Classical Mechanics

Introduction

These notes cover my first year course in classical mechanics. The notes assume familiarity with the following concepts:

1. Basic linear algebra (See Linear Algebra Notes: Physical Vectors).
2. Differential Calculus (See Calculus Notes).
3. Integral Calculus (See Calculus Notes).

1D Motion

• 1D motion is motion along a straight line.
• The principles and concepts of 1D motion will be later extended to 2 and 3D motion.

Position

• definition: position. Let ${p}$ be some particle. If ${p}$ exists in one-dimensional space ${\reals,}$ then ${p}$'s position is an ${x \in \reals,}$ relative to some reference point ${\mathscr{O}.}$ If ${p}$ exists in two-dimensional space ${\reals^2,}$ then ${p}$'s position is a pair ${(x,y) \in \reals^2,}$ again relative to some reference point ${\mathscr{O}.}$ If ${p}$ exists in three-dimensional space, then ${p}$'s position is a triple ${(x,y,z) \in \reals^3}$ (relative to some reference point ${\mathscr{O}.}$)

Displacement

• definition: displacement. Suppose a particle ${p}$ has an initial position ${x_1}$ and a final position ${x_2.}$ Then the change from position ${x_1}$ to ${x_2}$ is called the displacement of ${p,}$ denoted ${\Delta x,}$ where

• Displacement is a vector quantity — it has a magnitude and a direction.

Average Velocity

• definition. Average velocity, denoted ${v_{\text{avg}},}$ is the ratio of a particle's displacement ${\Delta x}$ over a particular time interval ${\Delta t.}$ $$v_{\text{avg}} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1}.$$ Above, ${t_2}$ is the time when the particle's position is at ${x_2,}$ and ${t_1}$ is the time when the particle's position is at ${x_1.}$
• In a graph of ${x}$ versus ${t,}$ where ${x}$ corresponds to position and ${t}$ to time, the slope of the secant line connecting two points on the ${x(t)}$ curve corresponds to ${v_{\text{avg}}.}$
• Like displacement, ${v_{\text{avg}}}$ is a vector quantity — it has a magnitude and a direction.
• ${v_{\text{avg}}}$ always has the same sign as ${\Delta x}$ because ${\Delta t}$ is always positive.

Average Speed

• definition. Average speed, denoted ${s_{\text{avg}},}$ is the ratio of the total distance a particle travels over a particular time interval ${\Delta t.}$
• Average speed is a scalar quantity — it has a magnitude but no direction (ergo, no algebraic sign).

Instantaneous Velocity

• definition. A moving particle's instantaneous velocity (or simply velocity), denoted ${v,}$ is $$v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{\text{d} x}{\text{d} t}.$$
• In physics, the term speed refers to the magnitude of ${v.}$
• A common unit for velocity is ${\text{m}/\text{s}}$ (meters per second).

Average Acceleration

• definition. An particle's average acceleration, denoted ${a_{\text{avg}},}$ is the ratio of a change in velocity ${\Delta v}$ to the time interval ${\Delta t}$ in which the change occurs: $$a_{\text{avg}} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 - v_1}{t_2 - t_1}$$ where ${v_2}$ is the particle's velocity at time ${t_2,}$ and ${v_1}$ at time ${t_1.}$
• ${a_{\text{avg}}}$ is a vector quantity — it has a magnitude and a direction.
• The algebraic sign of ${a_{\text{avg}}}$ indicates the direction of ${a_{\text{avg}}.}$

Instantaneous Acceleration

• definition. An particle's instantaneous acceleration (or simply acceleration), is $$a = \dfrac{\text{d} v}{\text{d} t} = \dfrac{\text{d}}{\text{d}t} \left( \dfrac{\text{d}x}{\text{d}t} \right) = \dfrac{\text{d}^2 x}{\text{d}t^2}.$$ That is, a particle's acceleration at any instant is the second derivative of its position ${x(t)}$ with respect to time.
• Acceleration is commonly measured in ${\text{m}/\text{s}^2}$ (meters per second squared).

g Units

• Large accelerations are sometimes expressed in ${\text{g}}$ units: $$1 \text{g} = 9.8 \text{m}/\text{s}^2.$$

Acceleration Signum

• If the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases.

