Chapter 9: Problem 9
Find the moment of inertia about an axis through its centre of a uniform hollow sphere of mass \(M\) and outer and inner radii \(a\) and \(b\). (Hint: Think of it as a sphere of density \(\rho\) and radius \(a\), with a sphere of density \(\rho\) and radius \(b\) removed.)
Short Answer
Expert verified
Answer: The moment of inertia of the hollow sphere is given by the formula: \(I = \frac{6M}{5} \frac{a^5 - b^5}{a^3 - b^3}\).
Step by step solution
01
Find the density of the hollow sphere
First, let's find the density of the hollow sphere. The volume of the outer sphere is \(V_a = \frac{4}{3}\pi a^3\), and the inner sphere has volume \(V_b = \frac{4}{3}\pi b^3\). The volume of the hollow sphere is their difference \(V = V_a - V_b = \frac{4}{3}\pi(a^3-b^3)\). Since we know the total mass of the hollow sphere is \(M\), we can find its density using \(\rho = \frac{M}{V} = \frac{3M}{4\pi(a^3-b^3)}\).
02
Setup the integral for the moment of inertia
We will use the formula for the moment of inertia for a small volume element, \(dI = r^2 dm\), where \(dm\) is the mass of this volume element. In spherical coordinates, the volume element is given by \(dV = r^2 \sin\theta dr d\theta d\phi\). Since \(dm = \rho dV\), we can rewrite the expression for \(dI\) as \(dI = r^2 (\rho dV)\). Now, we can set up the integral for moment of inertia by integrating over the volume of the hollow sphere:
\(I = \int_{b}^{a} \int_{0}^{\pi} \int_{0}^{2\pi} r^2 \rho r^2 \sin\theta dr d\theta d\phi\).
03
Evaluate the integral
First, we plug in the density \(\rho\) and simplify the expression inside the integral:
\(I = \int_{b}^{a} \int_{0}^{\pi} \int_{0}^{2\pi} r^4 \frac{3M}{4\pi(a^3-b^3)}\sin\theta dr d\theta d\phi\).
Now we can separate the variables and integrate each part:
\(I = \frac{3M}{4\pi(a^3-b^3)} \left(\int_{b}^{a} r^4 dr\right) \left(\int_{0}^{\pi} \sin\theta d\theta\right) \left(\int_{0}^{2\pi} d\phi\right)\).
Evaluating each integral, we have:
\(I = \frac{3M}{4\pi(a^3-b^3)} \left[\frac{1}{5}\left(a^5 - b^5\right)\right] \cdot 2 \cdot 2\pi\).
04
Simplify the expression for the moment of inertia
Finally, we can simplify the expression for the moment of inertia:
\(I = \frac{6M}{5} \frac{a^5 - b^5}{a^3 - b^3}\).
This is the moment of inertia of the hollow sphere with mass \(M\), outer radius \(a\) and inner radius \(b\), about an axis through its center.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hollow Sphere Inertia Calculation
Let's begin by understanding the concept of moment of inertia for a hollow sphere. The moment of inertia is a measure of how difficult it is to change the rotational motion of an object about an axis. For a hollow sphere, this calculation takes into account the mass distribution from the inner radius, b, to the outer radius, a. Unlike a solid sphere where the mass is uniformly distributed throughout its volume, the hollow sphere has all of its mass located between these two radial boundaries.
In our exercise, we account for the lack of mass in the inner portion by considering the sphere as a solid sphere of radius a with another solid sphere of radius b removed from its center. To find the inertia, we utilize the integral calculus, which allows us to sum up the tiny contributions to the total inertia from each infinitesimal mass element within the volume bounded by radii a and b.
These small mass elements are positioned at a distance r from the center, affecting inertia through the equation dI = r² dm, with dm representing the mass element. Therein lies the importance of integrating in spherical coordinates: it simplifies addressing the mass distribution within the spherical volume.
During the process, we must also consider the density of the hollow sphere, as it's an intrinsic property related to how the mass M is spread over the volume between the inner and outer surfaces. Approaching the problem stepwise ensures clarity and serves as a bridge towards the final formula that unveils the nature of rotational resistance for the given hollow sphere.
