Classical Mechanics Lecture 2: Newton's First and Second Laws

Review of Previous Concepts and Introduction to Dynamics

• Last time covered Newton's 3rd Law and a survey of various examples of forces.

• The focus of this lecture is to examine what happens when these forces are applied to physical objects.

Newton's First Law of Motion (Newton I)

• Statement of Newton's 1st Law: A body on which there is no net force (no unbalanced force) continues to move at constant velocity (that is, in a straight line at a constant speed).

• Scope and Applicability:

  • Newton's 1st Law does not always apply under all observation conditions.

  • It is strictly valid only when observing the object from an inertial reference frame.

• Reference Frame Definition:

  • A reference frame acts as an anchor for measuring physical quantities (position, velocity, acceleration).

  • It can be visualized as a set of coordinate axes nailed to a specific physical object.

• Inertial Frame Definition:

  • An inertial reference frame is a non-accelerating frame of reference.

  • If an observer is in an accelerating (non-inertial) reference frame, an unforced object will appear to "magically" accelerate relative to that observer, violating Newton's 1st Law.

• Statics vs. Dynamics:

  • Statics: The domain of mechanics where everything remains at rest ($\mathbf{v} = 0$), as opposed to dynamics.

  • In statics, Newton's 1st Law dictates that all forces acting on a body must be balanced ($\mathbf{F}_{net} = 0$).

  • Logical Caveat: The converse statement is false. Stating that forces are balanced ($\mathbf{F}_{net} = 0$) does not imply the object is at rest; it may be moving with a constant non-zero velocity.

Statics Example: Rope Wrapped Around a Cylindrical Pole

• Problem Setup:

  • A boat or hand pulls on a rope wrapped around a cylindrical post or pole.

  • The rope wraps around the pole through a total contact angle of $\alpha$ radians.

  • The coefficient of static friction between the rope and pole material is $\mu$.

  • Goal: Determine the maximum tension ratio or holding force $T$ achievable before the rope slips.

• Free-Body Diagram Analysis:

  • A free-body diagram isolates an object and explicitly labels all external forces acting on it (excluding forces exerted by the object on others).

  • Zoom in on an infinitesimal segment of the rope subtending a small angle $d\theta$.

  • Tension $T$ varies as a function of angular position $\theta$ ($T = T(\theta)$).

  • At position $\theta$, tension is $T(\theta)$; at position $\theta + d\theta$, tension is $T(\theta + d\theta)$.

• Radial (Normal) Direction Equilibrium:

  • In the inward radial direction, tension components from both ends balance the differential normal force $N\,d\theta$ exerted by the pole:     $N\,d\theta = 2 T(\theta) \sin\left(\frac{d\theta}{2}\right)$

  • Using the small-angle approximation $\sin(x) \approx x$ for small angles $x$ measured in radians:     $\sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2}$

  • Substituting this approximation into the radial balance:     $N\,d\theta = 2 T(\theta) \left(\frac{d\theta}{2}\right) = T(\theta)\,d\theta$

• Tangential Direction and Frictional Force Balance:

  • Frictional force $F\,d\theta$ acts along the surface, creating a difference in tension across the small chunk.

  • Maximum static friction force:     $F\,d\theta = \mu N\,d\theta = \mu T(\theta)\,d\theta$

  • Setting up the differential tension increment across segment $d\theta$:     $T(\theta + d\theta) = T(\theta) + \mu T(\theta)\,d\theta$     $T(\theta + d\theta) - T(\theta) = \mu T(\theta)\,d\theta$     $d T = \mu T(\theta)\,d\theta \implies \frac{d T}{d\theta} = \mu T$

• Differential Equation Solution:

  • Separating variables:     $\frac{d T}{T} = \mu\,d\theta$

  • Integrating both sides:     $\int \frac{d T}{T} = \int \mu\,d\theta$     $\ln(T) = \mu \theta + C$

  • Exponentiating with boundary condition $T(0) = T_0$ at $\theta = 0$ (integration constant $C = \ln(T_0)$):     $T(\theta) = T_0 e^{\mu \theta}$

  • Evaluated over total angle $\alpha$:     $T(\alpha) = T_0 e^{\mu \alpha}$

• Physical Insight and Positive Feedback:

  • The equation yields a positive exponential growth in holding tension.

