Problem 97 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

(a) The moment of inertia is \( \frac{3MR^2}{5} \). (b) It's smaller than for uniform density, due to mass distribution toward the outer regions.

Step 1: Understanding the Density Function

The problem states that the density function \[ \rho = \alpha r, \]where \( \rho \) is the density, \( \alpha \) is a constant, and \( r \) is the radial distance from the axis of the cylinder. This implies that the density of the material increases linearly as you move away from the center of the cylinder.

Step 2: Express Mass Element in Terms of Density

For a small disc element inside the cylinder with radius \( r \), thickness \( dr \), and length \( L \), the mass element \( dm \) can be expressed as:\[ dm = \rho \, dV, \]where \( dV \) is the volume element, expressed as:\[ dV = 2 \pi r L \, dr. \]Thus, \[ dm = \alpha r (2 \pi r L) \, dr = 2 \pi \alpha L r^2 \, dr. \]

Step 3: Compute Total Mass

Integrate \( dm \) over the entire volume from \( r = 0 \) to \( r = R \) to find the total mass \( M \):\[ M = \int_{0}^{R} 2 \pi \alpha L r^2 \, dr = \left[ \frac{2 \pi \alpha L r^3}{3} \right]_0^R = \frac{2 \pi \alpha L R^3}{3}. \]

Step 4: Express \( \alpha \) in Terms of \( M \)

Using the equation for total mass:\[ \frac{2 \pi \alpha L R^3}{3} = M, \]solve for \( \alpha \):\[ \alpha = \frac{3M}{2 \pi L R^3}. \]

Step 5: Calculate the Moment of Inertia

The moment of inertia \( I \) for a disc about the central longitudinal axis is given by:\[ dI = r^2 \, dm = r^2 (2 \pi \alpha L r^2 \, dr) = 2 \pi \alpha L r^4 \, dr. \]Now, integrate to find the total moment of inertia:\[ I = \int_{0}^{R} 2 \pi \alpha L r^4 \, dr = \left[ \frac{2 \pi \alpha L r^5}{5} \right]_0^R = \frac{2 \pi \alpha L R^5}{5}. \]

Step 6: Substitute \( \alpha \) into \( I \)

Substitute \( \alpha = \frac{3M}{2 \pi L R^3} \) into the equation for \( I \):\[ I = \frac{2 \pi L R^5}{5} \cdot \frac{3M}{2 \pi L R^3} = \frac{3MR^2}{5}. \]

Step 7: Compare with Uniform Density Cylinder

For a cylinder with uniform density, the moment of inertia \( I_{uniform} \) is:\[ I_{uniform} = \frac{MR^2}{2}. \]Comparing the moments:\[ \frac{3MR^2}{5} < \frac{MR^2}{2}, \]indicating the moment of inertia for the linearly increasing density is smaller.

Step 8: Explain the Result Qualitatively

The moment of inertia is smaller because more mass is concentrated farther from the axis, where the linear increase in density emphasizes the outer regions. This distribution pattern reduces the overall contribution to rotational inertia compared to a uniform density.

Key Concepts

Cylinder with Variable Density

Imagine a cylinder, like a solid rod, but with something special going on inside. Instead of having the same density throughout, its density actually changes as you move away from the center. This is what we call a cylinder with variable density. The density of this cylinder increases linearly, meaning if you trace a path from the center to the outer edge, the density gets higher bit by bit. Such a change in density impacts how the cylinder behaves when it spins around its center axis.

In this particular cylinder, the density is represented by the function \( \rho = \alpha r \). Here, \( \alpha \) is a constant that adjusts how fast the density changes, and \( r \) is the distance from the axis of the cylinder. It's like a slider that turns up the density as you extend the radius. This linear relationship means that at the center (where \( r = 0 \)) the density is zero, and it gradually increases as you head towards the edge of the cylinder.

Linear Density Function

The term 'linear density function' refers to how the density of the cylinder changes with distance from the center. This function is essential for understanding how mass is distributed in the cylinder. With a linear density function, as you move further from the center, each tiny piece of the cylinder contains just a bit more mass than the piece before it.

In mathematical terms, this is described by \( \rho = \alpha r \), where \( \rho \) is the density, \( \alpha \) is a constant we determined by comparing the total mass of the cylinder, and \( r \) is the radial distance. This function implies that the outer layers of the cylinder are heavier than the inner layers. It becomes crucial when we need to calculate properties like the moment of inertia, as it affects how the mass contributes to rotational dynamics. This distribution helps us systematically address changes in mass across the cylinder.

Rotational Dynamics

Rotational dynamics is the branch of physics that deals with the rotation of objects and the forces and torques that affect them. When we talk about a cylinder with variable density, its rotational dynamics are interesting because how the mass is distributed affects how easily it spins.

One key part of rotational dynamics is the moment of inertia. This tells us how much resistance an object has to changes in its rotational motion, much like how mass resists changes in linear motion. For this cylinder, the moment of inertia differs because its mass is not evenly spread. The formula we use, \( I = \int r^2 \, dm \), helps determine the moment of inertia by taking into account the variation in mass distribution because of the linear density function.

Calculating it precisely requires integrating across the entire radius, turning a uniform concept into one affected by the increased outer mass. This changes expectations and results of how the cylinder will respond to rotational forces.

Physics Problem Solving

Solving physics problems, especially those involving rotational dynamics and moments of inertia, can initially seem daunting. However, breaking them down into manageable steps helps. Let's take this cylinder as an example. When faced with a problem, the first key step is understanding the given parameters and functions, such as how the density varies and what the linear function is describing.

Once the basic understanding is established, one can proceed to express small parts of the object in terms of its known variables – like using \( dm = \rho \, dV \) to express tiny elements of mass. This approach leads to integrating over the entirety of the object to find values like total mass. From there, substituting known quantities into formulas allows for the derivation of larger characteristics, like moment of inertia. Finally, comparing these results to known values for similar uniform density objects provides insights into the impact of variable density.

This kind of systematic approach is very helpful for tackling complex physics problems effectively and achieving clearer understanding.