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I'm finally learning physics properly this summer, starting with U0 Physics (to prepare for classical mechanics in the Fall). I'm summarizing it here to keep track of things. This will be very boring for you if you have already taken U0 Physics (I'm working through the ``University Physics'' textbook), so please ignore and please don't laugh, I haven't taken a physics class since High School :)

Chapter 2: Motion Along a Straight Line

is the study of motion. Let \( x \) denote the position of a particle, \( x_0 \) denote the initial position, \(a \) denote the acceleration, \(v \) denote the velocity, and \(v_0\) denote the initial velocity. Then, we have the following , which describe the motion of a particle if we assume that the acceleration \( a \) is constant in time: \begin{align} v & = v_0 + a t \\ x & = x_0 + v_0 t + \frac{1}{2} at^2 \\ v^2 & = v_0^2 + 2a(x-x_0) \\ x-x_0 & = \frac{1}{2}(v_0 - v)t \end{align} is an idealized motion where we ignore the affects of (a) air resistance and (b) the curvature of the Earth. Thus, it's only suitable when the distance of the fall is small compared with the radius of the Earth (which allows us to assume a roughly constant acceleration, and apply the equations from 2.1). The only modification we make here is that we fix \( a = 9.80 m/s^2 \), which is the value for the acceleration due to gravity of a freely-falling body on the surface of the Earth. How can we get equations describing the motion of a particle moving in a straight line if acceleration is not constant; i.e., we're given an acceleration function \( a(t) \)? Let \( t_1 = 0 \) and let \( t_2 \) be some later time. We can obtain values for \( v(t) \) and \( x(t) \) for non-constant acceleration by integrating: \begin{align} v(t) & = v_0 + \int_0^t a(s) ds \\ x(t) & = x_0 + \int_0^t v(s) ds \end{align} For notation, we denote the by: \begin{align} \mathbf{r} := x \hat{\mathbf{i}} + y \mathbf{\hat{j}} + z \mathbf{\hat{k}} \end{align} where \(\hat{\mathbf{i}} \), \(\hat{\mathbf{j}} \), and \(\hat{\mathbf{k}} \) are the standard basis vectors of \( \mathbb{R}^3 \) The will describe changes in the velocity magnitude (speed) and changes in the direction of velocity. the acceleration vector \( \mathbf{a}(t) \) will be tangent to the path (and thus tangent to the velocity) \( \iff \) the particle is moving in a straight line.
We can resolve the acceleration vector into its parallel (\(a_{\parallel} \)) and perpendicular (\( a_{\perp}\)) components. The parallel component tells us about changes in the speed. The perpendicular component tells us about changes in the particle's direction of motion. In its fullest generality, \begin{align} a = a_{\parallel} + a_{\perp} \end{align} A is a body that is given an initial velocity and then follows a path entirely determined by the affects of gravitational acceleration \( g \) and air resistance. We call the path followed by a projectile a .
We study an idealized system where the projectile is represented as a particle with a constant acceleration due to gravity and where the effects of air resistance, curvature, and the rotation of the Earth are all neglected (for now). We specify an initial velocity by providing the initial magnitude (i.e., speed) [denoted by \( v_0 \)] and initial direction (denoted by \( \alpha_0 \)), which is the angle that the velocity vector makes with the \(x\)-axis. The equations of motion are hence given by: \begin{align} x(t) & = (v_{0} \cos(\alpha_0))t \\ y(t) & = (v_{0} \sin(\alpha_0))t - \frac{1}{2}gt^2 \\ v_x(t) &= v_0 \cos(\alpha_0) \\ v_y(t) &= v_0 \sin(\alpha_0) - gt \end{align} From the above set of equations, we can obtain the following pieces of information:

• The distance \(r \) from the origin at any time \( t \): \begin{align} r(t) = \sqrt{x(t)^2 + y(t)^2} \end{align}
• Projectile's speed at any time \( t \): \begin{align} || v(t) || = \sqrt{v_x(t)^2 + v_y(t)^2} \end{align}
• The direction (i..e, the angle \(\alpha \) the trajectory makes with the \(x\)-direction): \begin{align} \tan(\alpha) = \frac{v_y}{v_x} \end{align}

