Question

Here is a good exercise in vector identities and matrices, leading to some important general results: (a) For a rigid body made up of particles of mass \(m_{\alpha},\) spinning about an axis through the origin with angular velocity \(\omega,\) prove that its total kinetic energy can be written as $$ T=\frac{1}{2} \sum m_{\alpha}\left[\left(\omega r_{\alpha}\right)^{2}-\left(\omega \cdot \mathbf{r}_{\alpha}\right)^{2}\right] $$ Remember that \(\mathbf{v}_{\alpha}=\boldsymbol{\omega} \times \mathbf{r}_{\alpha} .\) You may find the following vector identity useful: For any two vectors a and b, $$ (\mathbf{a} \times \mathbf{b})^{2}=a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2} $$ (If you use the identity, please prove it.) (b) Prove that the angular momentum \(\mathbf{L}\) of the body can be written as $$ \mathbf{L}=\sum m_{\alpha}\left[\omega r_{\alpha}^{2}-\mathbf{r}_{\alpha}\left(\omega \cdot \mathbf{r}_{\alpha}\right)\right] $$ For this you will need the so-called \(B A C-C A B\) rule, that \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C})=\mathbf{B}(\mathbf{A} \cdot \mathbf{C})-\mathbf{C}(\mathbf{A} \cdot \mathbf{B})\) (c) Combine the results of parts (a) and (b) to prove that $$ T=\frac{1}{2} \omega \cdot \mathbf{L}=\frac{1}{2} \tilde{\omega} \mathbf{I} \omega $$ Prove both equalities. The last expression is a matrix product; \(\omega\) denotes the \(3 \times 1\) column of numbers \(\omega_{x}, \omega_{y}, \omega_{z},\) the tilde on \(\tilde{\omega}\) denotes the matrix transpose (in this case a row), and I is the moment of inertia tensor. This result is actually quite important; it corresponds to the much more obvious result that for a particle, \(T=\frac{1}{2} \mathbf{v} \cdot \mathbf{p} .\) (d) Show that with respect to the principal axes, \(T=\frac{1}{2}\left(\lambda_{1} \omega_{1}^{2}+\lambda_{2} \omega_{2}^{2}+\lambda_{3} \omega_{3}^{2}\right)\) as in Equation (10.68).

Short answer

The kinetic energy \( T \) can be derived using vector identities and is given by \( T = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{L} = \frac{1}{2} \tilde{\boldsymbol{\omega}} \mathbf{I} \boldsymbol{\omega} \).

Step-by-step solution

Understand the Problem Statement

We are given a rigid body and need to prove expressions for its kinetic energy and angular momentum in terms of angular velocity \( \omega \) and position vectors \( \mathbf{r}_{\alpha} \). We will use provided vector identities and matrix operations.

Expression for Velocity

Using the vector identity \( \mathbf{v}_{\alpha} = \boldsymbol{\omega} \times \mathbf{r}_{\alpha} \), express the magnitude squared of the velocity: \( \mathbf{v}_{\alpha}^2 = (\boldsymbol{\omega} \times \mathbf{r}_{\alpha})^2 \).

Prove Vector Identity for Cross Products

For any vectors \( \mathbf{a} \) and \( \mathbf{b} \), prove \( (\mathbf{a} \times \mathbf{b})^2 = a^2 b^2 - (\mathbf{a} \cdot \mathbf{b})^2 \). This identity is derived from expanding the cross product in terms of its components and applying the distributive properties of dot products.

Substitute and Simplify Kinetic Energy

Using the identity, compute \((\boldsymbol{\omega} \times \mathbf{r}_{\alpha})^2 = \omega^2 r_{\alpha}^2 - (\boldsymbol{\omega} \cdot \mathbf{r}_{\alpha})^2 \). Substitute this into the kinetic energy formula for each particle: \[T = \frac{1}{2} \sum m_{\alpha} (\omega^2 r_{\alpha}^2 - (\boldsymbol{\omega} \cdot \mathbf{r}_{\alpha})^2)\]

Apply BAC-CAB Rule for Angular Momentum

The BAC-CAB rule tells us:\[\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \cdot \mathbf{C}) - \mathbf{C} (\mathbf{A} \cdot \mathbf{B})\]Substitute for angular momentum \( \mathbf{L}_{\alpha} = \mathbf{r}_{\alpha} \times (m_{\alpha} \mathbf{v}_{\alpha}) \) to obtain:\[\mathbf{L} = \sum m_{\alpha} [\boldsymbol{\omega} r_{\alpha}^2 - \mathbf{r}_{\alpha} (\boldsymbol{\omega} \cdot \mathbf{r}_{\alpha})]\]

Connecting Kinetic Energy and Angular Momentum

Using results from steps 3 and 4, prove that:\[T = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{L}\]Since \[T\] can be expressed as \[\frac{1}{2} \sum m_{\alpha} [ \omega^2 r_{\alpha}^2 - (\boldsymbol{\omega} \cdot \mathbf{r}_{\alpha})^2 ]\], this matches the dot product form.

