Table of Contents
5.4 Center of mass and spin of extended objects
We consider a setting where there are only long distance force like gravity and no collisions between objects. The explicit calculation for the case of gravity in the previous section entails that in such a setting the force exerted by a planet on a point particle is identical to the one exerted by a mass point of identical mass that is located at the center of the planet (see also Problem 5.12 e). In the present section we therefore explore which effects the force of a point particle exerts on an extended body.
5.4.1 Evolution of the center of mass
The force on the body is described by an integral that takes exactly the same form as Equation 5.3.1,
where now $\mathbf q$ is a vector from the point particle to a volume element of the body.
The integral is best evaluated by introducing a coordinate frame $\hat{\boldsymbol e}_1(t), \dots , \hat{\boldsymbol e}_3(t)$
with orientation fixed in the rotating body and origin in the body's center of mass
$\mathbf Q = ( Q_x, Q_y, Q_z )$.
In immediate generalization of Equation 4.6.1 it is located at
\begin{align*}
\mathbf Q
= \frac{1}{M} \; \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: \mathbf q
\quad \Leftrightarrow \quad
\begin{pmatrix}
Q_x \\ Q_y \\ Q_z
\end{pmatrix}
=
\begin{pmatrix}
M^{-1} \; \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: q_x \\
M^{-1} \; \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: q_y \\
M^{-1} \; \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: q_z
\end{pmatrix}
\end{align*}
A given mass element will always have the same coordinates $(r_1, r_2, r_3)$ with respect to the body-fixed basis,
and in a stationary coordinate frame this position can be specified as
\begin{align*}
\mathbf q(t) = \mathbf Q(t) + \sum_{i=1}^3 r_i \: \hat{\boldsymbol e}_i(t)
\end{align*}
Remark 5.5.
The vector $\mathbf r$ describes the position $(r_1, \dots, r_3)$
in the body with respect to its center of mass.
When the body rotates $\mathbf r$ will evolve in time.
However, the coordinates $(r_1, \dots, r_3)$ are constant numbers describing the shape of the body
when they are calculated in a coordinate system
with base vectors $\{ \hat{\boldsymbol e}_1, \dots, \hat{\boldsymbol e}_3 \}$ fixed in the body and origin in its center of mass.
Hence,
\begin{align}
\mathbf r = \sum_{i=1}^3 r_i \: \hat{\boldsymbol e}_i(t)
\quad\text{and}\quad
\dot{\boldsymbol r} = \sum_{i=1}^3 r_i \: \dot{\hat{\boldsymbol e}}_i(t) \tag{5.4.1}
\end{align}
We note that this choice of coordinates entails \begin{align} M \, \mathbf Q &= \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: \mathbf q = \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: \bigl( \mathbf Q(t) + \sum_{i=1}^3 r_i \: \hat{\boldsymbol e}_i(t) \bigr) \nonumber \\ &= \mathbf Q(t) \: \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) + \sum_{i=1}^3 \hat{\boldsymbol e}_i(t) \: \int_{\mathbb R^3} \mathrm{d}^3 q \: \rho( \mathbf q ) \: r_i \nonumber \\ \Rightarrow\quad 0 &= \int_{\mathbb R^3} \mathrm{d}^3q \: \rho( \mathbf q ) \: r_i = \int_{\mathbb R^3} \mathrm{d}^3r \: \rho( \mathbf r ) \: r_i \tag{5.4.3} \end{align} The latter equality holds because a shift of the origin by $\mathbf Q$ and rotation of the coordinate axes do not affect the integration volume (i.e., the Jacobi determinant of the transformation is one). The acceleration $\ddot{\boldsymbol q}(t)$ takes the form
\begin{align*} \ddot{\boldsymbol q}(t) = \ddot{\boldsymbol q}(t) + \sum_{i=1}^3 r_i \: \ddot{\hat{\boldsymbol e}}_i(t) \end{align*} and the force on the spatially extended body results in \begin{align} \mathbf F_{\text{tot}} &= \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \ddot{\boldsymbol q}(t) \nonumber \\ &= \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho(\mathbf q) \; \left( \ddot{\boldsymbol q}(t) + \sum_{i=1}^3 r_i \: \ddot{\hat{\boldsymbol e}}_i(t) \right) \nonumber \\ &= M \: \ddot{\boldsymbol q} + \sum_{i=1}^3 \ddot{\hat{\boldsymbol e}}_i(t) \, \int_{\mathbb R^3} \mathrm{d}^3 r \: \rho( \mathbf r ) \: r_i = M \: \ddot{\boldsymbol q} \tag{5.4.4} \end{align} The overall force $\mathbf F_{\text{tot}}$ results in an acceleration of the center of mass that behaves exactly as for a point particle described in the previous chapter. Thus, we have justified the assumption of point particles adopted in Chapter 4.