Moreover, we can derive an equation for the shape of the trajectory in terms of \(x\) and \(y\) by eliminating \(t \). We'll obtain an expression of the form \( y = bx - cx^2 \) (i.e., a parabola). The expression is: \begin{align} y = ( \tan(\alpha_0))x - \frac{g}{2v_0^2 \cos (\alpha_0)} x^2 \end{align} is when a particle moves in a circle with a constant speed. Hence, the acceleration vector will be perpendicular to the path, and also directed inward. The magnitude of the acceleration, \( a_{\text{rad}} \), is given by: \begin{align} a_{\text{rad}} = \frac{v^2}{R} \end{align} We can also call this acceleration due to uniform circular motion . We define the of a motion to the time it takes for one revolution to occur. We can re-write the centripital acceleration in terms of the period: \begin{align} a_{\text{rad}} = \frac{4 \pi^2 R}{T^2} \end{align} If the speed varies, then acceleration will also have a tangential component: \begin{align} \mathbf{a} = a_{\text{rad}} + a_{\text{tan}} = \frac{d || \mathbf{v} ||}{dt} + \frac{v^2}{R} \end{align} The is the velocity of a particular observer relative to that observer. A is an observer equipped with a meter-stick and stop-watch. Notation: let \( P \) be a point, and let \( A \) and \( B \) be reference frames. Then, \( x_{P/A} \) will denote the position of \( P \) relative to the reference frame \( A \). Moreover, the position of the origin of \( B \) with respect to the origin of \( A \) is \( x_{B/A} \). We thus have: \begin{align} x_{P / A} & = x_{P / B} + x_{B/A} \\ v_{P/A} & = v_{P / B} + v_{B/A} \end{align} It's exactly what we would expect. Let \( \mathbf{r} \) denote the position vector. Then: \begin{align} \mathbf{r}_{P/A} & = \mathbf{r}_{P/B} + \mathbf{r}_{B/A} \\ \mathbf{v}_{P/A} & = \mathbf{v}_{P/B} + \mathbf{v}_{B/A} \end{align} We call the above transformation the . We also have the general rule that for all reference frames \( A \) and \( B \), \begin{align} \mathbf{v}_{A/B} = - \mathbf{v}_{B / A} \end{align} is the study of the relationships between motion and the forces which cause motion. The basis of this topic are , and this forms the foundation of , which is a very good description of reality so long as we are not working with relativistic speeds (speeds comparable to the speed of light) or small objects (atomic scale). A is an interaction between a body and its surroundings or between two objects. It's a vector quantity. Types of forces:

• Contact force
  : a force involving direct contact between two objects.
• Normal force
  : a force that's always perpendicular to the surface of contact between the objects.
• Friction force
  : a force that is parallel to the surface of contact between the objects that points in the opposite direction of motion.
• Tension force
  : a force that results from pulling on an object by something like a string. This is denoted \( \mathbf{T} \).
• Long-range forces
  : forces that do not requite direct contact between bodies, e.g., gravity or the electromagnetic force.

The principle of the states that any number of forces applied to a body have the same affect as a single force that is the sum of all the individual forces. This is called the acting on a body. We denote it by: \begin{align} \mathbf{R} = \sum_{i=1}^N \mathbf{F}_i \end{align} and we can find the magnitude and direction of \( \mathbf{R} \): \begin{align} || \mathbf{R} || = \sqrt{R_x^2 + R_y^2} \hspace{1cm} \tan (\theta) = \frac{R_y}{R_x} \end{align} An is a reference frame where Newton's First Law is valid. The Earth is approximately an inertial reference frame. Given one inertial reference frame \( A \), we can construct any other reference frame \( B \) by considering a reference frame that is moving with a constant velocity relative to \( A \). Many experiments show that for a given body, the ratio of the magnitude of the net force, \( || \sum_{i=1}^N \mathbf{F}_i || \), to the magnitude of the acceleration, \( || \mathbf{a} || \), is constant. Hence, we have a quantitative measure of inertia: \begin{align} m = \frac{|| \sum_{i=1}^N \mathbf{F}_i ||}{|| \mathbf{a} ||} \end{align} We define this ratio to be the . We can thus express Newtons (the unit of force) in terms of the standard units as follows: \begin{align} 1 N = 1 \text{kg} \cdot m/s^2 \end{align} Experiments also verify that what matters in terms of acceleration is the net force; these forces are what cause objects to accelerate. Four remarks on Newton's Second Law:

1. It's a vector equation, and so to analyze we break it down into component-form.
2. It only includes external forces
3. \(m \) must be constant; for systems where the mass varies, momentum is a more suitable concept.
4. As with Newton's First Law, it's only valid in inertial reference frames

The of an object is the gravitational force exerted on a body by the Earth. Hence, \begin{align} \mathbf{w} = m \mathbf{g} \end{align} We call the pair \( \mathbf{F}_{\text{ \(A \) on \( B \) }} \) and \( \mathbf{F}_{\text{ \( B \) on \( A \) }} \) an . Two forces in an action-reaction pair never act on the same body. diagrams are an important tool used to solve physics problems. Some things to keep in mind when solving physics problems:

1. Newton's first and second laws can only be applied once you choose the that you are going to analyze.
2. We only consider the forces acting on the body.
3. If there are multiple bodies, then you need to take the problem apart and draw a free-body diagram for each body.

When a body is on a surface, that surface exerts a single contact force on that body. We can resolve this force into its parallel and perpendicular components. The perpendicular component is the normal force \( \mathbf{n} \). The parallel component is the , denoted by \( \mathbf{f} \). The direction of the friction force is always opposite to the direction of the relative motion.

There are two types of friction force:

1. Kinetic friction force
   : denoted \( \mathbf{f}_k \), this is friction that acts when an object is moving along a surface. It has been experimentally verified that:
\begin{align} || \mathbf{f_k} || = \mu_k || \mathbf{n} || \end{align} The
coefficient of kinetic friction
, denoted \( \mu_k \), is the proportionality constant between the forces. The slippier the surface, the smaller \( \mu_k \) is. It's a dimensionless quantity.
2. Static friction force
   , denoted \(\mathbf{f}_s \), is the friction that acts on an object that is not moving relative to the surface. The maximum value that the static friction force can attain before the object starts moving -- \( ( \mathbf{f}_s )_{\text{max}} \) is proportional to \( || \mathbf{n} || \). \( \mu_s \) is the proportionality factor: \begin{align} || \mathbf{f}_s || \leq ( \mathbf{f}_s )_{\text{max}} = \mu_s || \mathbf{n} || \end{align} and we call it the
   Coefficient of static friction
   .

The coefficient of static friction is usually greater than the coefficient of kinetic friction. The , denoted \( \mu_T \), tells us how much easier it is to move something on wheels, as opposted to sliding it. It is given by: \begin{align} \mu_T = \frac{\text{horizontal force needed for a constant speed on a flat surface}}{\text{upward normal force exerted by the surface}} \end{align} is a force exerted by a fluid on a body which is moving through the fluid. The direction of the fluid resistance force is always opposite the direction of the body's velocity with respect to the fluid.

• For small objects moving at very low speeds (e.g. dust particles), the fluid resistance force is given by: \begin{align} \mathbf{f} = k \mathbf{v} \end{align} where \( k \) is a proportionality constant which depends on the shape and size of the body as well as the fluid's properties (e.g. viscosity).
• For larger objects (e.g. tennis ball), the proportionality constant is called
  air drag
  : \begin{align} || \mathbf{f} || = D || \mathbf{v} ||^2 \end{align} where \( D \sim kg/m \).

