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\begin{center}
{\bf Supplimentary Notes IV}\\
{\bf  Rotational Dynamics}
\end{center}

So far we have only considered the motion of "point" objects.
We have not allowed the objects to rotate, nor have we considered
how forces might produce a "twisting force".  In this last section,
the last 2 1/2 weeks of the quarter, we will investigate the laws
of physics pertaining to rotational motion.  As we shall see,
we can extend Newton's laws for translational motion to include
rotation, without introducing any new "laws of physics".  

We will first study the condition for objects not to rotate, then analyze rotations
about a fixed axis.  Finally, we will extend our ideas to cases in which
the axis of rotation can move, while keeping its direction constant.

\bigskip

\centerline{\bf Remembering Statics}

\vspace{3mm}
In the first part of the course, we showed that there are two conditions 
that need to be satisfied in order for an object to stay
at rest and not rotate:
the sum of the forces must be zero and the net torque about any axis must
also be zero:

\begin{eqnarray*}
   \sum_i \vec{F}_i & =  &0\\
   \sum_i \vec{\tau}_i & = & 0
\end{eqnarray*}

\noindent where the torque sum can be taken about any axis.  

In class we will do many examples of many objects in static equilibrium.  In evaluating
the torque sum, a proper choice of axis can reduce the mathematical complexity.

You might wonder why the twisting force equals the product of force times distance 
from the axis.  Consider a simple rigid object consisting of two
masses connected by a massless rod on the earth.  Suppose we consider balancing the object
about a point on the massless rod.  Let the object to the right have a mass
$m_1$ and be at a distance $x_1$ from the axis, and let the object to the left have
a mass of $m_2$ and be at a distance of $x_2$ from the axis.  If the axis is the
balance point then the rod will rotate with constant angular speed if slightly
pushed.  To see how the speed might change, we can use the work-energy theorum.
Let $\theta$ be the angle that the rod rotates.  Then $m_1$ moves a distance
$x_1 \theta$, and $m_2$ moves a distance $x_2 \theta$.  The work done by gravity
on the mass $m_1$ is $m_1 g x_1 \theta$, and the work done by gravity on the
mass $m_2$ is $-m_2 g x_2 \theta$.  Both objects sweep out the same angle $\theta$.
If the rod is to rotate with constant speed, then the Net Work done by gravity must
be zero: 

\begin{equation}
   m_1 g x_1 \theta - m_2 g x_2 \theta = 0
\end{equation}   

\noindent This equations yields $m_1 g x_1 = m_2 g x_2$.  Force times distance on one
side equals force times distance on the other when the rod is balanced.  We also verified
this condition for balance in class.  If $m_1 x_1$ were to be greater then $m_2 x_2$
then gravity would do more positive work on $m_1$ then negative work on $m_2$ because
the distance traveled by $m_1$ would be greater than $m_2/m_1$ times $x_2$.  There
would be net work done by gravity and the rod's "angular" speed would increase.
We will derive this result more generally using the vector cross product later.

\bigskip

\Large

\centerline{\bf Rotational Dynamics}

\normalsize

We will first discuss the dynamics of rotational motion about a fixed
axis, then treat the case when the axis of rotation can translate.\\

\bigskip
\large
\centerline {\bf Rotation of a Rigid Object about a fixed axis}
\normalsize
\vspace{4mm}

What is meant by a 
rigid object (or rigid body) is an object for which the separation between all pairs of points 
remains constant.  When we talk about an object in this section, we mean 
rigid object.  Let's first discuss
how we are going to describe rotation about an axis.

\bigskip

\vspace{3mm}
\noindent {\bf Describing rotations about a fixed axis}

\vspace{3mm}
If the axis of rotation is fixed, then the objects orientation
can be specified by an angle (with respect to a reference direction).
Specifying the angle in units of degrees is an earthly standard.
The assignment of $360^\circ$ for a complete rotation is probably
because the earth turns $365$ times about its axis for every
revolution around the sun.  One degree is approximately the angle that 
the earth "sweeps out" around the sun in one day.  A more universal
measure of angle is the radian.  The angle in radians is 
determined by drawing an arc of radius $r$ and measuring the distance
of the arc, $s$, that subtends the angle.  The angle $\theta$ in radians
is defined as $\theta \equiv s/r$.  This definition is gives the same
value for $\theta$ for any radius value $r$.  This is really a universal
definition which could be agreed upon by people living on any planet.

You might be wondering what kind of quantity angular displacement is.
Is it a scalar or a vector, or something else?  Angular displacement
has a magnitude, but is there a direction associated with the displacement?
There is an axis of rotation and a distinction can be made versus clockwise
and counter-clockwise displacements.  Thus, a direction and magnitude can
be associated with an angular displacement: the direction is along the
axis of rotation, with the right-hand rule defining positive displacements;
the magnitude equals the amount of angular displacement.  Although
angular displacements have a magnitude and a direction, angular displacements
do not add according to the laws of vector addition.  Try it by rotating a book with the
following example: $(\pi/2) \hat{i} + (\pi/2) \hat{j}$ does not equal
$(\pi/2) \hat{j} + (\pi/2) \hat{i}$.  Actually, $(\pi/2) \hat{i} + (\pi/2) \hat{j}
= (-\pi/2) \hat{k} + (\pi/2) \hat{i}$.  Rotations are not vectors, they
form what mathematicians call a group.