Constant Acceleration

Constant acceleration is a common case of movement. E.g., while stopped at a red light, you accelerate your car at an approximately constant rate when the light turns green.

The following equations mathematically model this case. The equations only apply to constant acceleration.

• equation. Where ${a}$ is constant, $$a = a_\text{avg} = \dfrac{v_1 - v_0}{t_1 - t_0},$$ where ${v_1}$ is the velocity at time ${t_1}$ and ${v_0}$ is the velocity at time ${t_0.}$
• equation. Under constant acceleration, $$v_{\text{avg}} = \dfrac{x_1 - x_0}{t_1 - t_0}.$$
• equation. Under constant acceleration, $$x_1 = x_0 + v_{\text{avg}}t.$$
• equation. Under constant acceleration, $$v_{\text{avg}} = \dfrac{1}{2}(v_0 + v_1).$$
• equation. Under constant acceleration, $$v_{\text{avg}} = v_0 + \dfrac{1}{2}at.$$
• equation. Under constant acceleration, $$x_1 - x_0 = v_0t + \dfrac{1}{2}at^2.$$
• equation. Under constant acceleration, $$v_1^2 = v_0^2 + 2a(x_1 - x_0).$$
• equation. Under constant acceleration, $$x_1 - x_0 = \dfrac{1}{2}(v_0 + v_1)t.$$
• equation. Under constant acceleration, $$x_1 - x_0 = vt - \dfrac{1}{2}at^2.$$

Free-fall Acceleration

• If we toss an object either up or down and eliminate the effects of air on the object's flight, we would find that the object accelerates downward at a certain constant rate.
• We call this rate free-fall acceleration, of magnitude ${g.}$
  • Free-fall acceleration is independent of the object's mass, density, or shape — it is the same for all objects.
  • At sea-level in Earth's midlatitudes, the value is roughly ${9.807 \text{m}/\text{s}^2.}$

2D & 3D Motion

Position & Displacement

• We can locate a particle with a position vector ${\overrightarrow{r}:}$ $$\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}.$$
• Suppose an object moves from position ${\overrightarrow{r}_1}$ to ${\overrightarrow{r}_2.}$ Then its displacement is: $$\Delta\overrightarrow{r} = \overrightarrow{r}_2 - \overrightarrow{r}_1.$$
  • Using unit-vector notation, we can rewrite this as: $$\Delta \overrightarrow{r} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j} + (z_2 - z_1) \hat{k}.$$ where coordinates ${(x_1, y_1, z_1)}$ correspond to position vector ${\overrightarrow{r}_1,}$ and coordinates ${(x_2, y_2, z_2)}$ correspond to position vector ${\overrightarrow{r}_2.}$
  • We may also write: $$\Delta \overrightarrow{r} = \Delta x \hat{i} + \Delta y \hat{j} + \Delta z \hat{k}.$$ where ${\Delta x = x_2 - x_1,}$ ${\Delta y = y_2 - y_1,}$ and ${\Delta z = z_2 - z_1.}$