In our exercise, we account for the lack of mass in the inner portion by considering the sphere as a solid sphere of radius a with another solid sphere of radius b removed from its center. To find the inertia, we utilize the integral calculus, which allows us to sum up the tiny contributions to the total inertia from each infinitesimal mass element within the volume bounded by radii a and b.
These small mass elements are positioned at a distance r from the center, affecting inertia through the equation dI = r² dm, with dm representing the mass element. Therein lies the importance of integrating in spherical coordinates: it simplifies addressing the mass distribution within the spherical volume.
During the process, we must also consider the density of the hollow sphere, as it's an intrinsic property related to how the mass M is spread over the volume between the inner and outer surfaces. Approaching the problem stepwise ensures clarity and serves as a bridge towards the final formula that unveils the nature of rotational resistance for the given hollow sphere.
Integrating in Spherical Coordinates
When we talk about integrating in spherical coordinates, we are referring to a mathematical approach tailored to objects with spherical symmetry, like our hollow sphere. Spherical coordinates define points in three-dimensional space based on a radial distance, r, an inclination angle, θ, and an azimuthal angle, φ.
The integration in spherical coordinates is essential for our problem as it aligns perfectly with the symmetry of a sphere. It breaks up the volume into concentric shells which we then subdivide further into ring-like elements using the inclination and azimuthal angles. This subdivision allows us to accumulate the moment of inertia contributions from each shell and ring.
Specifically, the volume element in these coordinates is dV = r² sin(θ) dr dθ dφ. When we calculate a volume integral, this expression helps to ensure that we're accounting for the varying effect that each point within the hollow sphere's volume has on the moment of inertia. Multiplying by the density yields the mass element dm, which is then inserted into our inertia differential equation, facilitating the process of integrating over the whole volume of the hollow sphere.
The integration in spherical coordinates is essential for our problem as it aligns perfectly with the symmetry of a sphere. It breaks up the volume into concentric shells which we then subdivide further into ring-like elements using the inclination and azimuthal angles. This subdivision allows us to accumulate the moment of inertia contributions from each shell and ring.
Specifically, the volume element in these coordinates is dV = r² sin(θ) dr dθ dφ. When we calculate a volume integral, this expression helps to ensure that we're accounting for the varying effect that each point within the hollow sphere's volume has on the moment of inertia. Multiplying by the density yields the mass element dm, which is then inserted into our inertia differential equation, facilitating the process of integrating over the whole volume of the hollow sphere.
Density of a Hollow Sphere
Considering the density of a hollow sphere is a fundamental step in calculating its moment of inertia. Density, defined as mass per unit volume (ρ = M/V), plays a critical role in how the mass is distributed across the sphere. For a homogenous hollow sphere, the density is consistent throughout the material.
With inner radius b and outer radius a, the sphere's volume — and thus its density — is affected by both. In our exercise, we subtract the volume of the smaller sphere from the larger to find the volume of the hollow region. Understanding that the given mass M is spread out in this hollow volume allows us to calculate the density ρ, which is constant and essential for setting up our integral for the inertia.
It's worth noting that the density directly influences the differential mass element dm which is integral to our calculations. Since our sphere is hollow, the mass is not located at the center but rather distributed along the volume between a and b. This understanding of density is crucial for students to fully grasp how it interplays with the physical configuration to impact the moment of inertia.
With inner radius b and outer radius a, the sphere's volume — and thus its density — is affected by both. In our exercise, we subtract the volume of the smaller sphere from the larger to find the volume of the hollow region. Understanding that the given mass M is spread out in this hollow volume allows us to calculate the density ρ, which is constant and essential for setting up our integral for the inertia.
It's worth noting that the density directly influences the differential mass element dm which is integral to our calculations. Since our sphere is hollow, the mass is not located at the center but rather distributed along the volume between a and b. This understanding of density is crucial for students to fully grasp how it interplays with the physical configuration to impact the moment of inertia.