An objects motion about a fixed axis is one of spinning.  We can quantify
how fast an object is spinning by how fast the angle $\theta$ is changing
in time.  We refer to the spinning speed as the {\bf angular velocity}, and
give it the symbol $\omega$ (omega).  The magnitude of the instaneous 
anglular velocity is the magnitude of the instaneous time rate of change of 
$\theta$:

\begin{equation}
 |\omega| \equiv |{{d \theta} \over {dt}}|
\end{equation}
 
Is the angular velocity a scalar or a vector?  As with displacement it has a magnitude 
and a direction.  The direction is the axis of rotation, with clockwise and
counter-clockwise determined by the right-hand rule.  Curl the fingers of your
right hand in the direction of the rotation, and your thumb points in the direction 
or $\vec{\omega}$.  As with angular displacement, angular velocities do not add
like vectors.  However it is still useful to associate a vector with angular velocity.
The usefulness arises when one wants to relate the velocity of a particle rotating
about an axis.

Consider an object rotating about an axis with angular 
velocity $|\vec{\omega}|$.  
Let the origin of your coordinate system be located on the axis of
rotation.  That is, the vector $\vec{\omega}$ goes through the
origin.  Define the
vector $\vec{r}$ as the position vector to the object that is rotating.  
The speed of the object is $|\vec{\omega}|$ times the perpendicular
distance of the object to $\vec{\omega}$:  
$|\vec{v}| = |\vec{\omega}| |\vec{r}| sin(\theta)$, where $\theta$ is the 
angle between
$\vec{\omega}$ and $\vec{r}$.  The direction of the particles
velocity is perpendicular to both $\vec{r}$
and the vector $\vec{\omega}$.  Thus, the velocity $\vec{v}$ of a 
particle at a position $\vec{r}$ can be written as 

\begin{equation}
   \vec{v} = \vec{\omega} \times \vec{r}  
\end{equation}

\noindent This is a succinct way of expressing the velocity of an object that is rotating, and
will be useful in deriving the equations of rotational motion.

Objects don't always rotate with a constant angular velocity.  As was done for the 
case of translational motion, acceleration is defined as the change in velocity.
For the case of rotation, we can define the angular acceleration as the change in
angular velocity.  

\begin{equation}
  |\alpha| \equiv |{{d \omega} \over {dt}}|
\end{equation}

\noindent In this class we will only consider situations in which the axis of
rotation does not change direction.  Thus, $\alpha$ will be in the same direction
as $\omega$.

For the special case of {\bf constant angular acceleration} the relationship
between the angular velocity $\omega$ and angular displacement $\theta$ is simple.
Since $\theta$, $\omega$, and $\alpha$ have the same mathematical relationship
to each other as $x$, $v$, and $a$ for one dimensional motion, the formulas
for constant angular acceleration are the same as those for constant
acceleration:

\begin{eqnarray*}
  \alpha & = & \alpha_0\\
  \omega & = & \alpha_0 t + \omega_0\\
  \theta & = & {{\alpha_0} \over 2} t^2 + \omega_0 t + \theta_0
\end{eqnarray*}

\noindent where $\omega_0$ is the initial angular velocity, and
$\theta_0$ is the initial angular displacement.  The angular acceleration is
$\alpha_0$, which is constant.  From the middle equation,
$t = (\omega - \omega_0)/\alpha_0$.  This expression for $t$ can
be substituted into the last equation to give:

\begin{equation}
  \omega^2 = \omega_0^2 + 2 \alpha (\theta - \theta_0)
\end{equation}

\noindent Remember, the above equations are only valid for the
special case of {\it constant} angular acceleration.

\bigskip

\begin{figure}
\includegraphics[width=14cm]{fig3nb.png}
\end{figure}

\centerline{\bf Energy of Rotation for a Rigid Object about a fixed axis}

\vspace{4mm}
When we analyzed translational motion, we started first with the relationship between
acceleration, force and mass.  Then we discussed the work-energy theorum and 
kinetic energy.  For rotational motion, it is easier to first discuss energy
considerations, then how torque changes angular velocity.

Let's consider the energy of motion (rotational kinetic energy) of a 
rigid object
rotating about a fixed axis.  We first start with the simple case of a rigid object
consisting of two objects of mass $m_1$ and $m_2$ connected by a massless rod.  Let
the axis of rotation be on the axis, and mass $m_1$ be located a distance of $r_1$
from the axis and mass $m_2$ be located a distance of $r_2$.  If the rod is rotating
about its axis with an angular velocity $\omega$, the speed $v_i$ of each
object is $v_i = r_i \omega$.  The energy of motion of the rigid object is

\begin{eqnarray*}
  K.E. & = & {m_1 \over 2} v_1^2 + {m_2 \over 2} v_2^2\\
       & = & {m_1 \over 2} (\omega r_1)^2 + {m_2 \over 2} (\omega r_2)^2\\
  K.E. & = & {{m_1 r_1^2 + m_2 r_2^2} \over 2} \omega^2  
\end{eqnarray*}

\noindent A nice property of rotational motion about a fixed axis 
is that the angular velocity $\omega$ is the same for both objects.
This allows a factorization of the kinematical quantity ($\omega$) 
from the inertial quantity $m_1 r_1^2 + m_2 r_2^2$.  As we shall
see, the combination of masses times distance squared always appears
as an inertia to changes in rotational motion.  It is a special 
combination, and deserves a special name, {\bf rotational inertial},
and is given the symbol $I_A$:

\begin{equation}
  I_A \equiv m_1 r_1^2 + m_2 r_2^2
\end{equation}

\noindent where $A$ denotes the axis about which the rigid object
is rotating.  The rotational inertial depends not only on the
amount of mass that is rotating, but also on the location of the
masses (i.e. the distance the masses are from the axis of rotation).
When dealing with rotational inertial, remember to always specify
the axis about which the inertial is determined.

