Problem 35 - Thomas Calculus

Question

The Parallel Axis Theorem Let \(L_{\mathrm{cm}}\) be a line through the center of mass of a body of mass \(m\) and let \(L\) be a parallel line \(h\) units away from \(L_{c . m .}\) The Parallel Axis Theorem says the moments of inertia \(I_{\mathrm{cm}}\) and \(I_{L}\) of the body about \(L_{\mathrm{cm}}\) and \(L\) satisfy the equation $$ I_{L}=I_{\mathrm{c.m.}}+m h^{2} $$ As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known. Proof of the Parallel Axis Theorem a. Show that the first moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the \(y z\) -plane. What does the formula \(\overline{x}=M_{y z} / M\) then tell you?) b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{c . m . \text { along the }}\) \(z\) z-axis and the line \(L\) perpendicular to the \(x y\) -plane at the point \((h, 0,0) .\) Let \(D\) be the region of space occupied by the body. Then, in the notation of the figure, $$ I_{L}=\iiint_{D}|\mathbf{v}-h \mathbf{i}|^{2} d m $$ Expand the integrand in this integral and complete the proof.

Step-by-Step Solution

Answer

The Parallel Axis Theorem states that \( I_L = I_{cm} + mh^2 \). The first moment about the center of mass is zero, confirming the theorem.

Step 1: Understanding the First Moment Condition

We need to show that the first moment of a body about any plane through the body's center of mass is zero. Place the body's center of mass at the origin, and let's choose the yz-plane. The formula for the x-coordinate of the center of mass is \( \overline{x} = \frac{M_{yz}}{M} \), where \( M_{yz} \) is the first moment of mass about the yz-plane. Since the center of mass is at the origin, \( \overline{x} = 0 \) implies \( M_{yz} = 0 \). Therefore, the first moment about the yz-plane is zero.

Step 2: Setting Up the Integral for Moment of Inertia

For step b, we need to calculate the moment of inertia \( I_L \) about the line \( L \). The line \( L_{cm} \) is along the z-axis and \( L \) is parallel but shifted by \( h \) units on the x-axis. The moment of inertia about \( L \) is given by \( I_L = \iiint_D |\mathbf{v} - h\mathbf{i}|^2 \ dm \), where \( \mathbf{v} \) is the position vector \((x, y, z)\) and \( \mathbf{i} \) is the unit vector in the x-direction.

Step 3: Expand and Simplify the Integrand

Expand the integrand: \( |\mathbf{v} - h\mathbf{i}|^2 = |(x - h)\mathbf{i} + y\mathbf{j} + z\mathbf{k}|^2 = (x - h)^2 + y^2 + z^2 \). This expansion is straightforward as we apply the distance formula in three-dimensional space.

Step 4: Express and Split into Known Components

The integral splits as follows: \[ I_L = \iiint_D (x^2 - 2hx + h^2 + y^2 + z^2) \ dm = \iiint_D x^2 \ dm + \iiint_D y^2 \ dm + \iiint_D z^2 \ dm - 2h \iiint_D x \ dm + h^2 \iiint_D \ dm \]. Notice that the middle term (\(-2h \iiint_D x \ dm\)) is zero due to step 1, as it represents the first moment about the yz-plane.

Step 5: Completing the Parallel Axis Theorem Proof

Given that the center of mass is at the origin and the first moment about planes through the center of mass is zero, we have: \[ I_L = I_{cm} + mh^2 \], where \( I_{cm} = \iiint_D (x^2 + y^2 + z^2) \ dm \) using the center of mass origin property. Thus, the expression for \( I_L \) is shown to be \( I_{cm} + mh^2 \), completing the proof.

Key Concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational mechanics. It describes how mass is distributed in a body and affects its resistance to rotational motion. Think of it as the rotational equivalent of mass in linear motion. This means objects with larger moments require more effort to rotate.
This physical property depends on the object's mass and how that mass is distributed relative to the axis of rotation. Mathematically, it is represented as:\[ I = \int r^2 \, dm \]where \( r \) is the distance from the axis of rotation, and \( dm \) represents an infinitesimal mass element.

• For a point mass, \( I = mr^2 \)
• For continuous objects, integration is used to sum up the contributions of all small mass elements

Studying moments of inertia helps understand why different bodies rotate differently and is key to solving problems using the Parallel Axis Theorem.

Center of Mass

The center of mass is the point in an object or system where the entire mass can be considered to be concentrated. It’s like the balance point of the body. For symmetric bodies with uniform distribution, it is located at the geometric center, but for irregular shapes, it might be away from the center.

The center of mass is crucial because it simplifies the analysis of motion. Instead of tracking each particle of a body, you can often track just this single point.
The mathematical expresion for center of mass in three dimensions is:\[ \overline{x} = \frac{1}{M}\int x \, dm, \quad \overline{y} = \frac{1}{M}\int y \, dm, \quad \overline{z} = \frac{1}{M}\int z \, dm \]where \( M \) is the total mass, and \( dm \) is an element of mass at position \( x, y, z \).

In solving physics problems, particularly with the Parallel Axis Theorem, knowing the center of mass allows you to more easily calculate moments of inertia around other axes.

Integral Calculus

Integral calculus is a branch of mathematics dealing with integrals and their properties. It is the continuous analogue to summation, allowing one to calculate areas, volumes, and more. In physics, it is often used to determine quantities when dealing with continuous distributions of mass or charge.

When applying calculus to find the moment of inertia, we consider the entire body's contribution by integrating small mass elements, each multiplied by the square of its distance from the axis of rotation. This gives more precise calculations for objects of irregular shapes.

• The basic idea is to sum up all tiny contributions (\( \int r^2 \, dm \)) across the body.
• This method ensures that all masses are accounted for, even those far from the center.

Understanding integral calculus is essential for being able to apply the Parallel Axis Theorem correctly.

Physics Application

The Parallel Axis Theorem is a practical application of physics concepts, essential for calculating the moment of inertia about different axes. This theorem states:\[ I_L = I_{\text{cm}} + m h^2 \]where \( I_L \) is the moment of inertia about a line parallel to and a distance \( h \) away from, the axis through the center of mass.

This theorem is especially useful in engineering and mechanics where machinery and structures often involve rotating parts.

• The theorem simplifies calculations when designing components like gears or when analyzing the rotation of objects like tires.
• It's not just theoretical, but also a powerful tool in understanding real-world phenomena.

By leveraging this theorem, engineers and physicists can predict how an object will behave when subjected to rotational forces. This understanding is critical in ensuring the safety and efficiency of many mechanical systems.