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Show that the inertia tensor is additive, in this sense: Suppose a body \(A\) is made up of two parts \(B\) and \(C\). (For instance, a hammer is made up of a wooden handle wedged into a metal head.) Then \(\mathbf{I}_{A}=\mathbf{I}_{B}+\mathbf{I}_{C} .\) Similarly, if \(A\) can be thought of as the result of removing \(C\) from \(B\) (as a hollow spherical shell is the result of removing a small sphere from inside a larger sphere), then \(\mathbf{I}_{A}=\mathbf{I}_{B}-\mathbf{I}_{C}\).

Short Answer

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The inertia tensor is additive: \( \mathbf{I}_{A} = \mathbf{I}_{B} + \mathbf{I}_{C} \) when composed, \( \mathbf{I}_{A} = \mathbf{I}_{B} - \mathbf{I}_{C} \) when removed.

Step by step solution

01

Understand Inertia Tensor

The inertia tensor of a rigid body is a mathematical representation describing how the mass of the object is distributed with respect to certain axes. It depends on both the mass and the geometry of the object.
02

Define Inertia Tensor Additivity

The question states that if a body \( A \) is composed of two parts \( B \) and \( C \), the inertia tensor of the whole body \( \mathbf{I}_{A} \) should be the sum of the inertia tensors of parts \( B \) and \( C \), symbolically expressed as \( \mathbf{I}_{A} = \mathbf{I}_{B} + \mathbf{I}_{C} \) and similarly for subtraction when one is removed from another.
03

Express Each Inertia Tensor

Each part of the body, \( B \) and \( C \), has its own inertia tensor, \( \mathbf{I}_{B} \) and \( \mathbf{I}_{C} \), respectively. For a composite body \( A \) that consists of \( B \) and \( C \), the inertia tensor is derived by considering their contributions to the overall mass distribution.
04

Sum of Inertia Tensors when Adding Components

Additivity means that the total inertia tensor for body \( A \), when composed of \( B \) and \( C \) together, is the mathematical addition of the individual inertia tensors: \( \mathbf{I}_{A} = \mathbf{I}_{B} + \mathbf{I}_{C} \). This is because the mass distribution of \( B \) and \( C \) independently contribute to the resistance to changes in rotational motion, and can be summed.
05

Subtract Inertia Tensors when Removing a Component

When considering the removal of \( C \) from \( B \), the resulting inertia tensor \( \mathbf{I}_{A} \) can be found by subtracting \( \mathbf{I}_{C} \) from \( \mathbf{I}_{B} \) as \( \mathbf{I}_{A} = \mathbf{I}_{B} - \mathbf{I}_{C} \). This accounts for the decrease in mass and redistribution that occurs when \( C \) is removed.
06

Apply Additivity to General Cases

These principles of additivity and subtractive combination can be applied to any body composed of simpler sub-bodies: adding components increases the inertia tensor and removing components decreases it, corresponding to the physical redistribution of the mass of how it resists rotational acceleration.

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Key Concepts

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

Additivity of Inertia Tensor
The concept of the inertia tensor being additive centers on understanding that the total inertia of a composite body can be determined by summing the inertia tensors of its individual parts. Imagine a hammer: it is composed of a metal head and a wooden handle. Here, the mathematical representation of its inertia, called the inertia tensor, can be found by simply adding the inertia tensors of both the head and the handle.
  • Combination of Parts: If you know the inertia of each part, simply add them together to get the whole body's inertia.
  • Breaking Down Complex Bodies: This approach can be used to break down complex bodies into known units and calculate the inertia easily.
For instance, if part A of a body is made up of parts B and C, the inertia tensor is given as \( \mathbf{I}_{A} = \mathbf{I}_{B} + \mathbf{I}_{C} \). Conversely, if A is essentially what's left when you take C out of B, then \( \mathbf{I}_{A} = \mathbf{I}_{B} - \mathbf{I}_{C} \). This is a significant idea because it applies to a variety of mechanical problems, particularly in engineering and physics.
Rigid Body Dynamics
Rigid body dynamics focuses on the motion and properties of bodies that do not deform. This means that during its motion, no point within the body changes position relative to its center of mass. It is within this realm that the inertia tensor is most relevant.
  • Predicting Motion: In rigid body dynamics, knowing the inertia tensor helps predict how an object will react when forces are applied, especially rotational forces.
  • Fixed Shapes: Since the object's shape and mass distribution do not change, calculations made using the inertia tensor remain valid across different scenarios of motion.
In rigid body dynamics, the inertia tensor helps us calculate the angular motion of bodies effectively. This is instrumental in fields like robotics and aeronautics where precise motion control is essential.
Mass Distribution
Mass distribution within a rigid body determines its resistance to changes in its rotational motion, which is mathematically captured by the inertia tensor.
  • Impact on Rotational Characteristics: How the mass of an object is spread out influences its rotational inertia—objects with mass far from the rotation axis have higher inertia.
  • Design Implications: Engineers must consider mass distribution to ensure optimal performance in machinery and vehicles.
The inertia tensor is more than just a number; it represents how mass is laid out in three-dimensional space, which affects how an object spins and responds to torque. By altering mass distribution, like adding weight to a different part of an object, its inertia tensor changes—impacting its ability to rotate or change orientation.
Rotational Motion
Rotational motion refers to the movement of a body around a fixed axis, which is a common form of motion for many objects. To understand this fully, the concept of the inertia tensor is crucial because it determines how the object resists.rotation.
  • Axis of Rotation: Any rotating body has an axis around which it rotates. The inertia tensor helps understand how the rotation occurs around different axes.
  • Angular Acceleration: The resistance to changes in rotational motion is vital to determining the angular acceleration when torque is applied.
Using the inertia tensor, physicists and engineers can calculate the torque needed to achieve desired angular accelerations, or predict how a spinning object will behave when forces are applied. It allows for precision in applications like designing turbines or balancing wheels in a vehicle.