Kinetic Energy as a Matrix Expression

Recognize the tensor form of the moment of inertia \( \mathbf{I} \). Using the matrix form, the kinetic energy expression can be re-written as: \(T = \frac{1}{2} \tilde{\boldsymbol{\omega}} \mathbf{I} \boldsymbol{\omega}\) by substituting the definition of \( \mathbf{L} \) in terms of \( \mathbf{I} \).

Express Kinetic Energy in Principal Axes

Transform the problem into principal axes where \( \mathbf{I} \) is diagonal. This reduces the expression to simple squared terms: \[T = \frac{1}{2} (\lambda_1 \omega_1^2 + \lambda_2 \omega_2^2 + \lambda_3 \omega_3^2)\] consistent with the principal moments of inertia.

Key concepts

Vector Identities

Vector identities are mathematical expressions that simplify the manipulation and analysis of vector equations. They emerge frequently in physics and engineering, particularly when dealing with rigid body dynamics. One important vector identity that helps in these contexts is the relation for the square of a cross product:

• Given any two vectors \( \mathbf{a} \) and \( \mathbf{b} \), an identity states that \((\mathbf{a} \times \mathbf{b})^2 = a^2 b^2 - (\mathbf{a} \cdot \mathbf{b})^2\).
• This identity helps simplify expressions involving motion of particles subject to forces, and is especially useful in problems involving angular velocities and positions in three-dimensional space.

The derivation of this identity involves expanding the cross product \( \mathbf{a} \times \mathbf{b} \) in terms of their components and recognizing the relation between dot products and magnitudes.
Understanding this identity can make it easier to derive key equations in rigid body dynamics, such as the kinetic energy of a system or its angular momentum. It's often coupled with another vector operation rule, known as the BAC-CAB rule, which deals with triple cross products:

• The rule states that \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B}) \).
• This identity further simplifies the computation of quantities like angular momentum, making it an invaluable tool in vector calculus.

Moment of Inertia

Moment of inertia is a fundamental concept in the study of rigid body dynamics that quantifies an object's resistance to changes in its rotational motion. Unlike mass, which deals with linear motion, the moment of inertia deals with rotation and involves the distribution of mass around an axis.
The moment of inertia \( \mathbf{I} \) of a rigid body is expressed through a tensor when extended to three-dimensional systems. It incorporates the mass distribution about the axis of rotation and can be expressed in various forms depending on the system and coordinate system chosen.

• For simplest cases, like symmetric objects rotating about their symmetry axis, the moment of inertia can often be found using simpler expressions such as \( I = \sum m_r r^2 \), where each mass element \( m_r \) is located at a distance \( r \) from the axis of rotation.
• For more complex objects, the calculation might involve integrating over the volume of the object.

The moment of inertia tensor includes all the components and interactions across multiple axes, leading to a more involved expression: \( \mathbf{I} \). This representation accounts for rotary motion effects in all directions.
In the principal axes form, this tensor becomes diagonal, simplifying its expression as the sum of squared angular velocities weighted by the principal moments of inertia: \( T = \frac{1}{2} (\lambda_1 \omega_1^2 + \lambda_2 \omega_2^2 + \lambda_3 \omega_3^2) \).

• This configuration assists in solving complex dynamics problems by reducing them into more familiar forms.

Angular Momentum

Angular momentum is a central concept in rotational dynamics, much like linear momentum in translational motion. It characterizes the quantity of rotation a body possesses and is influenced by the object's moment of inertia and its angular velocity.
The expression for the angular momentum \( \mathbf{L} \) of a rigid body rotating around an axis involves both the angular velocity \( \boldsymbol{\omega} \) and the distribution of mass around the rotation axis. It can be described by the formula:

• \( \mathbf{L} = \sum m_\alpha [\boldsymbol{\omega} r_\alpha^2 - \mathbf{r}_\alpha (\boldsymbol{\omega} \cdot \mathbf{r}_\alpha)] \)

This formulation emerges from applying vector identities such as the BAC-CAB rule, which rearranges components of triple products in vectors.
Angular momentum isn't only a measure of rotational influence but is conserved in isolated systems. This means that if no external torques act, the total angular momentum of a system remains constant.

• This conservation law plays a pivotal role in understanding systems like planetary orbits or gyroscopic effects.
• And it reminds us of the beauty and consistency that physical laws offer across different scales and cases.

Learning to calculate and interpret angular momentum is crucial for solving exercises related to rotational motion, allowing a glimpse into how objects behave when acting under rotational forces.