5.4.2 Angular momentum and particle spin
Let us now explore the angular momentum of a spatially extended particles. To this end we introduce the decomposition $\mathbf q = \mathbf Q + \mathbf r$ into the definition \begin{align*} \mathbf L_{\text{tot}} &= \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: ( \mathbf q \times \dot{\boldsymbol q} ) = \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \bigl( (\mathbf Q + \mathbf R) \times (\dot{\boldsymbol q} + \dot{\boldsymbol r}) \bigr) \\ &= \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \bigl( \mathbf Q \times \dot{\boldsymbol q} \bigr) + \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \bigl( \mathbf Q \times \dot{\boldsymbol r} \bigr) \\ & + \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \bigl( \mathbf r \times \dot{\boldsymbol q} \bigr) + \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \bigl( \mathbf r \times \dot{\boldsymbol r} \bigr) \\ &= M \: \mathbf Q \times \dot{\boldsymbol q} + \mathbf Q \times \frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \mathbf r(t) - \dot{\boldsymbol q} \times \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \mathbf r(t) \\ & + \int_{\mathbb R^3} \mathrm{d}^3 q \; \rho( \mathbf q ) \: \bigl( \mathbf r \times \dot{\boldsymbol r} \bigr) \end{align*} The first summand amounts to the angular momentum of the center of mass, $\mathbf L_{CM} = M \: \mathbf Q \times \dot{\boldsymbol q}$. The second and the third term vanish due to Equation 5.4.3. The forth term can be simplified by performing the integration in the comoving coordinate frame. The coordinate transformation involves a translation by $\mathbf Q$ and rotation. Hence, the Jacobi determinant is one, and the term only depends on the local coordinates $\mathbf r$. It is denoted as particle spin.
Definition 5.4 Particle Spin
The total angular momentum $\mathbf L_{\text{tot}}$ of a particle can be decomposed into
the angular momentum $\mathbf L_{CM}$ of its center-of-mass motion,
and its spin $\mathbf S$ around the center of mass, $\mathbf Q$,
\begin{align}
\mathbf L_{\text{tot}} &= \mathbf L_{CM} + \mathbf S \tag{5.4.5 a}
\\
\text{with}\qquad
\mathbf L_{CM} &= M \: \mathbf Q \times \dot{\boldsymbol q} \tag{5.4.5 b}
\\
\mathbf S &= \int_{\mathbb R^3} \mathrm{d}^3 r \: \rho( r_1. r_2. r_3 ) \: \mathbf r \times \dot{\boldsymbol r} \tag{5.4.5 c}
\end{align}
Remark 5.6. The decomposition of the total angular momentum has important consequences in collisions. For spatially extended objects the conservation of angular momentum only implies that the sum of the spin and the angular momentum of the center-of-mass motion are conserved. As a consequence, the incoming and outgoing angle can differ for a reflection at a wall, and the center of mass of the particle can even move in different planes before and after the collision. This will be demonstrated in the worked example in Section 5.6.
The discussion of particle spin can further be simplified by expressing the rotation of the body by the vector $\mathbf\Omega$ that indicates the rotation axis and angular speed $\left\lvert \mathbf\Omega \right\rvert$, and exploring that the relative positions of the mass elements in the body do not change upon rotation. Due to Equation 5.4.1 we have \begin{align*} \mathbf S &= \int_{\mathbb R^3} \mathrm{d}^3 r \: \rho( \mathbf r ) \: \mathbf r \times \dot{\boldsymbol r} \\ &= \sum_{ij=1}^3 \hat{\boldsymbol r}_i \times \dot{\hat{\boldsymbol r}}_j \; \int \mathrm{d}^3 r \: r_i \, r_j \; \rho( \mathbf r ) = \sum_{ij=1}^3 \hat{\boldsymbol r}_i \times \dot{\hat{\boldsymbol r}}_j \; t_{ij} \\ \text{with}\quad t_{ij} &= \int_{\mathbb R^3} \mathrm{d} r_1 \, \mathrm{d} r_2 \, \mathrm{d} r_3 \; r_i \, r_j \; \rho( r_1, r_2, r_3 ) \end{align*} Note that the coefficients $t_{ij}$ are properties of