Theoretical Mechanics IPSP

Jürgen Vollmer, Universität Leipzig

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book:chap5:5.4_center_of_mass_and_spin [2022/01/03 14:52] – created jvbook:chap5:5.4_center_of_mass_and_spin [2022/01/06 05:22] (current) – abril
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-FIXME draft with missing figures and references :!: 
- 
 ===== 5.4  Center of mass and spin of extended objects  ===== ===== 5.4  Center of mass and spin of extended objects  =====
-<WRAP id=section_particleExtension />+<WRAP #section_particleExtension></WRAP>
  
 We consider a setting where there are only long distance force like gravity We consider a setting where there are only long distance force like gravity
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 The explicit calculation for the case of gravity in the previous section entails  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 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\cref{quest:volIntegral-professor}e).+that is located at the center of the planet (see also [[book:chap5:5.7_problems #quest_volIntegral-professor |Problem 5.12 e)]].
 In the present section we therefore explore which effects the force of a point particle exerts on an extended body. In the present section we therefore explore which effects the force of a point particle exerts on an extended body.
  
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 ==== 5.4.1  Evolution of the center of mass  ==== ==== 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\cref{eq:professor-totForce},+The force on the body is described by an integral that takes exactly the same form as [[book:chap5:5.3_volume_integrals #eq_professor-totForce |Equation 5.3.1]],
 where now $\mathbf q$ is a vector from the point particle to a volume element of the body. 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 $\hat{\boldsymbol e}_1(t), \dots , \hat{\boldsymbol e}_3(t)$ 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 with orientation fixed in the rotating body and origin in the body's center of mass
 $\mathbf Q = ( Q_x, Q_y, Q_z )$.  $\mathbf Q = ( Q_x, Q_y, Q_z )$. 
-In immediate generalization of\cref{eq:defCenterMass} it is located at+In immediate generalization of [[book:chap4:4.6_the_center_of_mass_cm_inertial_frame #eq_defCenterMass |Equation 4.6.1]] it is located at
 \begin{align*}  \begin{align*} 
   \mathbf Q   \mathbf Q
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 with base vectors $\{ \hat{\boldsymbol e}_1, \dots, \hat{\boldsymbol e}_3 \}$ fixed in the body and origin in its center of mass. with base vectors $\{ \hat{\boldsymbol e}_1, \dots, \hat{\boldsymbol e}_3 \}$ fixed in the body and origin in its center of mass.
 Hence,  Hence, 
-\begin{align} \label{eq:spin-r-rdot}+<wrap #eq_spin-r-rdot></wrap> 
 +\begin{align}
     \mathbf r =  \sum_{i=1}^3 r_i \: \hat{\boldsymbol e}_i(t)      \mathbf r =  \sum_{i=1}^3 r_i \: \hat{\boldsymbol e}_i(t) 
     \quad\text{and}\quad     \quad\text{and}\quad
-    \dot{\boldsymbol r} =  \sum_{i=1}^3 r_i \: \dot{\hat{\boldsymbol e}}_i(t)+    \dot{\boldsymbol r} =  \sum_{i=1}^3 r_i \: \dot{\hat{\boldsymbol e}}_i(t) \tag{5.4.1}
     \\     \\
 \end{align} \end{align}
- \manimpossiblecube 
 </wrap> </wrap>
  