  • Positive exponentials represent positive feedback processes that amplify rapidly:

  • Wrapping the rope around an additional turn increases the contact surface and normal force.

  • Greater normal force yields greater maximum friction ($F\,d\theta = \mu N\,d\theta$).

  • Greater friction permits a higher tension $T$, which in turn increases normal force further.

Conceptual Meaning and Formulation of Newton's Second Law (Newton II)

• Interconnection between Newton I and II:

  • Newton's 1st Law can be viewed as providing the criterion for defining an inertial reference frame.

  • Because Newton's 2nd Law is valid strictly within inertial frames, Newton I serves as "permission to use Newton's 2nd Law".

  • If an observer verifies that unforced objects move with constant velocity (Newton I holds), they are in an inertial frame and are permitted to use Newton II when unbalanced forces are applied.

• Newton's Second Law Definition:

  • Unbalanced forces acting on an object cause its motion to change (specifically, cause it to accelerate).

  • Forces do not cause velocity itself; forces cause changes in velocity (acceleration).

  • Vector equation:     $\mathbf{a} = \frac{\mathbf{F}_{net}}{m} \quad \text{or} \quad \mathbf{F}_{net} = m \mathbf{a}$

  • Component equations:     $a_x = \frac{F_{x,net}}{m}, \quad a_y = \frac{F_{y,net}}{m}, \quad a_z = \frac{F_{z,net}}{m}$

  • Net force $\mathbf{F}_{net}$ is the vector sum of all individual forces acting on mass $m$:     $\mathbf{F}_{net} = \sum \mathbf{F}$

  • Vector Addition Example: A body subject to a $2\,\text{N}$ force to the left and a $3\,\text{N}$ force to the right experiences a net force of $1\,\text{N}$ to the right.

• Predictive Power:

  • Newton's 2nd Law becomes predictive when combined with independent force laws provided by nature (e.g., Hooke's Law $\mathbf{F} = -k \mathbf{x}$).

Free Fall, Equivalence of Mass, and Projectile Motion

• Free Fall Analysis:

  • Define vertical direction as $z$, choosing "up" as the positive direction.

  • In ideal free fall without air resistance, horizontal net forces are zero ($F_{x,net} = 0$, $F_{y,net} = 0$).

  • Vertical net force consists solely of gravity:     $F_{z,net} = -m g$

  • Acceleration in vertical direction:     $a_z = \frac{F_{z,net}}{m} = \frac{-m g}{m} = -g$

  • Result: All objects in free fall accelerate downwards at the exact same rate $g$, completely independent of their mass $m$.

• Inertial Mass vs. Gravitational Mass:

  • In $a_z = \frac{-m_{grav} g}{m_{inertial}}$, two distinct physical concepts of mass appear:

  • Inertial Mass ($m_{inertial}$): The mass in Newton II ($\mathbf{F} = m \mathbf{a}$) that measures an object's resistance to acceleration/changes in motion.

  • Gravitational Mass ($m_{grav}$): The mass in gravitational force ($\mathbf{F}_g = m \mathbf{g}$) that determines how strongly an object responds to gravity.

  • Their strict equality ($m_{inertial} = m_{grav}$) causes exact cancellation of $m$.

  • Historical context:

  • Demonstrated dramatically on Apollo 15 where an astronaut dropped a hammer and a feather in the lunar vacuum, observing them land simultaneously.

  • To Isaac Newton, this cancellation was a curious coincidence.

  • To Albert Einstein, this equivalence was a foundational postulate underlying General Relativity (the Equivalence Principle).

• Projectile Motion Derivations:

  • Projectile motion is free fall with an initial horizontal velocity, producing a parabolic trajectory.

  • Kinematic notation: Time derivatives of position are represented with dots (e.g., $\dot{x} = \frac{d x}{d t}$, $\ddot{x} = \frac{d^2 x}{d t^2}$).

  • Acceleration components:     $\ddot{x} = 0, \quad \ddot{y} = 0, \quad \ddot{z} = -g$

  • Integrating once for velocity components:     $\dot{x} = v_{0x}, \quad \dot{y} = 0, \quad \dot{z} = v_{0z} - g t$

  • Integrating twice for position components:     $x(t) = x_0 + v_{0x} t$     $z(t) = z_0 + v_{0z} t - \frac{1}{2} g t^2$

The Ball and Monkey Problem

• Problem Description:

  • A ball launcher at origin $(0, 0)$ aims directly at a monkey hanging from a tree at horizontal distance $d$ and initial height $h$.