The existence of fluid resistance means that objects falling through a fluid do not have constant acceleration, and hence the kinematic equations that we've been using all along are no longer valid. We observe that as speed increases, the resisting force increases until \( mg - kv_y = 0 \). At this point, acceleration becomes zero, and velocity no longer increases, and we attain what is called the . We denote this by \(v_t \), and it is given by: \begin{align} \mathbf{v}_t = \frac{m \mathbf{g}}{k} \end{align} We can obtain expressions for \( \mathbf{v}_y \), \( \mathbf{a}_y \), and \( y \) using Newton's second law: \begin{align} m \frac{d \mathbf{v}_y}{dt} = m \mathbf{g} - k \mathbf{v}_y \end{align} This is a differential equation in \( v_y \); solving for this, and then differentiating for acceleration and integrating for position, we obtain: \begin{align} \mathbf{v}_y & = \mathbf{v}_t \left[ 1 - e^{-(k/m)t} \right] \\ \mathbf{a}_y & = \mathbf{g} e^{(-k/m)t} \\ \mathbf{y} & = \mathbf{v}_t \left[ t - \frac{m}{k} \left( 1 - e^{-(k/m)t} \right) \right] \end{align} The terminal velocity in the case of \( || \mathbf{f} || = D || \mathbf{v} ||^2 \) is: \begin{align} || \mathbf{v}_t || = \sqrt{ \frac{m || \mathbf{g} ||}{D}} \end{align} Recall that the magnitude of the acceleration of an object in unoform circular motion is given by: \begin{align} a_{\text{rad}} = \frac{|| \mathbf{v} ||^2}{R} \end{align} or, equivalently, in terms of the period \( T \): \begin{align} a_{\text{rad}} = \frac{4 \pi^2 R}{T^2} \end{align} These forces and accelerations are governed by Newton's Second Law, since the particle is accelerating towards the centre, and hence \( \sum \mathbf{F} \) must be directed inwards. If this force is zero, then the particle will fly off in the tangent direction. In uniform circular motion, the magnitude \( || \sum \mathbf{F} || \) is constant since \( || \mathbf{a} || \) is constant. The magnitude of the force \( \mathbf{F}_{\text{net}} \) is hence: \begin{align} || \mathbf{F}_{\text{net}} || = m a_{\text{rad}} = m \frac{|| \mathbf{v} ||^2}{R} \end{align} Remark: these expressions are valid for any circular arc; we don't require the particle to complete a full circle.

1. Graviational Interactions
   : e.g. weight, planets in orbit \( \rightarrow \) Ch. 13
2. Electromagnetic Interaction
   : electric and magnetic fores, e.g. contact forces (e.g. normal force, friction, fluid reistance).
3. Strong Interaction
   : holds the nucleus of an atom together.
4. Weak Interaction
   : responsible for beta decay.

In the 1960s, the electromagnetic and weak interactions were unified into an ``.'' The total work done on a particle by all forces acting on the object will tell us the change in the particle's kinetic energy, a quantitty which is related to the particle's mass and speed. This rule even holds when the force varies. The SI unit for work is the joule \( \sim \) \( N \cdot m \). The of a particle with mass \( m \) and velocity \( \mathbf{v} \) is given by: \begin{align} K = \frac{1}{2} m || \mathbf{v} ||^2 \end{align} One can use Newton's Second Law + Kinematic Eqns to derive: Since this theorem is derived using Newton's Second law, it is only valid in intertial reference frames. The Work-Energy theorem is useful when one wants to relate a body's speed \( || \mathbf{v}_1 || \) at one point in the trajectory to its speed \( || \mathbf{v}_2 || \) at another point in its trajectory. The physical interpretation of kinetic energy or a particle is that it's the work required to bring that particle from rest to that speed. An equivalent way to look at kinetic energy is to see it as the work a particle can do as it's brought to rest fro a certain speed. Application: consider a spring which is stretched by an amount \( x \). Then, the force required to stretch the spring is: \begin{align} F_x = k x \end{align} where \( k \) is called the , \( k \sim N/m \). This observation is known as . We can now generalize the definition to include a force that varies in both magnitude and direction; here, work is defined as a line integral: We need a measure of how quickly work is done \( \Rightarrow \) power. The is the time rate at which work is done: \begin{align} P := \frac{dW}{dt} \end{align} The SI unit is the Watt, denoted \( W \); \( W \sim J/s \). We also have another definition of power: \begin{align} P := \mathbf{F} \cdot \mathbf{v} \end{align} : the potential energy associated with the position of a particle. One particular type is , which is the potential energy associated with the body's weight and height above the ground. It is given by: \begin{align} U_{\text{grav}} := mgy \end{align} The work done by the gravitational force on a particle is: \begin{align} W = - \Delta U_{\text{grav}} \end{align} If only the gravitational force does the work, then the total mechanical energy is conserved: \begin{align} K_1 + U_{\text{grav, 1}} = K_2 + U_{\text{grav, 2}} \end{align} We call the sum \( E := K + U_{\text{grav}} \) the of the system. Hence, when only gravity is doing the work, \begin{align} E = K + U_{\text{grav}} = c \end{align} for \( c \in \mathbb{R} \). Mechanical energy is conserved; this principle is called the . Let \( W_{\text{other}} \) be the work done by the force \( \mathbf{F}_{\text{other}} \), forces which are not the gravitational force. Then, one has: \begin{align} K_1 + U_{\text{grav, 1}} + W_{\text{other}} = K_2 + U_{\text{grav, 2}} \end{align} That is, \( W_{\text{other}} \) is the change in the system's total mechanical energy, \( E = K + U_{\text{grav}} \). The is the energy stored in a deformable body such as a string. An object which returns to its original size and shape after being defored is said to be . Elastic potential energy is given by: \begin{align} U_{\text{el}} := \frac{1}{2} kx^2 \end{align} where \( U_{\text{el}} \) denotes the elastic potential energy stored in a spring, \( k \) is the force constant, and \( x \) is the displacement from the spring's natural length. The work done by the elastic force is given by: \begin{align} W_{\text{el}} = U_{\text{el}, 1} - U_{\text{el}, 2} = - \Delta U_{\text{el}} \end{align} From the Work-Energy theorem, the total mechanical energy is conserved if only the elastic force is doing the work: \begin{align} K_1 + U_{\text{el}, 1} = K_2 + U_{\text{el}, 2} \end{align} A is a force where the sum of the kinetic and potential energy remains the same for all time. Exampels of conservative forces include gravity and the spring force. The work done by conservative forces obeys the following four properties:

1. Work can be expressed as the difference between initial and final values of a
   potential-energy
   function.
2. Reversible
3. Path-independence: work does not depend on the path the body takes.
4. When \( x_1 = x_2 \), the total work done is zero.

Mechanical energy is conserved if the only forces doing the work are conservative forces. A is a force that is not conservative; it cannot be represented by a potential-energy function. A sub-class of non-conservative forces are , which are non-conservative forces which cause mechanical energy to be lost or dissipated. An example of this is kinetic friction or fluid resistance. We can represent non-conservative forces in terms of , which is the energy associated with the change in the state of the materials. For example, raising the temperature of an object raises its internal energy. The following has been experimentally verified: The branch of physics that studies the relationship between internal energy and temperature changes, heat, and work is called thermodynamics. We have the following expression which gives us the force corresponding to a given potential-energy expression, of course, assuming that the force is a conservative force. Let \( F_x (x) \) be the x-component of the force and let \( U(x) \) be the associated potential energy function. Using the formula for work and taking the limit, we obtain that: \begin{align} F_x(x) = - \frac{ dU(x)}{dx} \end{align} The physical meaning behind this is that conservative forces always act to push the system towards the lowest potential energy. \begin{align} \mathbf{F} = - \left( \frac{\partial U}{\partial x} \mathbf{\hat{i}} + \frac{\partial U}{\partial y} \mathbf{\hat{j}} + \frac{\partial U}{\partial z} \mathbf{\hat{k}} \right) = - \nabla \mathbf{U} \end{align} An is a graph that shows the potential energy function \( U(x) \) and the energy of a particle \( E \) of a particle subjected to a force with a corresponding potential energy function \( U(x) \).

1. Stable equilibrium
   : a minimum of a potential-energy curve.
2. Unstable equilibrium
   : a maximum of a potential-energy curve.

