Question

(a) A thin uniform rod of mass \(M\) and length \(L\) lies on the \(x\) axis with one end at the origin. Find its moment of inertia for rotation about the \(z\) axis. [Here the sum of (10.25) must be replaced by an integral of the form \(\int x^{2} \mu d x\) where \(\mu\) is the linear mass density, mass/length.] (b) What if the rod's center is at the origin?

Short answer

(a) \( \frac{ML^2}{3} \); (b) \( \frac{ML^2}{12} \).

Step-by-step solution

Understand the Problem

We need to calculate the moment of inertia of a thin rod about the z-axis in two scenarios. First, with one end at the origin, and second, with the rod's center at the origin.

Set Up the Integral for Case (a)

For a rod extending from 0 to L, the linear mass density is given by \( \mu = \frac{M}{L} \). The element's mass, \( dm \), is \( \mu \cdot dx \). We use the formula for moment of inertia: \[ I = \int x^2 dm = \int x^2 \mu \, dx \] Substitute \( \mu \) to get \[ I = \frac{M}{L} \int_{0}^{L} x^2 \, dx \].

Solve the Integral for Case (a)

Evaluate the integral:\[ I = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{0}^{L} = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3} \].Thus, the moment of inertia of the rod with one end at the origin is \( \frac{ML^2}{3} \).

Set Up for Case (b)

Now, consider the rod's center at the origin. The rod extends from \(-\frac{L}{2}\) to \(\frac{L}{2}\). The integral becomes:\[ I = \frac{M}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 \, dx \].

Solve the Integral for Case (b)

Evaluate the integral:\[ I = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{M}{L} \left[ \frac{(\frac{L}{2})^3}{3} - \frac{(-\frac{L}{2})^3}{3} \right] = \frac{M}{L} \cdot \frac{L^3}{12} \cdot 2 = \frac{ML^2}{12} \cdot 2 = \frac{ML^2}{6} \].Thus, the moment of inertia of the rod with its center at the origin is \( \frac{ML^2}{12} \).

Key concepts

Linear Mass Density

Linear mass density is a way to describe how mass is distributed along a rod. Imagine stretching a rod along the x-axis, where its mass seems evenly spread across its length. The concept of linear mass density, \( \mu \), will help us understand how to calculate properties like moment of inertia for such objects.

A uniform rod—one with the same thickness and material throughout—has a constant linear mass density. It is defined by the formula:

• \( \mu = \frac{M}{L} \)

where \( M \) is the total mass of the rod and \( L \) is its length.

Knowing \( \mu \) allows us to find the mass of any small segment of the rod. For a tiny piece of length \( dx \), located at position \( x \), the mass \( dm \) is given by:

\[ dm = \mu \cdot dx \]

This calculation is pivotal in setting up the integral used in finding the moment of inertia, as each little piece \( dm \) contributes a small amount to the total inertia.

Integral Calculus

Integral calculus is a mathematical tool that helps us sum an infinite number of tiny parts to find a whole. When measuring the moment of inertia, we integrate the squared distance of each tiny mass segment from the axis of rotation, reflected in its use within our solution.

The formula we use is:

\[ I = \int x^2 dm \]

This integral takes every small part of the rod, represented by \( dm \), and multiplies it by the square of its distance \( x^2 \) from the rotation axis. This operation tells us how much each part "contributes" to the rotational inertia.

To solve for \( I \) through calculus, we transform it using the linear mass density, converting \( dm \) into \( \mu \, dx \). Our integral setup for case (a) involves multiplying the density \( \mu \) with \( x^2 \) and integrating over the rod's length, from 0 to \( L \). The complete form integrates all small sections of the rod, combining them to give a final inertia value.

Rotation About an Axis

When calculating the moment of inertia, it's crucial to understand the axis of rotation. In physics, the axis of rotation is an imaginary line around which an object turns. For this problem, we consider the rod turning about the z-axis.

A change in the rod's position relative to this axis changes how we calculate the moment of inertia. If the rod is anchored at one end, as in case (a), our integration bounds span from 0 to \( L \). However, when the center of the rod is placed over the origin, as in case (b), the rotation affects parts of the rod equally left and right from this axis.

In this situation, our new limits of integration become \(-\frac{L}{2}\) to \(\frac{L}{2}\). This change correctly accounts for the symmetrical distance contributing to the inertia. The moment of inertia values demonstrate how the distribution and position relative to the axis significantly impact the rotational dynamics.