  • Aim angle $\alpha$ satisfies $\tan(\alpha) = \frac{h}{d}$.

  • Ball is launched with initial speed $v_0$ at angle $\alpha$. At the exact instant of launch, the monkey drops from rest under gravity.

• Trajectory Equations:

  • Ball Trajectory:

  • Horizontal position: $x_{ball}(t) = (v_0 \cos(\alpha)) t$

  • Vertical position: $z_{ball}(t) = (v_0 \sin(\alpha)) t - \frac{1}{2} g t^2$

  • Monkey Trajectory:

  • Horizontal position: $x_{monkey}(t) = d$

  • Vertical position: $z_{monkey}(t) = h - \frac{1}{2} g t^2$

• Proof of Collision:

  • Determine time $t_{flight}$ when the ball reaches horizontal position $x = d$:     $d = (v_0 \cos(\alpha)) t_{flight} \implies t_{flight} = \frac{d}{v_0 \cos(\alpha)}$

  • Substitute $t_{flight}$ into the vertical position of the ball:     $z_{ball}(t_{flight}) = (v_0 \sin(\alpha)) \left(\frac{d}{v_0 \cos(\alpha)}\right) - \frac{1}{2} g t_{flight}^2$     $z_{ball}(t_{flight}) = d \tan(\alpha) - \frac{1}{2} g t_{flight}^2$

  • Substitute geometry condition $h = d \tan(\alpha)$ into the ball's position:     $z_{ball}(t_{flight}) = h - \frac{1}{2} g t_{flight}^2$

  • Compare with monkey's vertical position at $t_{flight}$:     $z_{monkey}(t_{flight}) = h - \frac{1}{2} g t_{flight}^2$

  • Conclusion: $z_{ball}(t_{flight}) = z_{monkey}(t_{flight})$. The ball will always hit the monkey, regardless of launch speed $v_0$ or gravity $g$.

Constrained Motion: Block on an Accelerated Incline

• Unconstrained vs. Constrained Motion:

  • Free fall and simple projectile motion are unconstrained problems.

  • Constrained motion occurs when objects are forced to move along specific paths or surfaces (e.g., roller coaster, block on an inclined plane).

• Problem Setup:

  • An inclined wedge with an angle of $45^\circ$ ($\sin(45^\circ) = \cos(45^\circ) = \frac{1}{\sqrt{2}}$) has height h$.\n - A block of mass m sits on the incline.\n - The wedge is accelerated horizontally to the left at a constant rate A$.

  • Position coordinates: Horizontal $x$ and vertical $y$ with appropriate sign conventions.

• Equations of Motion:

  • Horizontal equation of motion (Normal force component):     $m \ddot{x} = N \cos(45^\circ) = \frac{N}{\sqrt{2}} \quad \text{--- (1)}$

  • Vertical equation of motion (Normal force and gravity):     $\frac{N}{\sqrt{2}} - m g = m \ddot{y} \quad \text{--- (2)}$

• Constraint Analysis and Differentiation:

  • Geometric constraint along the $45^\circ$ incline profile:     $x = h - y$

  • Differentiating the constraint equation twice with respect to time:     $\ddot{x} = -\ddot{y}$

  • Incorporating external horizontal acceleration $A$ of the wedge:     $\ddot{x} = -\ddot{y} + A$

• Solving the Coupled System:

  • Substitute $\ddot{x} = A - \ddot{y}$ into Equation (1):     $\frac{N}{\sqrt{2}} = m (A - \ddot{y})$

  • Substitute this expression for $\frac{N}{\sqrt{2}}$ into Equation (2):     $m (A - \ddot{y}) - m g = m \ddot{y}$     $m A - m \ddot{y} - m g = m \ddot{y}$     $m A - m g = 2 m \ddot{y}$

  • Solving for vertical acceleration $\ddot{y}$:     $\ddot{y} = \frac{A - g}{2}$

  • Substituting $\ddot{y}$ back to solve for horizontal acceleration $\ddot{x}$:     $\ddot{x} = A - \frac{A - g}{2} = \frac{A + g}{2}$