the body. They characterize the mass distribution of the body, and do not depend on the motion. The situation simplifies further when one observes that the velocities $\dot{\hat{\boldsymbol r}}_k$ are unit vectors that must be orthogonal to $\hat{\boldsymbol r}_k$ and to $\Omega$.1) Hence, the velocities can be expressed as \begin{align*} \dot{\hat{\boldsymbol r}}_k = \hat{\boldsymbol r}_k \times \mathbf\Omega \end{align*} With this notations the $k$th component of $\mathbf S$ can be expressed as \begin{align*} S_k = \hat{\boldsymbol r}_k \cdot \mathbf S &= \sum_{ij=1}^3 \hat{\boldsymbol r}_k \cdot \bigl( \hat{\boldsymbol r}_i \times (\hat{\boldsymbol r}_j \times \mathbf \Omega) \bigr) \; t_{ij} \\ &= \sum_{ij=1}^3 \hat{\boldsymbol r}_k \cdot \bigl( \hat{\boldsymbol r}_j \: ( \mathbf\Omega \cdot \hat{\boldsymbol r}_i) - \mathbf\Omega \: (\hat{\boldsymbol r}_i \cdot \hat{\boldsymbol r}_j) \bigr) \; t_{ij} \\ &= \sum_{ij=1}^3 \bigl( \delta_{jk} \: \Omega_i - \Omega_k \: \delta_{ij} \bigr) \; t_{ij} \\ &= \sum_{i=1}^3 \Omega_i \; \sum_{j=1}^3 \delta_{jk} \: t_{ij} - \Omega_k \sum_{ij=1}^3 \delta_{ij} t_{ij} \\ &= \sum_{i=1}^3 \Omega_i \; \left( t_{ik} - \delta_{ik} \: \sum_j t_{jj} \right) \end{align*} This amounts to a multiplication of the vector $\mathbf\Omega$ written in terms of its components $\Omega_i$. We summarize this observation in the following definition
Definition 5.5 Tensor of Inertia
The rotation of a solid body with a fixed mass distribution $\rho(\mathbf r)$ can be described
by a vector $\Omega$ that defines the rotation axis and speed.
It is related to the spin $\mathbf S$ of the body by multiplication with the tensor or inertia
\begin{align*}
\mathbf S = \mathsf\Theta \, \mathbf \Omega \, ,
\end{align*}
i.e., a symmetric matrix with components
\begin{align*}
\Theta_{ij} = \int_{\mathbb R^3} \mathrm{d} r_1 \, \mathrm{d} r_2 \, \mathrm{d} r_3 \: \left( r_i \, r_j - \sum_{k=1}^3 r_k \, r_k \right)
\; \rho( r_1, r_2, r_3 )
\end{align*}
Example 5.4 Intertial tensor for a solid ball
For a ball of radius $R$ with uniform mass density $\rho$ the tensor of inertia has the following
entries for its diagonal elements
\begin{align*}
\Theta_{ii}
= \int_{\mathbb R^3} \mathrm{d} r_1 \, \mathrm{d} r_2 \, \mathrm{d} r_3 \: \left( r_i \, r_i - \sum_{k=1}^3 r_k \, r_k \right)
\; \rho( r_1, r_2, r_3 )
\end{align*}
We evaluate the integral in spherical coordinates with $r = \left\lvert \mathbf r \right\rvert$ and
$\theta$ denoting the angle with respect to the $i$-axis,
which we denote as $z$-axis in the following.
Hence,
\begin{align*}
(r_x,\, r_y,\, r_z) = r \, ( \sin\theta\, \cos\phi ,\, \sin\theta\, \sin\phi ,\, \cos\theta ) \, ,
\end{align*}
and
\begin{align*}
\Theta_{ii}
&= \rho \; \int_0^R \mathrm{d} r \, r^2 \int_{-1}^1 \mathrm{d}\cos\theta \, \int_0^{2\pi} \mathrm{d}\phi \:
( r_x^2 + r_y^2 )
\\
&= 2\,\pi \: \rho \; \int_0^R \mathrm{d} \rho \, r^2 \int_{-1}^1 \mathrm{d}\cos\theta \; r^2\, \sin^2\theta
\\
&= 2\,\pi \: \rho \;
\left( \int_0^R \mathrm{d} \rho \, r^4 \right) \;
\left( \int_{-1}^1 \mathrm{d}\cos\theta \; \left( 1 - \cos^2\theta \right) \right)
\\
&= 2\,\pi \: \rho \; \frac{R^5}{5} \; \left( 2 - \frac{2}{3} \right)
= \frac{2}{5} \: M \, R^2
\end{align*}
Moreover, for the off-diagonal element $\theta_{ik}$ we align the $k$-axis with $\phi=0$
and find
\begin{align*}
\Theta_{ik}
&= \rho \; \int_0^R \mathrm{d} r \, r^2 \int_{-1}^1 \mathrm{d}\cos\theta \, \int_0^{2\pi} \mathrm{d}\phi \:
r_x\, r_z
\\
&= \rho \; \int_0^R \mathrm{d} r \, r^2 \int_0^\pi \mathrm{d}\theta \: \sin\theta \, \int_0^{2\pi} \mathrm{d}\phi \:
r^2 \, \sin\theta \: \cos\theta
\\
&= 2\pi \, \rho \; \frac{R^5}{5} \int_0^\pi \mathrm{d}\theta \: \sin^2\theta \: \cos\theta = 0
\end{align*}
since $\sin^2\theta \: \cos\theta$ is antisymmetric with respect to $\pi/2$.