The expression for rotational inertial can be generalized to any
number of point particles that are "rigidly" connected.  If
$N$ point objects are connected so that their relative distances
do not change, the rotational inertial about the axis $A$ is

\begin{eqnarray*}
  I_A & = & m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + ...\\
      & = & \sum_{i=1}^N m_i r_i^2
\end{eqnarray*}

This expression can also be extended to a solid object.  To do
this, we divide the solid object up into $N$ small pieces.
We can denote $m_i$ as the mass of piece $i$, and $r_i$ as
the distance to the axis of rotation.  Note: the distance
$r_i$ is the shortest distance from $m_i$ to the axis (i.e.
the perpendicular distance).  The mass $m_i$ can be written
as $\rho \Delta V_i$.  If we take the limit as $N \rightarrow
\infty$ and $\Delta V_i \rightarrow 0$, the sum becomes an
integral:

\begin{equation}
   I_A = \int \rho r^2 dV
\end{equation}

\noindent In lecture we will calculate $I_A$ for a thin rod
of length $L$ and mass $M$ for axis of rotation $A$ about an
end of the rod and about the middle of the rod.  We will also
calculate $I_A$ for a uniform cylinder for rotations about
the center.

The kinetic energy of rotation (rotational kinetic energy) for
any rigid object rotating about the {\it fixed axis} $A$ is

\begin{equation}
  K.E. (of\: rotation) = {I_A \over 2} \omega_A^2
\end{equation}

\noindent where $I_A$ is the rotational inertia about the
axis $A$, and $\omega_A$ is the angular velocity about the
axis $A$.

\bigskip

\centerline{\bf Dynamics of a rigid object about a fixed axis}

\vspace{4mm}
Many engineering applications involve rigid objects rotating about fixed axes.  For this
reason we discuss briefly this special situation.  The simplest case is one particle of
mass $m$ attached by massless rod to an axis point $A$.  Let the position vector from the axis
to the particle be $\vec{r}$.  The particle is confined to rotate a distance 
$r \equiv |\vec{r}|$ away from the point $A$.
Suppose the particle is subject to the net force $\vec{F}_{net}$ at a particular time.  What kind
of motion will happen?  Only the component of the force $\vec{F}_{net}$ perpendicular to 
$\vec{r}$ will cause any change in the particles speed.  The component perpendicular
to $\vec{r}$ is $|\vec{F}_{net}| sin(\theta)$ where $\theta$ is the angle between $\vec{r}$
and $\vec{F}_{net}$.  This component of the force will cause an acceleration equal to $|\vec{F}_{net}| sin(\theta)/m$
along the path of the rotating particle.

\begin{equation}
  |\vec{F}_{net}| sin(\theta) = m |\vec{a}| 
\end{equation}

\noindent If we multiply both sides by r, we have

\begin{equation}
  r |\vec{F}_{net}| sin(\theta) = m r |\vec{a}|   
\end{equation}

\noindent Since the particle is always the same distance from $A$, the
particle's acceleration is along the circle confining the particle.  Thus,
the acceleration can be specified by the angular acceleration about the
axis point $A$, $|\vec{a}| = |\vec{\alpha}| r$

\begin{eqnarray*}
  r |\vec{F}_{net}| sin(\theta) & = & m r^2 {{|\vec{a}|} \over r}\\
                         & = & m r^2 |\vec{\alpha}|     
\end{eqnarray*}

\noindent The left side of the equation is recognized as the torque
about the axis $A$.  Since $\vec{\alpha}$ will point along the same
direction as $\vec{r} \times \vec{F}_{net}$, 

\begin{eqnarray*}
  \vec{r} \times \vec{F}_{net} & = & m r^2 \vec{\alpha}_A\\
         \tau_{net} & = & I_A \alpha_A
\end{eqnarray*}

\noindent where the subscript $A$ refers to the axis $A$: $I$ and $\vec{\alpha}$
are measured about the axis $A$.  From this derivation, one can see that the
end result is just Newton's second law for translational motion applied to
the motion in a circle that a particle must move if connected by a rigid
(massless) rod to an axis point.  As we will show, this expression also
applies in general to a solid rigid object rotating about a fixed axis.

Consider a rigid object confined to rotate about a fixed axis $A$.  Suppose
a force $\vec{F}$ is applied at a position $\vec{r}$ from the axis.  Let
$\Delta \theta$ be the (small) angle the object turns while the force is
applied.  The work done on the system $W$ by the force is the
component of the force perpendicular to $\vec{r}$ times $r \Delta \theta$,
which is the distance the force acts:

\begin{equation}
  W = F_{\perp} (\Delta \theta r)
\end{equation}

\noindent where $F_{\perp}$ is the component of $\vec{F}$ perpendicular to
$\vec{r}$.  This can also be written as  

\begin{equation}
  W = |\vec{r} \times \vec{F}| \Delta \theta
\end{equation}

\noindent The torque times the angular displacement is the work that the
torque does.  If this is the only torque that does work on the system, then
it is equal to a change in kinetic energy (work-energy theorum):

\vspace{2mm}
\centerline{Torque times angular displacement equals change in K.E.}

\vspace{2mm}
\noindent Letting $\Delta (K.E.)$ denote the change in kinetic energy,
we have

\begin{eqnarray*}
    \tau_A \Delta \theta & = & \Delta (K.E.)\\
                      & = & \Delta ({I_A \over 2} \omega_A^2)
\end{eqnarray*}

\noindent If the torque occurs over a small time interval $\Delta t$,
the equation becomes

\begin{equation}
    \tau_A {{\Delta \theta} \over {\Delta t}} = 
                            {{\Delta ({I_A \over 2} \omega_A^2)} \over {\Delta t}}
\end{equation}

\noindent Taking the limit as $\Delta t$ goes to zero results in derivatives with
respect to $t$:

\begin{equation}
  \tau_A {{d \theta} \over {dt}} = {{I_A} \over 2} {{d \omega^2} \over {dt}}
\end{equation}

\noindent or

\begin{equation}
  \tau_A \omega = I_A \omega {{d \omega} \over {dt}} 
\end{equation}

\noindent canceling the $\omega$ on both sides:

\begin{equation}
   \tau_A = I_A \alpha_A
\end{equation}

Thus, for a rigid object confined to rotate about a fixed axis, {\bf the net
torque about the axis equals the rotational inertial times the angular acceleration
about the axis}.  To generalize to situations where the axis of rotation is not
fixed and where the rotational inertial can change, it is useful to introduce a
new quantity: angular momentum.