  • Solving for normal force $N$:     $N = \sqrt{2} m \ddot{x} = \frac{m (A + g)}{\sqrt{2}}$

  • Physical Interpretation: The vertical acceleration $\ddot{y}$ can be positive, negative, or zero depending on the relative values of wedge acceleration $A$ and gravitational acceleration g$.\n\n\n# Variable Mass Systems and The Rocket Equation\n\n- **Generalized Form of Newton's Second Law:**\n - The expression \mathbf{F} = m \mathbf{a}$is valid only for systems with constant mass$m$.

  • Defining linear momentum:     $\mathbf{p} = m \mathbf{v}$

  • The complete, general form of Newton's 2nd Law states that net force equals the time rate of change of linear momentum:     $\mathbf{F}_{net} = \frac{d \mathbf{p}}{d t}$

  • When mass $m$ is constant, $\frac{d \mathbf{p}}{d t} = \frac{d(m \mathbf{v})}{d t} = m \frac{d \mathbf{v}}{d t} = m \mathbf{a}$.

• Rocket Propulsion Analysis:

  • Rockets represent variable mass systems because they continually eject propellant out the rear to propel themselves forward.

  • Let system momentum comprise Rocket ($r$) of mass $M_r$ and velocity $\mathbf{v}_r$, and ejected Fuel ($f$) of mass $m_f$ and velocity \mathbf{v}_f$.\n - Total system force equation:\n    \mathbf{F}{net} = \frac{d}{d t}(M_r \mathbf{v}_r + m_f \mathbf{v}_f)\n    \mathbf{F}{net} = M_r \frac{d \mathbf{v}_r}{d t} + \mathbf{v}_r \frac{d M_r}{d t} + m_f \frac{d \mathbf{v}_f}{d t} + \mathbf{v}_f \frac{d m_f}{d t}\n\n\n\n- **Simplification and Mass Conservation:**\n - After ejection, discarded fuel coasts freely with no additional forces, so \frac{d \mathbf{v}_f}{d t} = 0$.

  • By conservation of mass, fuel mass gained equals rocket mass lost:     $\frac{d m_f}{d t} = -\frac{d M_r}{d t}$

  • Substituting these relations simplifies net force to:     $\mathbf{F}_{net} = M_r \frac{d \mathbf{v}_r}{d t} + \mathbf{v}_r \frac{d M_r}{d t} - \mathbf{v}_f \frac{d M_r}{d t}$     $\mathbf{F}_{net} = M_r \frac{d \mathbf{v}_r}{d t} + (\mathbf{v}_r - \mathbf{v}_f) \frac{d M_r}{d t}$

• Formulation with Exhaust Relative Velocity:

  • Define exhaust velocity $\mathbf{u}$ relative to rocket:     $\mathbf{u} = \mathbf{v}_f - \mathbf{v}_r \implies \mathbf{v}_r - \mathbf{v}_f = -\mathbf{u}$

  • Relative velocity $\mathbf{u}$ is chosen because engine burn rate and exhaust speed are controlled relative to the rocket itself, not relative to an arbitrary external observer.

  • Dropping rocket subscript $r$ for brevity ($M_r \to M$, $\mathbf{v}_r \to \mathbf{v}$):     $\mathbf{F}_{net} = M \frac{d \mathbf{v}}{d t} - \mathbf{u} \frac{d M}{d t}$

  • This expression is known as The Rocket Equation.

• Consistency Check (Deep Space Firing):

  • Consider a rocket in deep space far from gravitational fields where net external force is zero ($\mathbf{F}_{net} = 0$):     $M \frac{d \mathbf{v}}{d t} = \mathbf{u} \frac{d M}{d t} \implies \frac{d \mathbf{v}}{d t} = \frac{\mathbf{u}}{M} \frac{d M}{d t}$

  • Sign verification:

  • Rocket loses mass during burn, so \frac{d M}{d t} < 0$.\n - Propellant is expelled out the back (opposite to motion), so relative exhaust velocity \mathbf{u} < 0$.

  • The product of two negative quantities yields a positive acceleration $\frac{d \mathbf{v}}{d t} > 0$, confirming forward acceleration as expected physically.