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I'd like to derive some laws of motions for rigid bodies only by considering a rigid body as a system of particles such that the distances from every particle to every other particle doesn't change with time.

The beautiful answer by ja72 was already posted before I wrote this one, but this one uses a different approach, so I decided to post it anyway. A distinguishing feature of this approach is that it works for $D$-dimensional space with arbitrary $D\geq 2$.

Setup

Notation: A rotation about the origin is described by a $D\times D$ matrix $R$ whose transpose is equal to its inverse and whose determinant is equal to $1$. A vector $\mathbf{x}$ can be represented by a matrix with $D$ components in a single column. With this notation, the result of applying a rotation $R$ to a vector $\mathbf{x}$ is the vector $R\mathbf{x}$, using ordinary matrix multiplication.

Consider a rigid body made of point particles. Let $m_n$ be the mass of the $n$th particle, and let $\mathbf{b}_n$ be its position in a coordinate system attached to the body, so $\mathbf{b}_n$ doesn't change as the body shifts or rotates. Choose the origin of the coordinate system so that $\sum_n m_n\mathbf{b}_n=0$, where $0$ denotes the zero vector.

Let $\mathbf{x}_n$ be time-dependent position of the $n$th particle in some inertial coordinate system. The assumption that the body is rigid means $$ \mathbf{x}_n = R\mathbf{b}_n+\mathbf{x} \tag{1} $$ where $R$ is a time-dependent rotation matrix and $\mathbf{x}$ is the time-dependent position of the center of mass: $$ \mathbf{x} := \frac{\sum_n m_n\mathbf{x}_n}{\sum_n m_n}. \tag{2} $$ The goal is to derive equations for the time-dependence of $R$ and $\mathbf{x}$ in terms of the forces applied to the particles.

The motion of the center of mass

Let $\mathbf{f}_n$ be the force applied to the $n$th particle, not including the inter-particle forces that keep the body rigid. This section derives the result $$ \sum_n \mathbf{f}_n = m\mathbf{\ddot x} \hskip1cm \text{with } \hskip1cm m:=\sum_n m_n \tag{3} $$ where each overhead dot denotes a derivative with respect to time. Equation (3) says that the net applied force $\sum_n \mathbf{f}_n$ is equal to the total mass $m$ times the acceleration $\mathbf{\ddot x}$ of the center of mass.

To derive (3), use the fact that the total force on the $n$th particle is $\mathbf{F}_n := \mathbf{f}_n+\sum_k\mathbf{f}_{nk}$ where $\mathbf{f}_{nk}$ is the force that the $k$th particle must exert on the $n$th particle to keep the body rigid. The total force on the whole body is then $$ \sum_n\mathbf{F}_n=\sum_n\mathbf{f}_n +\sum_{n,k}\mathbf{f}_{nk}. \tag{4} $$ Since the forces that the $k$th and $n$th particles exert on each other must be equal and opposite (because the body is rigid), the last term in equation (4) is zero, so $$ \sum_n\mathbf{F}_n=\sum_n\mathbf{f}_n. \tag{5} $$ Newton's law for each individual particle says $\mathbf{F}_n=m_n\mathbf{\ddot x}_n$, and using this on the left-hand side of equation (5) gives equation (3), remembering the definition (2) of $\mathbf{x}$. This completes the derivation of (3).

Rotational motion

The usual way of describing rotational motion in $3$-dimensional space involves some special conventions that only make sense in $3$-dimensional space. The generalization to $D$-dimensional space is easier if we use the different set of conventions that I'll introduce here. The only difficult part is relating these more-natural conventions to the more-traditional conventions that only work in $3$-dimensional space. I'll omit that difficult part, but a hint is given near the end of this post.

For the $n$th particle, we can combine the vectors $\mathbf{f}_n$ and $\mathbf{x}_n$ into a square ($D\times D$) matrix $\mathbf{f}_n\mathbf{x}_n^T$, where the superscript $T$ denotes transpose. We will be interested in the antisymmetric part of this matrix. Geometrically, the antisymmetric part of this matrix corresponds to the plane spanned by the vectors $\mathbf{f}_n$ and $\mathbf{x}_n$, with an associated magnitude that goes to zero whenever the two vectors are proportional to each other (because then they don't define a plane). This is the $D$-dimensional replacement for the traditional "cross product." More generally, for any square matrix $B$, we can construct the antisymmetric matrix $$ \Delta(B) := B-B^T. \tag{6} $$ Using this notation, the net torque applied to the body is defined to be the antisymmetric matrix $$ \Delta\left(\sum_n\mathbf{f}_n \mathbf{x}_n^T\right). \tag{7} $$ This is the $D$-dimensional generalization of the torque "vector," but one of the things we learn from this generalization is that torque is not a vector! Torque is a bivector, represented here by an antisymmetric matrix. In $3$-d space, we can get away with using a vector-like notation only because there is a unique line through the origin orthogonal to any given plane through the origin.