\bigskip

\large
\centerline{\bf Rotational Dynamics if the axis of rotation is moving}
\normalsize

\bigskip

\centerline{\bf Kinetic Energy for objects rotating and translating}

\vspace{4mm}
Before we discuss how torques affect rotational motion, it is 
instuctive to consider the energy of motion for objects
that are rotating and translating (moving).  The complicated
situation of rotation while the object is moving is made
simple by expressing 'translation of' and 'rotation about'
the center-of-mass of the object.  The expression for
the total energy of motion is

\begin{equation}
  Total \: K.E. = {{M_{total}} \over 2} V_{c.m.}^2 +
                    {{I_{c.m.}} \over 2} \omega_{c.m.}^2
\end{equation}

Let's derive this equation to understand why the center-of-mass
plays such an important role.  Consider a system of particles,
or a rigid object, that is subject to certain forces.  We will
analyze the system of particles in an inertial reference (lab frame) in which
the center of mass (c.m.) of the system is moving.  The velocity of
the c.m. for $N$ particles is given by

\begin{equation}
  \vec{V}_{c.m.} = {{\sum_{i=1}^N m_i \vec{v}_i} \over {M_{tot}}}
\end{equation}

\noindent where $M_{tot} = \sum_{i=1}^N m_i$ is the total mass of the
system.  Let $\vec{u}_i$ be the velocity of object $i$ with respect to
the center-of-mass reference frame.  That is:

\begin{equation}
  \vec{v}_i = \vec{V}_{c.m.} + \vec{u}_i
\end{equation}
 
\noindent where $\vec{v}_i$ is the velocity of particle $i$ as measured in
the inertial reference frame (lab frame), and $\vec{V}_{c.m.}$ is the velocity
of the c.m.  Note that in the c.m. frame:

\begin{equation}
  \sum_{i=1}^N m_i \vec{u}_i = 0
\end{equation}

\noindent The center of mass frame is also called the center of momentum frame,
since the total momentum of the system is zero in the c.m. frame.

The total kinetic energy is:

\begin{eqnarray*}
  Total \: K.E. & = & \sum_{i=1}^N {m_i \over 2} v_i^2\\
                & = & \sum_{i=1}^N {m_i \over 2} (\vec{v}_i \cdot \vec{v}_i)
\end{eqnarray*}

\noindent Substituting the expression for $\vec{v}_i$ above gives

\begin{eqnarray*}
  Total \: K.E. & = & \sum_{i=1}^N {m_i \over 2} (\vec{V}_{c.m.} + \vec{u}_i) \cdot 
                                   (\vec{V}_{c.m.} + \vec{u}_i)\\
               & = & \sum_{i=1}^N {m_i \over 2} (V_{c.m.}^2  
                              + 2 \vec{V}_{c.m.} \cdot \vec{u}_i + u_i^2)
\end{eqnarray*}

In the center of mass frame, the middle term sums to zero:

\begin{eqnarray*}
  \sum_{i=1}^N m_i \vec{V}_{c.m.} \cdot \vec{u}_i & = & \vec{V}_{c.m.}
                                   \cdot    \sum_{i=1}^N m_i \vec{u}_i\\
                                 & = & 0    
\end{eqnarray*}

\noindent since in the c.m. frame $\sum m_i \vec{u}_i = 0$.  Thus, the total 
K.E. separates into two parts:

\begin{eqnarray*}
  Total \: K.E. &  = &  \sum_{i=1}^N {m_i \over 2} (V_{c.m.}^2  + u_i^2)\\
                & = &  {{M_{total}} \over 2} V_{c.m.}^2 + {1 \over 2} 
                                  \sum_{i=1}^N m_i u_i^2
\end{eqnarray*}

\noindent The second term in the expression is the kinetic energy as measured in the
c.m. frame.  For the case of a rigid object, $\vec{u}_i = \vec{\omega} \times \vec{r}_i$.
The magnitude of $\vec{u}_i$ squared is $(\vec{\omega} \times \vec{r}_i)^2$, which after
summing over all particles (and some algebra) gives $(\sum m_i r_i^2) \omega_{c.m.}^2$.
For a rigid object, the above expression becomes

\begin{equation}
  Total \: K.E. = {{M_{total}} \over 2} V_{c.m.}^2 + {{I_{c.m.}} \over 2} \omega_{c.m.}^2
\end{equation}

Thus, {\bf the total energy of motion equals the translational kinetic energy of the
center of mass plus the rotational kinetic energy about the center of mass}.
The key for the separation of the translational energy of the system from the rotational energy
about an axis was to find a reference frame in which $\sum m_i u_i = 0$.  This is the definition
of the center-of-mass (or momentum) frame.  A complicated expression is made simple when viewed
from the right perspective.