Velocity in Space

• A particle moves from point ${A}$ to point ${B.}$ How fast does it move?
• If a particle moves through a displacement ${\Delta \overrightarrow{r}}$ in a time interval ${\Delta t,}$ then its average velocity, denoted ${\overrightarrow{v}_{\text{avg}},}$ is: $$\overrightarrow{v}_{\text{avg}} = \dfrac{\Delta \overrightarrow{r}}{\Delta t}.$$
  • We can rewrite this as: $$\begin{aligned} \overrightarrow{v}_{\text{avg}} &= \dfrac{\Delta x \hat{i} + \Delta y \hat{j} + \Delta z \hat{k}}{\Delta t} \\[1.5em] &= \dfrac{\Delta x }{\Delta t} \hat{i} + \dfrac{\Delta y}{\Delta t} \hat{j} + \dfrac{\Delta z}{\Delta t} \hat{k}. \end{aligned}$$
• The instantaneous velocity of the particle is given by: $$\overrightarrow{v} = \dfrac{\text{d}\overrightarrow{r}}{\text{d}t}.$$
  • The direction of the instantaneous velocity ${\overrightarrow{v}}$ of a particle is always tangent to the particle's path at the particle's position.
• We can extend this to three dimensions: $$\begin{aligned} \overrightarrow{v} &= \dfrac{\text{d}}{\text{d}t} (x \hat{i} + y \hat{j} + z \hat{k}) \\[1.5em] \dfrac{\text{d}x}{\text{d}t} \hat{i} + \dfrac{\text{d}y}{\text{d}t} \hat{j} + \dfrac{\text{d}z}{\text{d}t} \hat{k}. \end{aligned}$$
  • This equation can be simplified to: $$\overrightarrow{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k},$$ where:
    $$v_x = \dfrac{\text{d}x}{\text{d}t}$$$$v_y = \dfrac{\text{d}y}{\text{d}t}$$$$v_z = \dfrac{\text{d}z}{\text{d}t}.$$

Acceleration in Space

• When a particle's velocity changes from ${\overrightarrow{v}_1}$ to ${\overrightarrow{v}_2}$ over a time interval ${\Delta t,}$ its average acceleration during ${\Delta t}$ is given by: $$\overrightarrow{a}_{\text{avg}} = \dfrac{\overrightarrow{v}_2 - \overrightarrow{v}_1}{\Delta t} = \dfrac{\Delta \overrightarrow{v}}{\Delta t}.$$
• As ${\Delta t \to 0,}$ the limit ${\overrightarrow{a}_{\text{avg}}}$ aproaches the instantaneous acceleration ${\overrightarrow{a}:}$ $$\overrightarrow{a} = \dfrac{\text{d}\overrightarrow{v}}{\text{d}t}.$$
• If a particle's velocity changes in either magnitude or direction (or both), then the particle has an acceleration.
• Instantaneous acceleration can be written in unit-vector form as: $$\begin{aligned} \overrightarrow{a} &= \dfrac{\text{d}}{\text{d}t} (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \\[1.5em] &= \dfrac{\text{d}v_x}{\text{d}t} \hat{i} + \dfrac{\text{d}v_y}{\text{d}t} \hat{j} + \dfrac{\text{d}v_z}{\text{d}t} \hat{k}. \end{aligned}$$
  • We can rewrite this as: $$\overrightarrow{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k},$$ where
    $$a_x = \dfrac{\text{d}v_x}{\text{d}t}$$$$a_y = \dfrac{\text{d}v_y}{\text{d}t}$$$$a_z = \dfrac{\text{d}v_z}{\text{d}t}.$$

Projectile Motion

• A projectile is a particle that moves in a vertical plane with some initial velocity ${\overrightarrow{v}_0}$ but its acceleration is always the free-fall acceleration ${\overrightarrow{g},}$ which points downward.
• The projectile's initial velocity can be written as: $$\overrightarrow{v}_0 = v_{0_x} \hat{i} + v_{0_y} \hat{j}.$$
  • If we know the angle ${\theta_0}$ (the angle between ${\overrightarrow{v}_0}$ and the positive ${x}$-direction), we have: $$\begin{aligned} v_{0_x} &= v_0 \cos \theta_0 \\[1.5em] v_{0_y} &= v_0 \sin \theta_0. \end{aligned}$$
• During its projectile motion:
  • the particle's position vector ${\overrightarrow{r}}$ and velocity ${\overrightarrow{v}}$ change continuously.
  • But, its acceleration vector ${\overrightarrow{a}}$ is constant and always directed vertically downward.
  • There is no horizontal acceleration.
• In projectile motion, the horizontal motion and the vertical motion are independent of one another; that is, neither motion affects the other.

