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A cylinder with radius \(R\) and mass \(M\) has density that increases linearly with distance \(r\) from the cylinder axis, \(\rho=\alpha r\) where \(\alpha\) is a positive constant. (a) Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of \(M\) and \(R .\) (b) Is your answer greater or smaller than the moment of inertia of a cylinder of the same mass and radius but of uniform density? Explain why this result makes qualitative sense.

Short Answer

Expert verified
The moment of inertia for the given cylinder is \(\frac{3}{5} MR^2\), which is larger than \(\frac{1}{2} MR^2\) for a uniform distribution.

Step by step solution

01

Relationship between Density and Volume

The density of the cylinder is given as \( \rho(r) = \alpha r \), where \( r \) is the distance from the axis. The volume element for the cylinder in cylindrical coordinates is \( dV = 2\pi r \, dr \, dz \).
02

Find Differential Mass Element

The mass element \( dm \) can be expressed as \( dm = \rho(r) \, dV = \alpha r(2\pi r \, dr \, dz) = 2\pi \alpha r^2 \, dr \, dz \).
03

Integral for Total Mass

Integrate to find the total mass \( M \) of the cylinder.\[ M = \int_0^L \int_0^R 2\pi \alpha r^2 \, dr \, dz \]
04

Evaluate Total Mass Integral

Evaluate the inner integral with respect to \( r \).\[ M = \int_0^L \left(2\pi \alpha \int_0^R r^2 \, dr \right) dz \] The inner integral evaluates to \( \frac{R^3}{3} \). So,\[ M = 2\pi \alpha \frac{R^3}{3} L \]
05

Express \(\alpha\) in Terms of Known Quantities

From \( M = 2\pi \alpha \frac{R^3}{3} L \), solve for \( \alpha \).\[ \alpha = \frac{3M}{2\pi R^3 L} \]
06

Moment of Inertia Formula

The moment of inertia about the axis is \( dI = r^2 \, dm = r^2(2\pi \alpha r^2 \, dr \, dz) = 2\pi \alpha r^4 \, dr \, dz \).
07

Integral for Moment of Inertia

Calculate the moment of inertia \( I \).\[ I = \int_0^L \int_0^R 2\pi \alpha r^4 \, dr \, dz \]
08

Evaluate Moment of Inertia Integral

Evaluate the inner integral with respect to \( r \).\[ I = \int_0^L \left(2\pi \alpha \int_0^R r^4 \, dr \right) dz \] The inner integral evaluates to \( \frac{R^5}{5} \). So,\[ I = 2\pi \alpha \frac{R^5}{5} L \]
09

Substitute \(\alpha\) and Simplify

Substitute \( \alpha = \frac{3M}{2\pi R^3L} \) into the expression for \( I \) to obtain \[ I = 2\pi \left(\frac{3M}{2\pi R^3 L}\right) \frac{R^5}{5} L \] Simplifying gives \[ I = \frac{3}{5} MR^2 \]
10

Compare with Uniform Cylinder

The moment of inertia of a uniform cylinder is \( \frac{1}{2} MR^2 \). The calculated moment of inertia is \( \frac{3}{5} MR^2 \), which is larger. This is because the density increases with \( r \), resulting in more mass positioned further from the axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Distribution
In physics, density distribution describes how the mass of an object varies from one point to another. This is crucial when calculating properties like the moment of inertia. When we have a variable density, such as in this cylinder, density is not uniform but increases linearly with the distance from the axis. This characteristic is expressed mathematically as \( \rho = \alpha r \), where \( r \) is the distance from the center, and \( \alpha \) is a positive constant indicating the rate of increase. Because denser materials contribute more to the moment of inertia, the linear increase means more mass is distributed further from the axis, thus potentially increasing the moment of inertia compared to a uniformly dense object.
Cylindrical Coordinates
Cylindrical coordinates are used to describe the positions of points in a space where symmetry around an axis is involved, like in cylindrical shapes. They are particularly useful when computing integrals over cylindrical volumes, as they align well with the cylindrical geometry and simplify calculations. Three components describe the position of a point:
  • \( r \): the radial distance from the center axis.
  • \( \theta \): the angle around the axis, which we don't need for symmetry in this problem.
  • \( z \): the height along the axis.
For a volume element \( dV \) within a cylinder, we use the expression \( dV = 2\pi r \, dr \, dz \). This accounts for a small slice of the cylinder, where \( dr \) is the radial thickness and \( dz \) is the height.
Integral Calculus
Integral calculus lets us calculate quantities like mass and moment of inertia by summing infinitesimally small contributions over a continuous object. When the density distribution is not uniform, integrals efficiently handle these calculations. For example, to determine the total mass \( M \), we integrate the mass element \( dm = \rho(r) \, dV \), considering the whole volume. This becomes:\[ M = \int_0^L \int_0^R 2\pi \alpha r^2 \, dr \, dz \]Likewise, the moment of inertia, which fundamentally measures how mass is distributed with respect to an axis, requires an integral approach. In this problem, we calculate the moment of inertia with the integral:\[ I = \int_0^L \int_0^R 2\pi \alpha r^4 \, dr \, dz \]These operations illustrate the power of calculus to derive essential physical properties from foundational principles.
Rotational Motion
Rotational motion refers to an object's movement around an axis, central to understanding dynamics in physics. A critical feature of rotational motion is the moment of inertia, which is a measure of how resistant an object is to changes in its rotation. This is analogous to mass in linear motion. In our specific problem, the linear increase in density with distance from the axis means more mass is concentrated further out. Therefore, the moment of inertia is calculated to be \( \frac{3}{5} MR^2 \), larger than that of a uniformly dense cylinder, \( \frac{1}{2} MR^2 \). This larger inertia makes sense intuitively; the more mass you have farther from the axis, the harder it is to rotate the object. Hence, understanding the moment of inertia helps predict how objects will behave under rotational forces.

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