Solving

Solving physics problems often requires a mix of theoretical knowledge and practical grasp of concepts. This problem specifically illustrates applying several principles in a sequence to find a direct answer, demonstrating the blend of understanding and calculation work required in physics.

Here's a breakdown of the approach:

• Comprehend the problem context, identifying it as a physical pendulum scenario.
• Calculate the fundamental parameters like the moment of inertia using relevant theorems.
• Integrate these parameters into broader formulas, like the one for period calculation.

Understanding these steps and how they connect fosters a deeper comprehension of wider physics concepts, such as oscillations and rotational dynamics, preparing students for tackling more complex topics.