Problem 85 - University Physics with Modern Physics

Question

A sphere with radius \(R =\) 0.200 m has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(r =\) 3.00 \(\times\) 103 kg/m\(^3 -\) (9.00 \(\times\) 103 kg/m\(^4\))\(r\). (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

Step-by-Step Solution

Answer

Mass: Calculate using volume integration; Moment of inertia: Use density integral.

Step 1: Understand the Density Function

The density function is given as a function of distance, \( r \), from the center: \[ \rho(r) = 3.00 \times 10^3 \text{ kg/m}^3 - (9.00 \times 10^3 \text{ kg/m}^4) \cdot r \]This signifies that density decreases linearly from the center of the sphere.

Step 2: Set up the Mass Integral

To find the total mass, integrate the density function over the sphere's volume. The differential element of mass is:\[ dm = \rho(r) \cdot dV = \rho(r) \cdot 4\pi r^2 dr \]The volume of a spherical shell is \( dV = 4\pi r^2 dr \).

Step 3: Integrate to Find Total Mass

Substitute the density function into the mass differential and integrate from \( r = 0 \) to \( R = 0.200 \, \text{m} \):\[M = \int_0^{0.200} (3.00 \times 10^3 - 9.00 \times 10^3 \cdot r) \cdot 4\pi r^2 dr\]Calculate the integral:\[M = 4\pi \left[ \frac{3.00 \times 10^3}{3} r^3 - \frac{9.00 \times 10^3}{4} r^4 \right]_0^{0.200}\]

Step 4: Calculate the Values

Evaluate the above expression:\[M = 4\pi \left( \frac{3.00 \times 10^3}{3} \times (0.200)^3 - \frac{9.00 \times 10^3}{4} \times (0.200)^4 \right)\]\[M = 4\pi (26.67 \times 10^{-3} - 36.00 \times 10^{-4})\]Complete this calculation to obtain the mass.

Step 5: Set up the Moment of Inertia Integral

The differential moment of inertia, \( dI \), about a diameter is:\[ dI = r^2 dm \]Thus, substituting \( dm \),\[ dI = r^2 \cdot \rho(r) \cdot 4\pi r^2 dr = \rho(r) \cdot 4\pi r^4 dr \]

Step 6: Integrate for Moment of Inertia

Solve the integral to calculate the moment of inertia \( I \):\[I = \int_0^{0.200} 4\pi r^4 (3.00 \times 10^3 - 9.00 \times 10^3 \cdot r) dr\]This expands to:\[I = 4\pi \left[ \frac{3.00 \times 10^3}{5} r^5 - \frac{9.00 \times 10^3}{6} r^6 \right]_0^{0.200}\]

Step 7: Calculate Moment of Inertia Values

Evaluate the expression for moment of inertia:\[I = 4\pi \left( \frac{3.00 \times 10^3}{5} \times (0.200)^5 - \frac{9.00 \times 10^3}{6} \times (0.200)^6 \right) \]Perform the calculation to get the moment of inertia.

Key Concepts

Density Function

In the given problem, the density function is an essential concept. It describes how mass is distributed within an object.

The density function provided in the problem is a mathematical expression that changes with the distance from the center of the sphere. The formula given is:\[\rho(r) = 3.00 \times 10^3 \text{ kg/m}^3 - (9.00 \times 10^3 \text{ kg/m}^4) \cdot r.\]This means that at the very center of the sphere (where \(r = 0\)), the density is highest at \(3.00 \times 10^3\) kg/m\(^3\).

As you move away from the center, the density decreases linearly with \(r\). This indicates a non-uniform distribution of mass, which is quite common in natural and engineered materials.

• The term \(3.00 \times 10^3\) kg/m\(^3\) represents the density at the center.
• The negative term \((9.00 \times 10^3 \text{ kg/m}^4) \cdot r\) reduces the density depending on how far you move from the center.

Understanding this function is critical to solving many physics problems involving mass and geometry.

Mass Integral

The mass of an object is the total amount of matter it contains. For objects with changing density, like our sphere, we calculate mass using a mass integral.

The mass integral combines the density function with the volume element to sum up the total mass.To compute this, we break the sphere into infinitesimally small spherical shells and sum up their masses:\[dm = \rho(r) \cdot dV = \rho(r) \cdot 4\pi r^2 dr.\]Here, \(dV = 4\pi r^2 dr\) represents the volume of a thin spherical shell.

The total mass \(M\) is then found by integrating the differential mass \(dm\) from the center of the sphere (\(r = 0\)) to the outer edge (\(r = 0.200\) m):\[M = \int_0^{0.200} (3.00 \times 10^3 - 9.00 \times 10^3 \cdot r) \cdot 4\pi r^2 dr.\]This calculation effectively adds up the mass of each concentric layer of the sphere by accounting for both the geometry and the varying density.

Spherical Coordinates

Spherical coordinates are a system that allows us to express points in three-dimensional space using three values: the radial distance, polar angle, and azimuthal angle.

For our exercise, the key component is the radial distance because we are calculating properties of a sphere. The radial distance, \(r\), is very useful when assessing spherical objects as it directly relates to the radius of the sphere.

In spherical coordinates:

• \(r\) is the distance from the origin (center of the sphere) to a point in space.
• The volume element in spherical coordinates is \(dV = 4\pi r^2 dr\) for radial symmetry.

These coordinates simplify operations like integration for volume and mass calculations of symmetrical objects, making them perfect when working with equations involving spheres.

Integral Calculus

Integral calculus is the branch of mathematics used to calculate quantities like areas, volumes, and other relevant physical properties when the data is continuous.

In this problem, we use integral calculus to determine the mass and moment of inertia, two important physical properties.For calculating these properties, integrals help to sum up small quantities to get the total.

### Mass CalculationThe mass integral is a direct application of integral calculus. It adds up the infinitesimal mass contributions from each spherical shell:\[M = \int_0^{0.200} (3.00 \times 10^3 - 9.00 \times 10^3 \cdot r) \cdot 4\pi r^2 dr.\]

### Moment of InertiaTo find the moment of inertia, we use another integral:\[I = \int_0^{0.200} 4\pi r^4 (3.00 \times 10^3 - 9.00 \times 10^3 \cdot r) dr,\]which calculates how mass is distributed relative to an axis. It involves raising the radius to a higher power.

The practice of solving these integrals involves setting up, simplifying, and evaluating them, which underscores the power of integral calculus in physics.