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A triangular prism (like a box of Toblerone) of mass \(M,\) whose two ends are equilateral triangles parallel to the \(x y\) plane with side \(2 a,\) is centered on the origin with its axis along the \(z\) axis. Find its moment of inertia for rotation about the \(z\) axis. Without doing any integrals write down and explain its two products of inertia for rotation about the \(z\) axis.

Short Answer

Expert verified
Moment of Inertia \(I_z = \frac{1}{9} M a^2\); Products of inertia are zero.

Step by step solution

01

Understanding the Problem

We need to find the moment of inertia of a triangular prism about the z-axis. The two ends of the prism are equilateral triangles with side length \(2a\), aligned parallel to the xy-plane. The mass of the prism is \(M\) and it is centered on the origin.
02

Moment of Inertia Formula for Prism

The moment of inertia \(I_z\) for rotation about the z-axis for a solid object is given by \(I_z = \int_V (x^2 + y^2) \, dm\). Because no integrals should be done, we can use known results for standard shapes. For a triangular prism, the moment of inertia about the axis passing through the center and aligned with the height (z-axis) can be re-used from standard formulas.
03

Using Known Results

For an equilateral triangle of side \(2a\), centered at the origin, the moment of inertia \(I_{z\text{-triangle}}\) about its centroid perpendicular to its plane is \(\frac{1}{36} M (2a)^2 = \frac{1}{9} M a^2\). For a triangular prism of height \(h\), the moment of inertia about an axis along its length is \(I_z = \frac{1}{18} M (2a)^2 = \frac{1}{9} M a^2\). Here, \(h\) cancels out because the height of the prism is irrelevant for this axial rotation.
04

Calculating Products of Inertia

By symmetry, since the triangular prism is centered on the origin and symmetric about the z-axis, the products of inertia \(I_{xy}\), \(I_{yz}\), and \(I_{zx}\) with respect to z-axis are zero. Products of inertia indicate the distribution of mass in the off-diagonal elements of the inertia tensor, and for symmetric bodies about the axis considered, they are zero.

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Key Concepts

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

Triangular Prism
A triangular prism is a three-dimensional geometric shape. It consists of two parallel ends shaped like triangles, and three rectangular faces that connect these ends. This shape is much like a Toblerone chocolate bar, designed with equilateral triangles as its base.

In the context of this problem, the triangular prism has its axis aligned along the z-axis. This is essential to understanding its geometric configuration for calculating moments and products of inertia. The prism's center aligns with the origin of a coordinate system, simplifying calculations as symmetry properties can greatly reduce complexity.
  • Two ends: Equilateral triangles
  • Axis: Along the z-axis
  • Center: At the origin
Equilateral Triangles
Equilateral triangles have all three sides of equal length. In this exercise, each side is measured as \(2a\). These triangles form the base and top of the triangular prism and influence the mass distribution for inertia calculations.

The property of equilateral triangles that plays a crucial role here is symmetry. Each triangle's center of mass is at its centroid. For rotational inertia, this leads to symmetry-driven simplifications, meaning many inertia components like products of inertia can equal zero. Understanding this symmetry helps in deriving formulas without engaging complex integrations.
  • All sides equal: \(2a\)
  • Center of mass: At the centroid
  • Use of symmetry: Simplifies calculations
Products of Inertia
The products of inertia measure the coupling between different rotational axes. They reflect how mass distribution impacts rotational dynamics. In mathematical terms, these are represented as off-diagonal elements in the inertia tensor.

For a symmetric shape like our triangular prism centered at the origin, the products of inertia are zero. By aligning it on the z-axis, the symmetry ensures no preference for rotation around the x or y axes, vis-à-vis the z-axis. This simplifies tensor calculations dramatically, as off-diagonal terms do not contribute.
  • Reflect mass distribution between axes
  • Off-diagonal elements in inertia tensor
  • Zero for symmetric objects centered at origin
Z-axis Rotation
Z-axis rotation refers to spinning an object around the z-axis, perpendicular to the base triangles in our prism. The moment of inertia for this axis reveals how resistant the object is to spinning.

For the given triangular prism, the key is in using symmetry and known formulae for standard shapes to determine inertia without performing direct integrals. Instead of integrating mass elements, we rely on the configuration's simplicity; the tri-prism acts like multiple equilateral triangles stacked along the z-axis.
  • Axis of rotation: Z-axis
  • Using symmetry and known results
  • Determined without integrals for standard shapes

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Most popular questions from this chapter

Consider a rigid plane body or "lamina," such as a flat piece of sheet metal, rotating about a point \(O\) in the body. If we choose axes so that the lamina lies in the \(x y\) plane, which elements of the inertia tensor \(\mathbf{I}\) are automatically zero? Prove that \(I_{z z}=I_{x x}+I_{y y}\).

An axially symmetric space station (principal axis \(\mathbf{e}_{3},\) and \(\lambda_{1}=\lambda_{2}\) ) is floating in free space. It has rockets mounted symmetrically on either side that are firing and exert a constant torque \(\Gamma\) about the symmetry axis. Solve Euler's equations exactly for \(\omega\) (relative to the body axis) and describe the motion. At \(t=0\) take \(\omega=\left(\omega_{10}, 0, \omega_{30}\right)\)

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 \(\mathbf{I}_{A}=\mathbf{I}_{B}-\mathbf{I}_{C}\).

A thin, flat, uniform metal triangle lies in the \(x y\) plane with its corners at \((1,0,0),(0,1,0),\) and the origin. Its surface density (mass/area) is \(\sigma=24\). (Distances and masses are measured in unspecified units, and the number 24 was chosen to make the answer come out nicely.) (a) Find the triangle's inertia tensor I. (b) What are its principal moments and the corresponding axes?

Consider an arbitrary rigid body with an axis of rotational symmetry, which we'll call \(\hat{\mathbf{z}}\). (a) Prove that the axis of symmetry is a principal axis. (b) Prove that any two directions \(\hat{\mathbf{x}}\) and \(\hat{\mathbf{y}}\) perpendicular to \(\hat{\mathbf{z}}\) and each other are also principal axes. (c) Prove that the principal moments corresponding to these two axes are equal: \(\lambda_{1}=\lambda_{2}\).

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