A key part of Newtonian mechanics is the conservation of momentum; this allows us to study situations that may be very difficult to analyze with Newton's Laws. In particular, the conservation of momentum is valid in situations where Newton's Laws are not valid. We use this principle when studying collision problems and problems with changing masses. We'll see another useful way to re-state \( \mathbf{F} = m \mathbf{a} \), as we did by previously re-stating it as the Work-Energy theorem. We can write Newton's second law as, \begin{align} \sum \mathbf{F}_i = \frac{d}{dt} [ m \mathbf{v} ] \end{align} The or of a particle is defined as: \begin{align} \mathbf{p} = m \mathbf{v} \end{align} It's SI unit is \( kg \times m/s \). We can re-formulate Newton's Second Law in terms of momentum: \begin{align} \sum \mathbf{F}_i = \frac{d \mathbf{p}}{dt} \end{align} This is only valid in intertial reference frames. Question: What is the physical difference between momentum and kinetic energy? We can answer this question by introducing the concept of impulse. The of a constant net force \( \sum \mathbf{F}_i \) during some time interval \( \Delta t := [t_1, t_2] \) is defined as: \begin{align} \mathbf{J} := \sum \mathbf{F} ( t_2 - t_1) = \sum \mathbf{F} \Delta t \end{align} Its SI unit is \( N \cdot s \sim kg \times m/s \). We can also use the average net force: \begin{align} \mathbf{J} = \mathbf{F}_{\text{avg}} (t_2 - t_1) \end{align} On one hand, we have the impulse-momentum theorem, \begin{align} \mathbf{J} = \mathbf{p}_1 - \mathbf{p}_2 \end{align} which gives us the changes in a particle's momentum due to impulse. This depends on the time over which the net force acts. On the other hand, we have the work-energy theorem, \begin{align} W_{\text{tot}} = K_2 - K_1 \end{align} which gives us changes in the particle's kinetic energy due to work. This depends on the distance over which the net force acts. The physical meaning behind all of this is:

• Momentum is the impulse that accelerated a particle from rest to its present speed
• Kinetic energy is the total work done to a particle to accelerate it from rest

The impulse-momentum theorem and the work-energy theorem are examples of integral principles, principles which relate the motion of two different times separated by some finite time-interval. Conversely, Newton's Second Law is an example of a differential principle, which relates the rate of change of the velocity or momentum to the force. are forces that the particles in a system exert on each other. are forces exerted on any part of the system by some object outside of it. An is one where there are no external forces. Denote by \( \mathbf{F}_{\text{B on A}} \) the net force on a particle \( A \) and denote by \( \mathbf{F}_{\text{A on B}} \) the net force on a particle \( B \). Then, for an isolated system, we have \begin{align} \mathbf{F}_{\text{B on A}} = \frac{d \mathbf{p}_A}{dt} \hspace{1cm} \mathbf{F}_{\text{A on B}} = \frac{d \mathbf{p}_B}{dt} \end{align} By Newton's Third law, \begin{align} \mathbf{F}_{\text{B on A}} + \mathbf{F}_{\text{A on B}} = \frac{d \mathbf{p}_A}{dt} + \frac{d \mathbf{p}_B}{dt} = \frac{d(\mathbf{p}_A + \mathbf{p}_B)}{dt} \end{align} We define the , \( \mathbf{P} \), as the vector sum of the momenta of the individual particles of the system. Hence, the above formula is saying that the total momenta of an isolated system is constant. This gives us the first form of the Principle of the Conservation of Momentum:

1. Elastic Collision
   : when the forces acting on the bodies are conservative forces, then by the conservation of mechanical energy, no mechanical energy is lost, and hence kinetic energy remains the same.
2. Inelastic Collision
   : collision in which the total kinetic energy after the collision is less than the collision
1. A sub-class:
   Completely Inelastic Collision
   : inelastic collision where the bodies collide into one body and then move together.

Note. For any collision where the external forces are negligible, then the total momentum is conserved; for any collision that is elastic, the kinetic energy is conserved. One can use the conservation of momentum to prove that kinetic energy is always lost in a completely inelastic collision.

1. Kinetic energy conserved \( \rightarrow \) elastic.
2. Kinetic energy decreases \( \rightarrow \) inelastic.
3. Two bodies have a common final velocity \( \rightarrow \) completely inelastic.