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Most popular questions from this chapter

Suppose that you have found three independent principal axes (directions \(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\) ) and corresponding principal moments \(\lambda_{1}, \lambda_{2}, \lambda_{3}\) of a rigid body whose moment of inertia tensor \(\mathbf{I}\) (not diagonal) you had calculated. (You may assume, what is actually fairly easy to prove, that all of the quantities concerned are real.) (a) Prove that if \(\lambda_{i} \neq \lambda_{j}\) then it is automatically the case that \(\mathbf{e}_{i} \cdot \mathbf{e}_{j}=0\) (It may help to introduce a notation that distinguishes between vectors and matrices. For example, you could use an underline to indicate a matrix, so that \(\underline{\mathbf{a}}\) is the \(3 \times 1\) matrix that represents the vector a, and the vector scalar product a \(\cdot \mathbf{b}\) is the same as the matrix product \(\tilde{\mathbf{a}} \mathbf{b}\) or \(\underline{\mathbf{b}}\) a. Then consider the number \(\tilde{\mathbf{e}}_{i} \mathbf{I} \mathbf{e}_{j},\) which can be evaluated in two ways using the fact that both \(\mathbf{e}_{i}\) and \(\mathbf{e}_{j}\) are eigenvectors of I.) (b) Use the result of part (a) to show that if the three principal moments are all different, then the directions of three principal axes are uniquely determined. (c) Prove that if two of the principal moments are equal, \(\lambda_{1}=\lambda_{2}\) say, then any direction in the plane of \(\mathbf{e}_{1}\) and \(\mathbf{e}_{2}\) is also a principal axis with the same principal moment. In other words, when \(\lambda_{1}=\lambda_{2}\) the corresponding principal axes are not uniquely determined. (d) Prove that if all three principal moments are equal, then any axis is a principal axis with the same principal moment.

The moment of inertia of a continuous mass distribution with density \(\varrho\) is obtained by converting the sum of (10.25) into the volume integral \(\int \rho^{2} d m=\int \rho^{2} \varrho d V\). (Note the two forms of the Greek "rho": \(\rho=\) distance from \(z\) axis, \(\varrho=\) mass density.) Find the moment of inertia of a uniform circular cylinder of radius \(R\) and mass \(M\) for rotation about its axis. Explain why the products of inertia are zero.

Five equal point masses are placed at the five corners of a square pyramid whose square base is centered on the origin in the \(x y\) plane, with side \(L,\) and whose apex is on the \(z\) axis at a height \(H\) above the origin. Find the CM of the five-mass system.

To illustrate the result (10.18), that the total KE of a body is just the rotational KE relative to any point that is instantaneously at rest, do the following: Write down the KE of a uniform wheel (mass M) rolling with speed \(v\) along a flat road, as the sum of the energies of the CM motion and the rotation about the CM. Now write it as the energy of the rotation about the instantaneous point of contact with the road and show that you get the same answer. (The energy of rotation is \(\frac{1}{2} I \omega^{2} .\) The moment of inertia of a uniform wheel about its center is \(I=\frac{1}{2} M R^{2} .\) That about a point on the rim is \(I^{\prime}=\frac{3}{2} M R^{2}\).)

Find the inertia tensor for a uniform, thin hollow cone, such as an ice-cream cone, of mass \(M,\) height \(h,\) and base radius \(R,\) spinning about its pointed end.

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