Problem 74 - University Physics with Modern Physics

Question

A sphere consists of a solid wooden ball of uniform density 800 \(\mathrm{kg} / \mathrm{m}^{3}\) and radius 0.20 \(\mathrm{m}\) and is covered with a thin coating of lead foil with area density 20 \(\mathrm{kg} / \mathrm{m}^{2} .\) Calculate the moment of inertia of this sphere about an axis passing through its center.

Step-by-Step Solution

Answer

The total moment of inertia is approximately 1.612 kg m².

Step 1: Find the Volume of the Sphere

Calculate the volume of the solid wooden ball using the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). Given that \( r = 0.2 \) m, we have \[ V = \frac{4}{3} \pi (0.2)^3 = \frac{4}{3} \pi \times 0.008 = 0.0335 \pi \approx 0.0335 \times 3.1416 = 0.105 \text{ cubic meters.} \]

Step 2: Calculate the Mass of the Wooden Ball

The mass of the wooden ball can be found using its volume and density: \( m = \rho V \). With \( \rho = 800 \) kg/m\(^3\) and \( V = 0.105 \) m\(^3\), the mass is \[ m = 800 \times 0.105 = 84 \text{ kg.} \]

Step 3: Calculate the Mass of the Lead Coating

Find the surface area of the sphere: \( A = 4 \pi r^2 \). With \( r = 0.2 \) meters, we have \( A = 4 \pi (0.2)^2 = 0.16 \pi \approx 0.502 \text{ square meters.} \) Multiply this by the area density to find the mass of the coating, \( m_{lead} = 20 \times 0.502 \approx 10.04 \) kg.

Step 4: Calculate the Moment of Inertia of Wooden Ball

Use the formula for the moment of inertia of a solid sphere: \( I = \frac{2}{5} m r^2 \). Substitute the mass of the wooden ball (84 kg) and its radius (0.2 m): \[ I = \frac{2}{5} \times 84 \times (0.2)^2 = \frac{2}{5} \times 84 \times 0.04 = 1.344 \text{ kg m}^2. \]

Step 5: Calculate the Moment of Inertia of Lead Coating

The lead coating is considered as a thin spherical shell. Use its moment of inertia formula: \( I = \frac{2}{3} m r^2 \). With \( m = 10.04 \) kg, we have \[ I = \frac{2}{3} \times 10.04 \times (0.2)^2 = \frac{2}{3} \times 10.04 \times 0.04 \approx 0.2677 \text{ kg m}^2. \]

Step 6: Total Moment of Inertia

Add the moments of inertia from the wooden ball and lead coating: \[ I_{total} = I_{wood} + I_{lead} = 1.344 + 0.2677 = 1.6117 \text{ kg m}^2. \]

Key Concepts

Solid Sphere

A solid sphere is a three-dimensional object with every point on its surface equidistant from its center. Imagine a perfectly round ball where not only the outer surface is solid, but everything within it is also uniformly filled with material. This makes it different from a shell, which is hollow inside. Plenty of examples in real life can be categorized as solid spheres, like a marble or a basketball filled with material.

• The key property of a solid sphere is uniformity. This means that the material's distribution inside is the same throughout.
• Such uniformity leads to predictable behaviors, such as the calculation of moment of inertia, which we'll touch upon later.

Understanding the solid sphere concept is crucial when dealing with the physics of rotations and moments of inertia.

Spherical Shell

A spherical shell, unlike a solid sphere, is essentially hollow. It is like the outer layer of an onion with nothing but emptiness inside. Despite this difference, both the solid sphere and spherical shell share similar basic properties, like their round shape.

• The main difference lies in mass distribution. A spherical shell's mass is concentrated on its surface, resulting in different calculations for physical properties.
• Its applications include things like domes or thin-walled containers. Understanding this concept is essential when working with rotational dynamics and inertia calculations.

The importance of differentiating between a solid sphere and a spherical shell becomes evident in the calculation of their moments of inertia, which vary due to their mass distributions.

Volume Calculation

Volume calculation is crucial when determining the physical properties of three-dimensional objects like spheres. For spheres, the formula is quite standard: \[ V = \frac{4}{3}\pi r^3 \]This formula calculates how much space a sphere occupies based on its radius, a key measurement from its center to any point on its surface.

• A larger radius means a larger volume, which affects an object's mass and other properties.
• The formula emphasizes the cubic relationship of volume to the radius, showing that small changes in radius can create large changes in volume.

Correct volume calculation is vital, especially in physics and engineering, where the volume is foundational to further calculations involving mass or density.

Density

Density is a measure of how much mass is packed into a unit volume of a substance. It's like figuring out how tightly stuffed a suitcase is.

• Mathematically, it is expressed as \( \rho = \frac{m}{V} \), with \( m \) being mass and \( V \) being volume.
• A consistent density implies uniform distribution of material, important in predicting an object’s mechanical properties.

In our context, knowing the density allows us to compute the mass from the given volume, which is essential in evaluating the moment of inertia. The concept of density pops up in various scientific fields, signifying the importance of material consistency.

Mass Calculation

Calculating mass involves understanding both the object's volume and its density—two sides of the same coin. Essentially, the mass is determined by multiplying an object's density by its volume:\[ m = \rho V \]Each sphere, regardless of it being solid or a shell, will have mass depending on its specific attributes.

• For a solid sphere, this means combining its entire volume with its uniform density.
• For a spherical shell, focus only on the mass distributed across its surface, like a coat of paint.

Effective mass calculation is crucial in physics, influencing the calculation of other physical properties such as inertia and gravitational force.