Elastic collisions occur when the forces between colliding bodies is conservative. For elastic collisions, we have also the conservation of kinetic energy. This property, plus the conservation of momentum, allows us to find the final velocityes \( v_{A2} \) and \( v_{B2} \). Let's consider the special case when \( v_{B1} = 0 \) (for straight-line motion). Then, \begin{align} v_{A2} & = \frac{m_A - m_B}{m_A + m_B} v_{A1} \\ v_{B2} & = \frac{2m_A}{m_A + m_B} v_{A1} \end{align} In a straight-line elastic collision of two bodies, the relative velocities before and after the collision have the same magnitude but the opposite sign. When this condition is satisfied, the total kinetic energy is also conserved. Suppose that we have several particles with masses \( m_1, m_2, ... \) with position vectors \( \mathbf{r}_1, \mathbf{r}_2, ... \). Then, the of the system of particles is given by \begin{align} \mathbf{r}_{cm} := \frac{\sum_{i} m_i \mathbf{r}_i}{\sum_{i} m_i} \end{align} We call this the mass-weighted average position of the particles. Again, let \( m_1, m_2, ... \) denote the masses of the particles of the system. Let \( M \) denote the total mass of the system, \begin{align} M := m_1 + m_2 + m_3 + ... \end{align} Then, the is given by \begin{align} \mathbf{v}_{cm} := \frac{m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2}{m_1 + m_2+ ...} \end{align} We can re-write this to obtain the total momentum of the system, \begin{align} M \mathbf{v}_{cm} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 + ... = P \end{align} Hence, the total momentum \( P \) of a system equals the total mass times the velocity of the centre of mass. Observe: for a system of particles with zero net force, then we know by Newton's Second Law that this implies that \( P \) is constant. By the above, this implies that the velocity of the centre of mass is constant. Similarly, the acceleration of the centre of mass is given by, \begin{align} M \mathbf{a}_{\text{cm}} = m_1 \mathbf{a}_1 + m_2 \mathbf{a}_2 + ... \end{align} By Newton's Second Law (\( M \mathbf{a}_{\text{cm}} = \sum \mathbf{F} \)) and by Newton's Third Law, one has: \begin{align} \sum \mathbf{F}_{\text{ext}} = M \mathbf{a}_{\text{cm}}. \end{align} Here, \( \sum \mathbf{F}_{\text{ext}} \) is the net external force on a body of a collection of particles, \( M \) is the total mass of the collection of particles, and \( \mathbf{a}_{\text{cm}} \) is the acceleration of the centre of mass. What this means: when a body or a collection of paticles is acted upon by external forces, the centre of mass moves as if all the mass is concentrated at the centre of mass and as if the net force equal to the sum of the sum of the external forces acts on the centre of mass. We've been using this property all along, and we will extensively use it in Chapter 10 when studying motion of rigid bodies. Hence, for an extended body or a system of particles, \begin{align} \sum \mathbf{F}_{\text{ext}} = \frac{d\mathbf{P}}{dt} \end{align} Motivation: we need to develop methods for analyzing the motion of a rotating body. A is an idealized body that has a perfectly definite and unchanging shape and size, i.e., we ignore the fact that forces can deform real-world bodies. We will first study a rigid body that rotates about a fixed axis. A fixed axis is one that is at rest in some inertial reference frame and does not change direction relative to that frame. We'll use a more convenient coordinate system (rather than the Cartesian coordinate system) in this segment: we'll specify the location of a particle with the angle \( \theta \) that the position vector of the rotation body makes with the positive x-axis. We call this the and it uniquely describes the motion. The most natural unit for \( \theta \) is not in degrees but rather than radians. For an angle \( \theta \) subtended by an arc of length \( s \) on a circle of radius \( r \), we have: \begin{align} \theta = \frac{s}{r} \end{align} for \( \theta \) measured in radians. We have the following conversion between radians and degrees: \( 1 \) rad \( = 57.3 \) degrees. The for a body rotating about the z-axis, denoted \( \omega_{\text{z, avg}}) \) is the ratio of the angular displacement to \( \Delta t \): \begin{align} \omega_{\text{avg, z}} := \frac{\Delta \theta}{\Delta t} \end{align} The is defined as: \begin{align} \omega_z := \frac{d \theta}{d t} \end{align} We have the following conversions: \begin{align} 1 \text{rev/s} = 2 \pi \text{rad/s} \hspace{1cm} 1 \text{rev/min} = 1 \text{rpm} = \frac{2 \pi}{60} \text{rad/sec} \end{align}