\bigskip


\vspace{4mm}
We would like to generalize the result for rigid object rotating about
a fixed axis to the case of an axis of rotation that is not fixed 
as well as the case when $I$ can change.  We will consider a system
made up of $N$ particles.  If the system is a rigid body, then
the distances between the particles is fixed and $N$ approaches
infinity as the particle size goes to zero.  Let $\vec{r}_i$ be the
position vector, $\vec{v}_i$ be the velocity, and $m_i$ be the mass of
particle $i$.  Note that $\vec{r}_i$ and $\vec{v}_i$ both depend on time.
The key to a succinct formalism is the following vector identity which
applied to every particle:

\begin{equation}
  {{d (\vec{r} \times \vec{p})} \over {dt}} = ({{d \vec{r}} \over {dt}} \times \vec{p})
                            + (\vec{r} \times {{d \vec{p}} \over {dt}})
\end{equation}

\noindent where $\vec{p}$ is the momentum of the particle.  For a single particle, the
momentum is $\vec{p} = m \vec{v}$.  So, the first term on the right hand side is

\begin{eqnarray*}
  {{d \vec{r}} \over {dt}} \times \vec{p} & = & \vec{v} \times (m \vec{v})\\
                                          & = & m (\vec{v} \times \vec{v})\\
                                          & = & 0
\end{eqnarray*}

\noindent Thus, we have

\begin{equation}
  {{d (\vec{r}_i \times \vec{p}_i)} \over {dt}} = \vec{r}_i \times {{d \vec{p}_i} \over {dt}}
\end{equation}

\noindent for each particle $i$.  This is a wonderful vector equation.  Now we can put in
the physics: $\vec{F}_{i-net} = (d \vec{p}_i)/(dt)$, where $\vec{F}_{i-net}$ is the net
force on particle $i$.  Substituting gives:

\begin{equation}
  {{d (\vec{r}_i \times \vec{p}_i)} \over {dt}} = \vec{r}_i \times \vec{F}_{i-net}  
\end{equation}

\noindent  The expression on the right side is recognized as the net torque on particle
$i$ about the origin, $\tau_{i-net}$:

\begin{equation}
  {{d (\vec{r}_i \times \vec{p}_i)} \over {dt}} = \tau_{i-net}
\end{equation}

\noindent Thus, the torque on a particle changes the quantity $\vec{r} \times \vec{p}$.
What is $\vec{r} \times \vec{p}$?  It's magnitude is $|\vec{r}| |\vec{p}| sin{\theta}$,
where $\theta$ is the angle between $\vec{r}$ and $\vec{p}$.  $|\vec{p}| sin(\theta)$ is
the component of momentum perpendicular to $\vec{r}$.  It is in essence the "rotational"
momentum of the particle about the origin.  The name given to $\vec{r} \times \vec{p}$
is the {\bf angular momentum} of the particle about the origin.  Note that {\it the
particle does not have to be rotating to have angular momentum}.  A particle moving
in a straight line can have an angular momentum about a point (or origin).\\

\centerline{\bf Angular Momentum}
\bigskip

The symbol for angular momentum is usually $\vec{L}$.  For particle "i",
$\vec{L}_i \equiv \vec{r}_i \times \vec{p}_i$.  Note that the angular
momentum of a particle will depend on the frame of reference,
since the position vector $\vec{r}$ is measured from the origin 
of a particular reference frame.  For a single particle, the above
equation says that the time rate of change of a particles angular 
momentum equals the net torque on the particle.

\begin{equation}
  {{d \vec{L}_i} \over {dt}} = \tau_{i-net}
\end{equation}

\noindent where the angular momentum and the torque are evaluated
about the same axis.  Note, that if the net torque on a particle
equals zero, then the angular momentum of the particle about the
axis is a constant of the motion.  Forces that do not produce a
torque are forces that lie along the direction $\vec{r}$.

For a collection of particles, we can define the total angular momentum
of the system as the vector sum of the angular momentum of the particles:

\begin{equation}
  \vec{L}_{tot} \equiv \sum_{i=1}^N \vec{r}_i \times \vec{p}_i
\end{equation}

\noindent Since $\vec{r}$ and $\vec{p}$ are true vectors, the
angular momentum $\vec{L}$ and $\vec{L}_i$ will also have
the properties of a vector.  Note that $\vec{L}_{tot}$ will
depend on the choice of origin.  

Rotational dynamics is most easily handled by considering the
time rate of change of the total angular momentum:

\begin{eqnarray*}
  {{d \vec{L}_{tot}} \over {dt}} & = & \sum_{i=1}^N {{d (\vec{r}_i \times \vec{p}_i)} \over {dt}}\\
                        & = & \sum_{i=1}^N {{d \vec{r}_i} \over {dt}} \times \vec{p}_i +
                                             \vec{r}_i \times {{d \vec{p}_i} \over {dt}}
\end{eqnarray*}

\noindent The first term on the right side equals zero as before:

\begin{eqnarray*}
   {{d \vec{r}_i} \over {dt}} \times \vec{p}_i & = & \vec{v}_i \times \vec{p}_i\\
                               & = & \vec{v}_i \times m_i \vec{v}_i\\
                               & = & 0
\end{eqnarray*}

\noindent since $\vec{v}_i \times \vec{v}_i$ equals zero.  Thus, we have the same
result for a system of particles as we did with one particle:

\begin{eqnarray*}
  {{d \vec{L}_{tot}} \over {dt}} & = & \sum_{i=1}^N \vec{r}_i \times {{d \vec{p}_i} \over {dt}}\\
                                 & = & \sum_{i=1}^N \vec{r}_i \times \vec{F}_{i-net}\\
                                 & = & \vec{\tau}_{net}
\end{eqnarray*}

\noindent The time rate of change of the total angular momentum of a system of particles equals
the net torque on the system.  Note that this equation is valid for any axis point about
which the torque and angular momentum are calculated (or measured).  