 We note that this choice of coordinates entails We note that this choice of coordinates entails
 +<wrap #eq_vanishing-relative-CM></wrap>
 \begin{align}  \begin{align} 
   M \, \mathbf Q   M \, \mathbf Q
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   \Rightarrow\quad   \Rightarrow\quad
   0 &=  \int_{\mathbb R^3} \mathrm{d}^3q \: \rho( \mathbf q ) \: r_i   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 +      = \int_{\mathbb R^3} \mathrm{d}^3r \: \rho( \mathbf r ) \: r_i \tag{5.4.3}
-      \label{eq:vanishing-relative-CM}+
 \end{align} \end{align}
 The latter equality holds because a shift of the origin by $\mathbf Q$ The latter equality holds because a shift of the origin by $\mathbf Q$
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 (i.e., the Jacobi determinant of the transformation is one).  (i.e., the Jacobi determinant of the transformation is one). 
 The acceleration $\ddot{\boldsymbol q}(t)$ takes the form The acceleration $\ddot{\boldsymbol q}(t)$ takes the form
 +
 +<wrap #eq_body-Ftot></wrap>
 +
 \begin{align*}  \begin{align*} 
   \ddot{\boldsymbol q}(t) = \ddot{\boldsymbol q}(t) + \sum_{i=1}^3 r_i \: \ddot{\hat{\boldsymbol e}}_i(t)   \ddot{\boldsymbol q}(t) = \ddot{\boldsymbol q}(t) + \sum_{i=1}^3 r_i \: \ddot{\hat{\boldsymbol e}}_i(t)
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   \nonumber \\   \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} + \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} +    =  M \: \ddot{\boldsymbol q} \tag{5.4.4}
-    \label{eq:body-Ftot}+
 \end{align} \end{align}
 The overall force $\mathbf F_{\text{tot}}$ results in an acceleration of the center of mass 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. that behaves exactly as for a point particle described in the previous chapter.
-Thus, we have justified the assumption of point particles adopted in\cref{chapter:EOM}.+Thus, we have justified the assumption of point particles adopted in [[book:chap4:eom-ode|Chapter 4]].
  
  
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 \end{align*} \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 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  \cref{eq:vanishing-relative-CM}.+The second and the third term vanish due to  [[#eq_vanishing-relative-CM |Equation 5.4.3]].
 The forth term can be simplified by performing the integration in the comoving coordinate frame. 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. 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 $\mathbf r$.  Hence, the Jacobi determinant is one, and the term only depends on the local coordinates $\mathbf r$. 
 It is denoted as particle spin. It is denoted as particle spin.
 +
 <WRAP box round>**Definition 5.4** <wrap em>Particle Spin</wrap> \\  <WRAP box round>**Definition 5.4** <wrap em>Particle Spin</wrap> \\ 
 The //total angular momentum// $\mathbf L_{\text{tot}}$ of a particle can be decomposed into The //total angular momentum// $\mathbf L_{\text{tot}}$ of a particle can be decomposed into
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 and its //spin// $\mathbf S$ around the center of mass, $\mathbf Q$, and its //spin// $\mathbf S$ around the center of mass, $\mathbf Q$,
 \begin{align}  \begin{align} 
-    \mathbf L_{\text{tot}} &= \mathbf L_{CM} + \mathbf S+    \mathbf L_{\text{tot}} &= \mathbf L_{CM} + \mathbf S \tag{5.4.5 a}
     \\     \\
     \text{with}\qquad     \text{with}\qquad
-    \mathbf L_{CM} &= M \: \mathbf Q \times \dot{\boldsymbol q}+    \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}+    \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} \end{align}
 </WRAP> </WRAP>
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 As a consequence, the incoming and outgoing angle can differ for a reflection at a wall, 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. 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 \cref{section:workedExample-ballReflections}.  +This will be demonstrated in the worked example in [[book:chap5:5.6_worked_example|Section 5.6]]. 
- +
 </wrap> </wrap>
  
