Rigid Body motion

Some definitions

Inertial frame

Non accelerating, non-rotating frame of reference.

Omega, $ω$

Denotes angular velocity. unit : rad/s.
It is a property of the rigid body itself and is frame independent (for an inertial frame).
Omega about any two points in such a frame is always same.

Alpha, $α$

Denotes angular acceleration. unit : $rad / s^{2}$

Angular momentum, $L$

L = \sum_{i} r_{i} \times m v_{i}

For a system of particles.
For a rigid body, $$\mathbf{L} = \mathbf{I}\mathbf{\omega}$$
where $I$ is a $3 \times 3$ matrice and $ω$ is a vector.

Moment or torque $M$ or $τ$

M = \frac{d}{d t} L

Pure rotation

\begin{aligned} \vec{L} & = I ω \\ \vec{τ} & = I α = \sum_{i} r_{i} \times F_{i} \\ K & = \frac{1}{2} I ω^{2} \end{aligned}

Rotation plus translation

Considering point $O$ as the reference point and $C$ as the center of mass.
Point $O$ is a fixed point in an inertial frame.

\begin{aligned} L_{tot} & = L_{com} + L_{about com} \\ L_{o} & = I_{c} ω + r \times m v_{c} \\ M_{o} & = \frac{d L_{o}}{d t} \\ = I_{c} α + ω \times (I_{c} ω) + r \times (m a_{c}) \end{aligned}

This is for general motion in 3 dimensions.

The term $ω \times (I_{c} ω) = ω \times L$ is not zero, even though one may think $ω$ and $L$ are parallel, because $I$ is not a scalar, but rather a tensor, which changes the direction of $I_{c} ω = L$ .

In a general case also, if we consider a mass rotating about a point such that $r$ and $ω$ are not parallel or perpendicular, then $L$ has a different direction than $ω$ .

That term disappears when the motion is planar and rotation is about a fixed axis.

Last equation can also be written about $O$ , $$\mathbf{M}{o} = \mathbf{I}\alpha + \omega \times (\mathbf{I}{o} \omega)$$
where $$\mathbf{I} = \mathbf{I}{c} + m[(r^2)\mathbf{I - rr}^T]$$

Angular orientation

An angular orientation of a body cannot be represented by a vector of the form $$\vec{\theta} = (\theta_{x}\hat{x} + \theta_{y}\hat{y} + \theta_{z}\hat{z})$$
This is because changes in angular orientation do not commute, essentially meaning that, $$(\theta_{x}\hat{x} + \theta_{y}\hat{y} ) \neq (\theta_{y}\hat{y} + \theta_{x}\hat{x})$$
Order in which the rotation happens matters.

But however, angular velocity can be represented as a vector $$\vec{\omega} = (\omega_{x}\hat{x} + \omega_{y}\hat{y} + \omega_{z}\hat{z})$$

Torque scenes

Torque and force are intimately related and are always mutually perpendicular
There can be torque on a system and no force, conversely, there can be force on a system and no torque.

Conservation of angular momentum

\begin{aligned} L & = r \times p \\ \dot{L} & = \dot{r} \times p + r \times \dot{p} \\ = v \times p + r \times F \end{aligned}

When angular momentum is conserved, $\dot{L} = 0$ , therefore net torque on system is 0
However angular momentum is not conserved when it's direction is changing, but magnitude remains same. In this case, there is no external force on the body, but there is a torque necessary to change the angular momentum vector itself.

Moment of inertia as a tensor

Deriving angular momentum according to it's definition, $$\mathbf{L} = \sum_{j}[\mathbf{r}{j} \times m( \omega \times \mathbf{r}_{j})]$$for the j-th particle.