To derive an equation that describes the rotational motion of the rigid body, start with the definition of torque, equation (7). Using equation (4) with $\mathbf{f}_{kn}+\mathbf{f}_{nk}=0$ and with the fact that $\mathbf{f}_{kn}$ is directed along the line between the $k$th and $n$th particles, we can derive $$ \Delta\left(\sum_n\mathbf{f}_n \mathbf{x}_n^T\right)= \Delta\left(\sum_n\mathbf{F}_n \mathbf{x}_n^T\right). \tag{8} $$ This says that the applied torque (the left-hand side) is equal to the total torque (the right-hand side), where the total torque includes that due to the internal forces that keep the body rigid. Now use $\mathbf{F}_n=m_n\mathbf{\ddot x}_n$ in the right-hand side of (8), and then use equation (1). Some terms cancel due to the antisymmetry, leaving the result $$ \Delta\left(\sum_n\mathbf{f}_n \mathbf{x}_n^T\right)= m\Delta\big(\mathbf{\ddot x}\mathbf{x}^T\big) +\Delta\big(\ddot RM_b R^T\big) \tag{9} $$ with $$ M_b := \sum_n m_n\mathbf{b}_n\mathbf{b}_n^T. \tag{10} $$ Equation (9) relates the applied torque to the time-dependence of the body's rotation $R$ and of the body's center-of-mass $\mathbf{x}$.

The first term on the right-hand side of equation (9) has an easy interpretation: the matrix $\Delta\big(\mathbf{\ddot x}\mathbf{x}^T\big)$ is the angular acceleration of the body's center of mass. Geometrically, this has an orientation represented by the plane spanned by $\mathbf{x}$ and $\mathbf{\ddot x}$.

The second term on the right-hand side of equation (9) describes the rotation of the body about its center of mass. The matrix (10) is the special combination of particle-masses and their body-fixed positions that controls how the body responds to torque. The subscript $b$ on $M_b$ means "body-fixed." The second term on the right-hand side of equation (9) may be rewritten using the identity $$ \Delta\big(\ddot RM_b R^T\big) =\frac{d}{dt}L \tag{11} $$ where $L$ is the angular momentum bivector $$ L := \Delta\big(WM\big) \tag{12} $$ where $M := RM_b R^T$ is just $M_b$ expressed in the inertial coordinate system, and $W$ is the angular velocity bivector $$ W := \dot R R^T. \tag{13} $$ This matrix is already antisymmetric, so we don't need to use $\Delta$ here. (Proof: Take the time-derivative of both sides of the identity $R R^T=1$.)

Relationship to traditional notation when $D=3$

This is the tricky part, and this post is already long, so I'll omit the details. Here's a hint: For $D=3$, a bivector is represented by a $3\times 3$ antisymmetric matrix. Such a matrix has only $3$ independent components, because the components below the diagnal are the negatives of those above the diagonal, and the diagonal components are zero. By arranging these $3$ components into a "vector" and re-writing the preceding equations in that vector-like notation, we can recover the traditional formulation for $3$-dimensional space.

The messiest part is showing how the matrix (10) is related to the thing that is traditionally called the "moment of intertia tensor." The matrix (10) is actually simpler than the traditional moment of inertia tensor, and it conveys the same information, so we could have called (10) the "momentum of inertia tensor" instead... but history has already taken its course, and I won't try to change it here.

The easist case: $D=2$

In this case, a bivector (antisymmetric matrix) has only one indepenent component, which makes things relatively easy: all of the equations involving $\Delta$ reduce to essentially scalar equations. Actually, "pseudoscalar" is a better name for it: a pseudoscalar changes sign when any one direction is reflected, but a scalar in the strict sense does not. In $D=2$, a bivector is the same thing as a pseudoscalar. (In $D=3$, a bivector is the same thing as a "pseudovector.")