For a system of particles, the above equation can be somewhat simplified.  One can separate
the force on each particle into a part that comes from other particles {\it in the system} or
internal forces, $\vec{F}_{i-int}$, and another part originating from sources {\it outside 
the system} or external forces, $\vec{F}_{i-ext}$:

\begin{equation}
  \vec{F}_{i-net} = \vec{F}_{i-int} + \vec{F}_{i-ext}
\end{equation}

\noindent The internal forces are forces between the particles in the system.  Newton's
third law states that $\vec{F}_{ij} = - \vec{F}_{ji}$.  If these action-reaction forces
act along the line joining particle $i$ and $j$, then the sum of the torques due to the
internal forces will add to zero.  Thus, the net torque on the system, $\vec{\tau}_{net}$,
will equal the sum of the torques due to external forces, or net external torque,
$\vec{\tau}_{net-ext}$:

\begin{equation}
   {{d \vec{L}_{tot}} \over {dt}} = \vec{\tau}_{net-ext}  
\end{equation}

This equation has an important consequence for a system of particles.  {\bf If
the net torque on a system is zero, then the total angular momentum is
conserved!}  We can add another conserved quantity to our list: total momentum
of a system, total angular momentum of a system, mechanical energy for conservative
forces.  The symmetry responsible for the conservation of angular momentum is rotational
symmetry.  Angular momentum and its conservation play a very important role in our 
understanding of atomic, nuclear and particle physics.

The above equation is quite general.  In this class we will consider two special
cases: 1) rotation of a rigid body about a fixed axis, and 2) the case where the angular
momentum vector does not change direction (although the axis of rotation can translate).
These cases are important in engineering applications.

\bigskip

\noindent {\it Angular momentum of a rigid object about a fixed axis}

\vspace{4mm}
The expression for the angular momentum of a rigid object rotating about a fixed axis 
takes on a simple form.  Consider a flat rigid object that is confined to rotate 
about an axis perpendicular to its surface.  Divide the object up into N small pieces.
Let the mass of piece $i$ be $m_i$, and let the mass be located a distance $\vec{r}_i$
from the axis.  Let the angular velocity about the axis be $\omega$.  The speed of each
piece is $\omega |\vec{r}_i|$, and the direction of the velocity is perpendicular to
$\vec{r}_i$.  The angular momentum will lie along the direction of $\vec{\omega}$
with magnitude:

\begin{eqnarray*}
  |\vec{L}| & = & \sum_{i=1}^N r_i m_i v_i\\
            & = & \sum_{i=1}^N r_i m_i \omega r_i\\
            & = & (\sum_{i=1}^N m_i r_i^2) \omega\\ 
  |\vec{L}| & = & I \omega
\end{eqnarray*} 

\noindent where the direction of $\vec{l}$ is in the same direction as $\vec{\omega}$, and all quantities
are calculated about the same axis, $A$.  Written more precisely:

\begin{equation}
  \vec{L}_A = I_A \vec{\omega}_A
\end{equation}

The relationship between torque and angular momentum is

\begin{eqnarray*}
  \vec{\tau}_{A-net} & = & {{d \vec{L}_A} \over {dt}}\\
               & = & I_A {{d \vec{\omega}_A} \over {dt}}\\
  \vec{\tau}_{A-net} & = & I_A \vec{\alpha}_A
\end{eqnarray*}

\noindent where $\vec{\tau}_{A-net}$ is the net torque about the axis $A$.
You might remember that this is the same result we obtained before using
energy considerations: the net work done by the torque equals the change
in the rotational energy of the object.  

\bigskip

\noindent {\it Translation and rotation of a system of particles}

\vspace{4mm}
The translational motion and rotational motion of a rigid object (or a system of particles) 
separate if the center-of-mass is used as a reference point.  The position of the center of mass for a
system of $N$ particles, $\vec{R}_{cm}$ is:

\begin{equation}
  \vec{R}_{cm} \equiv {{\sum_{i=1}^N m_i \vec{r}_i} \over {\sum_{i=1}^N m_i}} 
\end{equation}

\noindent Letting $M_{tot} \equiv \sum_{i=1}^N m_i$, we can also write the
center of mass position as

\begin{equation}
  \vec{R}_{cm} \equiv {{\sum_{i=1}^N m_i \vec{r}_i} \over {M_{tot}}}    
\end{equation}

\noindent The velocity of the center of mass, $\vec{V}_{cm}$ is found by 
differentiating the expression for $\vec{R}_{cm}$ with respect to time:

\begin{equation}
  \vec{V}_{cm} \equiv {{\sum_{i=1}^N m_i \vec{v}_i} \over {M_{tot}}}      
\end{equation}

The center of mass is a special point for a system of particles, or a
rigid object.  For a reference frame that {\it moves such that the
center of mass is at the origin}, we have $\sum_{i=1}^N m_i \vec{r}_i = 0$
and $\sum_{i=1}^N m_i \vec{v}_i = 0$.  This reference frame is called
the "center-of-mass" frame, or "center-of-momentum" frame.  By analyzing
the dynamics from this frame, the translational motion of the center-
of-mass of a system (or rigid object) separates from the 
rotational motion about the center-of-mass.  This can be seen as follows:

\vspace{4mm}
\noindent Consider a frame of reference at rest with respect to the earth.
We will refer to this frame of reference as the lab frame.  Let $\vec{R}_i$
and $\vec{V}_i$ be the position and the velocity of particle $i$ as measured 
by the lab frame.  Let $\vec{r}_i$ and $\vec{v}_i$ be the position and
velocity of particle $i$ as measured in the center of mass frame described
above.  If $\vec{R}_{cm}$ and $\vec{V}_{cm}$ are the position and velocity
of the center of mass of the system (or rigid object), we have

\begin{equation}
  \vec{R}_i = \vec{R}_{cm} + \vec{r}_i
\end{equation}

\noindent for the position of particle $i$, and

\begin{equation}
  \vec{V}_i = \vec{V}_{cm} + \vec{v}_i
\end{equation}

\noindent for the velocity of particle $i$.  The angular momentum of
the system of particles in the lab reference frame is:

\begin{equation}
  \vec{L}_{lab} = \sum_{i=1}^N \vec{R}_i \times m_i\vec{V}_i
\end{equation}

\noindent where $m_i$ is the mass of particle $i$, and we have
written the momentum as mass times velocity, $m_i \vec{V}_i$.
If we express these variables in terms of the position and
velocity as measured in the center of mass frame, we have

\begin{equation}
  \vec{L}_{lab} = \sum_{i=1}^N (\vec{R}_{cm} + \vec{r}_i)
                    \times m_i(\vec{V}_{cm} + \vec{v}_i)
\end{equation}

\noindent After expanding the product on the right side of the
equation, we have four terms:

\begin{eqnarray*}
  \vec{L}_{lab} & = & \sum_{i=1}^N (\vec{R}_{cm} \times m_i\vec{V}_{cm})\\
             &  & + \sum_{i=1}^N (\vec{R}_{cm} \times m_i\vec{v}_{i})\\
             &  & + \sum_{i=1}^N (\vec{r}_{i} \times m_i\vec{V}_{cm})\\
             &  & + \sum_{i=1}^N (\vec{r}_{i} \times m_i\vec{v}_{i})
\end{eqnarray*}

\noindent However, the middle two terms are zero due to the properties of the center of mass
reference frame.  This can be seen as follows.  The second term is

\begin{eqnarray*}
  \sum_{i=1}^N (\vec{R}_{cm} \times m_i\vec{v}_{i}) & = & \vec{R}_{cm}
                                     \times \sum_{i=1}^N m_i \vec{v}_i\\
                      & = & 0
\end{eqnarray*}

\noindent since $\sum_{i=1}^N m_i \vec{v}_i$ is zero in the center of mass frame.
The third term is also zero,

\begin{eqnarray*}
 \sum_{i=1}^N (\vec{r}_{i} \times m_i \vec{V}_{cm}) & = & (\sum_{i=1}^N m_i \vec{r}_i)
                                   \times \vec{V}_{cm}\\
                    & = & 0
\end{eqnarray*}

\noindent since $\sum_{i=1}^N m_i \vec{r}_i$ is zero in the center of mass frame.
So the angular momentum measured in the lab frame separates into two pieces:

\begin{eqnarray*}
  \vec{L}_{lab} & = & \sum_{i=1}^N (\vec{R}_{cm} \times m_i\vec{V}_{cm})
                   + \sum_{i=1}^N (\vec{r}_{i} \times m_i\vec{v}_{i})\\
          & = &    (\vec{R}_{cm} \times m_{tot}\vec{V}_{cm})
                   + \sum_{i=1}^N (\vec{r}_{i} \times m_i\vec{v}_{i})
\end{eqnarray*}

\noindent The first term on the right is the angular momentum as if all the mass
were at the center of mass point, and the second term is the angular momentum
about the center of mass.

The torque in the lab frame is

\begin{eqnarray*}
  \vec{\tau}_{lab} & = & \sum_{i=1}^N \vec{R}_i \times \vec{F}_i\\
                   & = & \sum_{i=1}^N (\vec{R}_{cm} + \vec{r}_i) \times \vec{F}_i\\
                   & = & (\vec{R}_{cm} \times \sum_{i=1}^N \vec{F}_i) + 
                              (\sum_{i=1}^N \vec{r}_i \times \vec{F}_i)\\
                   & = & (\vec{R}_{cm} \times \vec{F}_{net}) + 
                              (\sum_{i=1}^N \vec{r}_i \times \vec{F}_i) 
\end{eqnarray*}

\noindent We showed that Newton's laws of motion applied to rotations gives
$\vec{\tau}_{lab} = d \vec{L}_{lab} / dt$.  Equating the two expressions
above gives:

\begin{eqnarray*}
 (\vec{R}_{cm} \times \vec{F}_{net}) + (\sum_{i=1}^N \vec{r}_i \times \vec{F}_i) 
    & = & {{( d (\vec{R}_{cm} \times m_{tot}\vec{V}_{cm})
                   + \sum_{i=1}^N (\vec{r}_{i} \times m_i\vec{v}_{i}))} \over {dt}}\\
    & = & \vec{R}_{cm} \times {{d m_{tot} \vec{V}_{cm}} \over {dt}} 
                   + \sum_{i=1}^N {{(\vec{r}_{i} \times m_i\vec{v}_{i})} \over {dt}}\\
    & = & (\vec{R}_{cm} \times \vec{F}_{net}) 
                   + \sum_{i=1}^N {{(\vec{r}_{i} \times m_i\vec{v}_{i})} \over {dt}}\\
   \sum_{i=1}^N \vec{r}_i \times \vec{F}_i & = & 
                   \sum_{i=1}^N {{(\vec{r}_{i} \times m_i\vec{v}_{i})} \over {dt}}\\
   \vec{\tau}_{c.m.} & = & {{d \vec{L}_{c.m.}} \over {dt}}
\end{eqnarray*}

\noindent Thus, {\bf eventhough the center of mass is moving, the net torque about the center of
mass equals the time rate of change of the angular momentum about the center of mass}.
This is a very nice result: analyzing the motion about the center of mass greatly simplifies
our analysis.