If we expand this according to the definition of cross product, we finally obtain $$\mathbf{L} = \begin{bmatrix} \hat{x} & \hat{y} & \hat{z} \ x_{j} & y_{j} & z_{j} \ m_{j}(\omega_{y}z_{j} - y_{j}\omega_{z}) & -m_{j}(\omega_{x}z_{j} - x_{j}\omega_{z}) & m_{j}(\omega_{x}y_{j} - x_{j}\omega_{y})
\end{bmatrix}$$
Therefore, we have $$L_{x} = m_{j}(x_{j}^2 + y_{j}^2)\omega_{x} - m_{j}x_{j}y_{j}\omega_{y} - m_{j}x_{j}z_{j}\omega_{z}$$
Extrapolating this,
Recall product of inertia from ME1201

We define $$\begin{align} I_{xx} &= \int m_{j}(y_{j}^2 + z_{j}^2) \ I_{xy} &= -\int m_{j}x_{j}y_{j} \ I_{xz} &= -\int m_{j}x_{j}z_{j}\end{align}$$

L_{x} = I_{x x} ω_{x} + I_{x y} ω_{y} + I_{x z} ω_{z}

Extrapolating this, we can rewrite $L = I ω$ as a matrix equation, $$\begin{pmatrix} \mathbf{L}{x} \ \mathbf{L} \ \mathbf{L}{z}\end{pmatrix} = \begin{pmatrix} I & I_{xy} & I_{xz} \ I_{yx} & I_{yy} & I_{yz} \ I_{zx} & I_{zy} & I_{zz} \end{pmatrix}\cdot \begin{pmatrix} \omega_{x} \ \omega_{y} \ \omega_{z} \end{pmatrix}$$
Thus, $L$ about one axis, depends on all three axes.
If we differentiate this, we can replace $L$ with $τ$ and $\vec{ω}$ with $\vec{α}$ , thus obtaining a similar relationship between torque and angular acceleration, ALSO implying that they are not unidirectional.

Notice that in the equation, $I_{x z} = I_{z x}$ and so on, implying that it is a symmetric matrix.

For continuous medium, $$\mathbf{I} = \begin{bmatrix} \int(y^2 + z^2)dm & -\int xydm & -\int xzdm \ -\int xydm & \int(x^2 + z^2)dx & -\int yzdm \ -\int xzdm & -\int yzdm & \int(x^2 + y^2)dm \end{bmatrix}$$
We already know that the product of inertia measures how disbalanced the object is, how the mass is distributed around the COM.

Important

Principal axes of inertia are those three mutually perpendicular axes along which the product of inertia all disappear. (often passes through COM)
When body rotates about it's principal axes, it's angular momentum is parallel to angular velocity.
To simplify the moment of inertia matrix, we can always diagonalize it in order to find the principal moments of inertia of the body.
The moment of inertia in this new diagonalized tensors are principal moments of inertia, corresponding to eigen vectors of original inertia tensors

Tensors

Tensor represents a physical quantity that is characterized by magnitude and multiple directions simultaneously.

The rank of a tensor is defined as the number of axes required to describe a single component of it.
For example, a 3D tensor with 3 axes has rank 3, since we need 3 indices (therefore, it spreads in 3 axes) to describe a component.

Scalars are rank 0, vectors are rank 1, matrices are rank 2, and so on...

The dimensionality of a tensor is a touchy topic, which has different interpretations in different fields.

The dimensions of a tensor talks about the range of each axes. For example a $(5, 5, 5)$ tensor has dimensionality 5, if we consider the physicist approach of modelling dimensionality.
Similarly, in this same model, a tensor of $(a, b, c)$ has different dimensions in each different axes, we can consider the same as choosing a vector from $R^{a}, R^{b}, R^{c}$ , ie these three independent vector spaces model the tensor.

In the linear algebra way of modelling or describing dimensionality, it defines the underlying dimensionality of the tensor space is defined as the dimensionality of the total space, ie, $\dim (T) = a b c$

Some more lectures that talk about diagonalization of the inertia matrix to isolate principal axes, and solve some questions based on it (lec 21, 22).

Gyroscope

Gyroscopic precession is what we are concerned with.

When the flywheel has a spin angular momentum and we let it go, the downward torque produced by the unbalanced weight actually translates into changing $L_{s}$ of the spin. This in turn causes the angular momentum vector to change direction progressively, and the entire system rotates about a vertical axis with a different angular velocity $Ω$ .

The precession formula dictates $$\Omega = \frac{lW}{I_{0}\omega_{s}}$$
where $W$ is the weight, $l$ is the length of arm (from pivot to center of disk), $I_{0}$ is the MOI of the disk, ie, $L_{s} = I_{o} ω_{s}$ .

It can be shown that the precession angular velocity is independent of inclination of gyroscope disk with the vertical. While the discussion above was done for $α = π / 2$ , the formula holds for any $α$ .

Eulers Equations of motion