The Horizontal Motion

• There is no acceleration in the horizontal direction, so the horizontal component ${v_x}$ of the projectile's velocity remains unchanged from its initial value ${v_{0_x}}$ throughout the motion.
• At any time ${t,}$ the projectile's horizontal displacement ${x_t - x_0}$ (where ${x_t}$ is the projectil's position at time ${t}$) from an initial position ${x_0}$ is given by: $$x_t - x_0 = v_{0_x} t.$$
• Since ${v_{0_x} = v_0 \cos \theta_0,}$ we have: $$x_t - x_0 = (v_0 \cos \theta_0) t.$$

The Vertical Motion

• The vertical motion of a projectile is free-fall.
• This means that the acceleration is constant.
• Thus, we may use the kinematic equations, with ${a = g:}$ $$y_t - y_0 = v_{0_y} t - \dfrac{1}{2}gt^2,$$ wherer ${y_t}$ is the projectile's position at time ${t,}$ ${y_0}$ is its initial position, ${v_{0_y}}$ is the ${y}$-component of ${v_0,}$ and ${g}$ is the acceleration due to gravity.
• Since ${v_{0_y} = v_0 \sin \theta_0,}$ we have: $$y - y_0 = (v_0 \sin \theta_0)t - \dfrac{1}{2}gt^2.$$
• Replacing ${a = g}$ with the other kinematic equations, we have: $$\begin{aligned} v_y &= v_0 \sin \theta_0 - gt \\[1.5em] v_y^2 &= (v_0 \sin \theta_0)^2 - 2g(y_t - y_0). \end{aligned}$$

The Equation of the Path

• The equation of the projectile's path (its trajectory) is given by: $$y = (\tan \theta_0)x - \dfrac{gx^2}{2(v_0 \cos \theta_0)^2}.$$
• For simplicity, we let ${x_0}$ and ${y_0 = 0.}$
• Because ${g,}$ ${\theta_0,}$ and ${v_0}$ are constants, the trajectory equation is of the form ${y = ax + bx^2,}$ where ${a}$ and ${b}$ are constants. Thus, the projectile's trajectory is parabolic.

The Horizontal Range

• The horizontal range of the projectile, denoted ${R,}$ is the horizontal distance the projectile traveled when it returns to its initial height.
• Where ${x_f - x_0 = R}$ and ${y_f - y_0 = 0,}$ (where ${x_f}$ is the final ${x}$-position and ${y_f}$ is the final ${y}$-position) we have: $$R = (v_0 \cos \theta_0) t$$ and $$0 = (v_0 \sin \theta_0) t - \dfrac{1}{2} g t^2.$$ If we eliminate ${t}$ in both equations, we get: $$R = \dfrac{2v_0^2}{g} \sin \theta_0 \cos \theta_0.$$ Using the identity ${2 \theta_0 = 2 \sin \theta_0 \cos \theta_0,}$ we get: $$R = \dfrac{v_0^2}{g} \sin 2 \theta_0.$$
• This equation does not give the horizontal distance travelled by a projectile when the final height is not the launch height.
• Notice: The ${R}$ is at its maximum when ${\sin 2 \theta_0 = 1.}$ This corresponds to ${2 \theta_0 = 90 \degree}$ or ${\theta_0 = 45 \degree.}$
  • But, when the launch and landing heights differ, a launch angle of ${45 \degree}$ does not yield the maximum horizontal distance.

The Effects of Air

• In the preceding discussion, we assumed that the effects of air are negligible.
• In reality, air resistance can have a significant impact on the projectile motion: Air resistance reduces both height and range.