The finding that the off-diagonal elements of the tensor of inertia vanish is no coincidence. In Section 5.7 we will show that this happens whenever the mass distribution features a symmetry in the $ik$ plane. Moreover, the …theorem of linear algebra states that one can always choose coordinates where all off-diagonal elements of the tensor of inertia vanish2). The particular axes where this happens are called the axis of inertia of a body.
Definition 5.6 Axis of inertia
For each solid body there is a choice of internal coordinate axes $\hat{\boldsymbol r}_i$, $i=1,\cdots,3$
where the tensor of inertia takes a diagonal form.
The directions selected by the axis are called axes of inertia,
and the related diagonal entry of the matrix of inertial is denoted as moment of inertia.
Remark 5.7. If the mass distribution of the body obeys reflection or rotation symmetry, the axes of inertia are invariant under the symmetry transformation.
5.4.3 Time evolution of angular momentum and particle spin
The angular momentum $\mathbf L_{CM}$ of its center-of-mass motion behaves in exactly the same way as for point particles. The spin changes in time according to the differential equation \begin{align*} \dot{\boldsymbol S} &= \int_{\mathbb R^3} \mathrm{d}^3 r \: \rho( r_1, r_2, r_3 ) \: \mathbf r \times \ddot{\mathbf r} \\ &= \int_{\mathbb R^3} \mathrm{d}^3 r \: \rho( r_1, r_2, r_3 ) \: \mathbf r \times \ddot{\boldsymbol q} = \int_{\mathbb R^3} \mathrm{d}^3 r \; \mathbf r \times \mathbf F( \mathbf Q + \mathbf r ) \end{align*} In order to arrive at the second line, we noted that $\ddot{\mathbf r} = \ddot{\boldsymbol q} - \ddot{\boldsymbol q}$, and that the integral for the $\ddot{\boldsymbol q}$ contribution vanishes because $\int_{\mathbb R^3} \mathrm{d}^3 r \: \rho( r_1, r_2, r_3 ) \: \mathbf r = 0$. Moreover, it is understood that the force $\mathbf F$ is zero for coordinates $\mathbf r$ outside the body.
Definition 5.7 Particle Torque
When the part $\mathbf r$ of a body is subjected to force $\mathbf F$
then its spin $\mathbf S$ is changing due to a torque $\mathbf M$
\begin{align} \label{eq:particleTorque}
\dot{\boldsymbol S}
= \mathbf M
= \int_{\text{body}} \mathrm{d}^3 r \; \mathbf r \times \mathbf F( \mathbf Q + \mathbf r ) \tag{5.4.6}
\end{align}
Remark 5.8. Note that the torque is denoted by the letter capital M that is also frequently used for the mass. Nevertheless, there is no immediate risk that they are mixed up: The torque, $\mathbf M$, is a vector, while the mass, $M$, is a scalar. To further reduce the risk we will denote masses by a small letter $m$, when mass and torque appear in a problem.
In general the force $\mathbf F( \mathbf Q + \mathbf r )$ can only be evaluated after the CM motion has been determined. From the point of view of the rotating body it is a time-dependent force. This renders the motion of a particle in an inhomogeneous force field to be a very complex problem. However, the gravitational force on small spatial distances, where the gravitational acceleration $\mathbf g$ takes a constant value, forms a noticeable exception.
Theorem 5.3 Spinning motion and gravity
When an extended body moves subject to a spatially uniform acceleration $\mathbf g$,
then its center of mass follows a free-flight parabola
and its spin is preserved.
Proof. The statement about the center-of-mass motion follows from Equation 5.4.4. Conservation of the spin is due to \begin{align*} \int \mathrm{d}^3 r \; \mathbf r \times \bigl( \rho(r_1, r_2, r_3) \: \mathbf g \bigr) &= \left( \int\mathrm{d}^3 r \; \rho(r_1, r_2, r_3) \; \mathbf r \right) \times \mathbf g \\ &= \mathbf 0 \times \mathbf g = \mathbf 0 \\ \end{align*} qed
Rather than in the free flight of a body, one also often encounters a spinning body that is fixed at some point. Besides gravity there is one additional force acting on the body that is constraining its motion. When this force acts only on the center of mass, then it has no effect on the spin and only changes the evolution of the center of mass. When it acts on another point on the body, then the total angular momentum is no longer conserved. This happens for instance for a spinning top where one fixes a point on its axis.