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 that indicates the rotation axis and angular speed $\left\lvert \mathbf\Omega \right\rvert$, 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.  and exploring that the relative positions of the mass elements in the body do not change upon rotation. 
-Due to \cref{eq:spin-r-rdot} we have+Due to [[#eq_spin-r-rdot |Equation 5.4.1]] we have
 \begin{align*}  \begin{align*} 
   \mathbf S   \mathbf S
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 The situation simplifies further when one observes The situation simplifies further when one observes
 that the velocities $\dot{\hat{\boldsymbol r}}_k$ are unit vectors that the velocities $\dot{\hat{\boldsymbol r}}_k$ are unit vectors
-that must be orthogonal to $\hat{\boldsymbol r}_k$ and to $\Omega$. +that must be orthogonal to $\hat{\boldsymbol r}_k$ and to $\Omega$.((Recall that $\hat{\boldsymbol r}_k \cdot \hat{\boldsymbol r}_k = 1$
-\footnote{ +
-Recall that $\hat{\boldsymbol r}_k \cdot \hat{\boldsymbol r}_k = 1$+
 such that $2 \, \hat{\boldsymbol r}_k \, \cdot  \dot{\hat{\boldsymbol r}}_k = 0$, such that $2 \, \hat{\boldsymbol r}_k \, \cdot  \dot{\hat{\boldsymbol r}}_k = 0$,
-and by construction the motion of mass elements is orthogonal to the axis of rotation. +and by construction the motion of mass elements is orthogonal to the axis of rotation.))
-}+
 Hence, the velocities can be expressed as Hence, the velocities can be expressed as
 \begin{align*}  \begin{align*} 
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 The finding that the off-diagonal elements of the tensor of inertia vanish is no coincidence. The finding that the off-diagonal elements of the tensor of inertia vanish is no coincidence.
-In \cref{exercise:selfTest-inertia-symmetry} we will show+In [[book:chap5:5.7_problems|Section 5.7]] we will show
 that this happens whenever the mass distribution features a symmetry in the $ik$ plane. that this happens whenever the mass distribution features a symmetry in the $ik$ plane.
 Moreover, the ...theorem Moreover, the ...theorem
 of linear algebra states of linear algebra states
 that one can always choose coordinates that one can always choose coordinates
-where all off-diagonal elements of the tensor of inertia vanish. +where all off-diagonal elements of the tensor of inertia vanish((For a general matrix this is not true.
-\footnote{For a general matrix this is not true.+
 It is a consequence of the fact that $\mathsf\Theta$ is symmetric, i.e.,  It is a consequence of the fact that $\mathsf\Theta$ is symmetric, i.e., 
-$\Theta_{ij}=\Theta_{ji}$ for all its entries.}+$\Theta_{ij}=\Theta_{ji}$ for all its entries.)).
 The particular axes where this happens are called the axis of inertia of a body. The particular axes where this happens are called the axis of inertia of a body.
 <WRAP box round>**Definition 5.6** <wrap em>Axis of inertia</wrap> \\  <WRAP box round>**Definition 5.6** <wrap em>Axis of inertia</wrap> \\ 
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 If the mass distribution of the body obeys reflection or rotation symmetry, If the mass distribution of the body obeys reflection or rotation symmetry,
 the axes of inertia are invariant under the symmetry transformation. the axes of inertia are invariant under the symmetry transformation.
- \manimpossiblecube 
 </wrap> </wrap>
  
  
 ==== 5.4.3  Time evolution of angular momentum and particle spin  ==== ==== 5.4.3  Time evolution of angular momentum and particle spin  ====
-<WRAP id=ssection_spinEvolution />+<wrap #ssection_spinEvolution></wrap>
  
 The angular momentum $\mathbf L_{CM}$ of its center-of-mass motion The angular momentum $\mathbf L_{CM}$ of its center-of-mass motion
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     \dot{\boldsymbol S}     \dot{\boldsymbol S}
     = \mathbf M     = \mathbf M
-    = \int_{\text{body}} \mathrm{d}^3 r \;  \mathbf r \times \mathbf F( \mathbf Q + \mathbf r )+    = \int_{\text{body}} \mathrm{d}^3 r \;  \mathbf r \times \mathbf F( \mathbf Q + \mathbf r ) \tag{5.4.6}
 \end{align} \end{align}
 </WRAP> </WRAP>
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 where the gravitational acceleration $\mathbf g$ takes a constant value,  where the gravitational acceleration $\mathbf g$ takes a constant value, 
 forms a noticeable exception. forms a noticeable exception.
-<WRAP box round>**Theorem 5.3 <wrap hi> Spinning motion and gravity </wrap>** \\ + 
 +<WRAP box round>**Theorem 5.3 <wrap em> Spinning motion and gravity </wrap>** \\ 
 When an extended body moves subject to a spatially uniform acceleration $\mathbf g$, When an extended body moves subject to a spatially uniform acceleration $\mathbf g$,
 then its center of mass follows a free-flight parabola then its center of mass follows a free-flight parabola
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 //Proof.// //Proof.//
-The statement about the center-of-mass motion follows from \cref{eq:body-Ftot}.+The statement about the center-of-mass motion follows from [[#eq_body-Ftot |Equation 5.4.4]].
 Conservation of the spin is due to Conservation of the spin is due to
 \begin{align*}  \begin{align*} 
book/chap5/5.4_center_of_mass_and_spin.1641217963.txt.gz · Last modified: 2022/01/03 14:52 by jv