\bigskip

\centerline{\bf Summary of Rotational Dynamics}

\bigskip

This concludes the discussion on rotational motion for our first 
year physics course.  The relationships that we developed on rotational
dynamics were derived from Newton's Laws of motion.  We did not need to
introduce any "new physics" the past 3 weeks.  We applied the
laws for translation motion to rotations.  We realized that the rotational
part of the motion is the part perpendicular to the position vector, $\vec{r}$.
If $\vec{A}$ is a translational quantity, then
$|\vec{r}| |\vec{A}| sin\theta$ is the corresponding rotational quantity.
Here $\vec{A}$ can be the
force, momentum, velocity or acceleration vectors, and $\theta$ is
the angle between $\vec{r}$ and $\vec{A}$.  The vector cross product is
the mathematical operation that facilates our analysis.  Thus starting
with Newton's second law for particle $i$ in a system of particles:

\begin{equation}
   \vec{F}_{i-net} = {{d \vec{p}_i} \over {dt}} 
\end{equation}

\noindent we can take the vector product of both sides with $\vec{r}_i$:

\begin{eqnarray*}
\vec{r}_i \times \vec{F}_{i-net} & = & \vec{r}_i \times {{d \vec{p}_i} \over {dt}}\\
                   & = & {{d (\vec{r}_i \times \vec{p}_i) } \over {dt}}
\end{eqnarray*}

\noindent where the second equation equals the first since $\vec{v}_i \times 
m_i \vec{v}_i = 0$.  Thus, we are motivated to define a 
"twisting force" vector (torque) as $\vec{r} \times \vec{F}$, and a
"rotational momentum" vector (angular momentum) as $\vec{r} \times
\vec{p}$.  With these definitions, we have:

\begin{equation}
    \vec{\tau}_{i-net} = {{d \vec{L}_i } \over {dt}}
\end{equation}

\noindent Summing over the $N$ particles in a system gives:

\begin{eqnarray*}
 \sum_{i=1}^N \vec{\tau}_{i-net} & = & \sum_{i=1}^N {{d \vec{L}_i } \over {dt}}\\
      \vec{\tau}_{net} = {{d \vec{L}_{net}} \over {dt}}  
\end{eqnarray*}

\noindent From this equation, we showed how translation and rotation can be
separated from each other via the center-of-mass reference frame.

\bigskip

\centerline{\bf Summary of Vector Operations}

\bigskip

We have covered all the vector operations that we will need for 
our first year physics series.  Let's summarize them:

\bigskip
\noindent {\it Vector Addition}:  When two vectors are added,
the result is another vector.  Vector addition is commutative,
$\vec{A} + \vec{B} = \vec{B} + \vec{A}$.  The "tail-to-tip" method
satisfies the commutative property.  In terms of components,
$(\vec{A} + \vec{B})_x = A_x + B_x$,
$(\vec{A} + \vec{B})_y = A_y + B_y$, and
$(\vec{A} + \vec{B})_z = A_z + B_z$.  If physical quantites are vectors,
addition of vectors is used to find the NET sum of the quantites
(e.g. the net force).

\bigskip
\noindent {\it Multiplication of a scalar and a Vector}:  Multiplication of a
vector by a scalar $c$ changes the magnitude of the vector by the factor
$|c|$.  If $c>0$, then the direction remains the same, but if $c<0$ the
direction is reversed.  Multiplication by a scalar enables one to
define unit vectors, whose linear combinations span the vector space.

\bigskip
\noindent {\it Scalar product of two vectors}: The scalar product of one
vector with another vector yields a scalar.  Since two vectors 
are involved, the scalar product must
depend on the magnitudes of each vector and the relative direction of 
one to the other.  If $\theta$ is the angle between the vectors, the
scalar product could depend on either $sin(\theta)$ or $cos(\theta)$.
Since the scalar product should be commutative, $cos(\theta)$ is the
only choice since $cos(-\theta)=cos(\theta)$.  Thus in order to
be commutative, the scalar product must be $\vec{A} \cdot \vec{B}
=|\vec{A}| |\vec{B}| cos(\theta)$.  In terms of components,
$\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.  In physics,
the scalar product represents the component of one vector in the
direction of another (times its magnitude).  It allowed us to express
the work done by a force in acting in a certain direction easily 
($\vec{F} \cdot \Delta \vec{r})$.  

\bigskip
\noindent {\it Vector product of two vectors}: The vector product of
one vector with another yields a vector.  Since two vectors are 
involved here also, the vector product must depend on the magnitudes
of each vector and the relative direction of one to the other.  Since the
result must be a vector, there is only one special direction: the
direction perpendicular to $\vec{A}$ and $\vec{B}$.  Should
$sin(\theta)$ or $cos(\theta)$ be used?  Theta equal to zero
helps us determine the proper choice.  If $\vec{A}$ is
parallel to $\vec{B}$ then there is no special direction, so the
vector product must be zero in this case.  The trig function that
is zero for $\theta = 0$ is $sin(\theta)$.  Thus, the only way to define
a vector product of two vectors is to use the sin function as described
above.

These vector operations will be very important in describing the electro-
magnetic interaction.  This is because the magnetic part of the interaction
depends on the velocity of the particles.  With so many vectors involved, 
$\vec{F}$, $\vec{r}$, and $\vec{v}$, the scalar and
vector products are important.  It will also be necessary to consider
derivatives of vectors, i.e. vector calculus.  Here we will consider rotational motion,
and we will see the usefulness of the vector product.  

\bigskip


\end{document}
