Physics Mechanics 100% Free Open Access
Chapter 3 • Theory & Derivations

Kinematics and Particle Dynamics

Inertial and non-inertial reference frames, Galilean relativity, fictitious forces, Frenet-Serret tangential and normal acceleration, projectile motion, circular dynamics, and dry friction.

§3.1 Frames of Reference, Galilean Invariance, and Non-Inertial Fictitious Forces

Kinematics describes the geometry of motion, which is fundamentally relative to an observer's choice of reference frame.

1. Inertial Frames and the Principle of Galilean Relativity

An **inertial frame** is a reference frame in which Newton's first law holds: an isolated body free from external forces moves with constant velocity in a straight line. If $S$ is an inertial frame and $S'$ moves relative to $S$ with constant translational velocity $\mathbf{V}_0$: $$\mathbf{r}'(t) = \mathbf{r}(t) - \mathbf{V}_0 t, \quad t' = t$$ Differentiating with respect to time: $$\mathbf{v}' = \mathbf{v} - \mathbf{V}_0, \quad \mathbf{a}' = \frac{d\mathbf{v}'}{dt} = \frac{d\mathbf{v}}{dt} = \mathbf{a}$$ Because acceleration is identical in all inertial frames, Newton's second law $\mathbf{F} = m\mathbf{a}$ retains the exact same mathematical form in all inertial frames. This is the **Principle of Galilean Relativity**.

2. Linearly Accelerating Reference Frames

If frame $S'$ has translational acceleration $\mathbf{A}_0(t)$ relative to inertial frame $S$: $$\mathbf{a}' = \mathbf{a} - \mathbf{A}_0$$ Multiplying by particle mass $m$: $$m \mathbf{a}' = m \mathbf{a} - m \mathbf{A}_0 = \mathbf{F}_{\text{real}} + \mathbf{F}_{\text{fictitious}}$$ where $\mathbf{F}_{\text{fictitious}} = -m\mathbf{A}_0$ is an inertial (pseudo) force that must be added to preserve Newton's second law in the accelerating frame.

3. Rotating Reference Frames: Centrifugal and Coriolis Forces

If frame $S'$ rotates with angular velocity $\boldsymbol{\omega}$ about an axis passing through the origin of inertial frame $S$: The time derivative of any vector $\mathbf{Q}$ relates by the operator identity: $$\left(\frac{d\mathbf{Q}}{dt}\right)_S = \left(\frac{d\mathbf{Q}}{dt}\right)_{S'} + \boldsymbol{\omega} \times \mathbf{Q}$$ Applying this twice to the position vector $\mathbf{r}$ yields the exact transformation of accelerations: $$\mathbf{a}_S = \mathbf{a}_{S'} + 2(\boldsymbol{\omega} \times \mathbf{v}_{S'}) + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) + \dot{\boldsymbol{\omega}} \times \mathbf{r}$$ The effective equation of motion in the rotating frame is: $$m \mathbf{a}_{S'} = \mathbf{F}_{\text{real}} + \mathbf{F}_{\text{coriolis}} + \mathbf{F}_{\text{centrifugal}} + \mathbf{F}_{\text{euler}}$$ where:
  • Coriolis Force: $\mathbf{F}_{\text{coriolis}} = -2m(\boldsymbol{\omega} \times \mathbf{v}_{S'})$. Acts perpendicular to velocity; deflects winds rightward in the Northern Hemisphere (cyclones).
  • Centrifugal Force: $\mathbf{F}_{\text{centrifugal}} = -m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) = m\omega^2 \mathbf{r}_\perp$. Acts radially outward from the rotation axis.
  • Euler Force: $\mathbf{F}_{\text{euler}} = -m(\dot{\boldsymbol{\omega}} \times \mathbf{r})$. Arises only when angular acceleration is non-zero.

§3.2 Frenet-Serret Coordinates: Tangential and Normal Acceleration

For general curvilinear motion along an arbitrary planar trajectory, intrinsic coordinates based on arc length $s(t)$ separate speed changes from directional changes.

1. Intrinsic Unit Basis Vectors

Let $\hat{\mathbf{t}}$ be the unit tangent vector pointing along velocity $\mathbf{v}$, and let $\hat{\mathbf{n}}$ be the principal unit normal vector pointing toward the local center of curvature: $$\mathbf{v} = v \hat{\mathbf{t}}, \quad v = \frac{ds}{dt} = \dot{s}$$ Differentiating with respect to time: $$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{dv}{dt} \hat{\mathbf{t}} + v \frac{d\hat{\mathbf{t}}}{dt}$$ Using the geometric Frenet-Serret relation $\frac{d\hat{\mathbf{t}}}{ds} = \frac{1}{\rho} \hat{\mathbf{n}}$, where $\rho$ is the local radius of curvature: $$\frac{d\hat{\mathbf{t}}}{dt} = \frac{d\hat{\mathbf{t}}}{ds}\frac{ds}{dt} = \left(\frac{1}{\rho}\hat{\mathbf{n}}\right) v = \frac{v}{\rho} \hat{\mathbf{n}}$$ Substituting into the total acceleration: $$\mathbf{a} = a_t \hat{\mathbf{t}} + a_n \hat{\mathbf{n}} = \left( \frac{dv}{dt} \right) \hat{\mathbf{t}} + \left( \frac{v^2}{\rho} \right) \hat{\mathbf{n}}$$

2. Dynamical Interpretation

  • Tangential Component ($a_t = \dot{v} = \ddot{s}$): Governs changes in the *magnitude* of velocity (speed). Caused by net forces parallel to trajectory: $F_t = m a_t$.
  • Normal (Centripetal) Component ($a_n = \frac{v^2}{\rho}$): Governs changes in the *direction* of velocity. Caused by net lateral forces perpendicular to trajectory: $F_n = m \frac{v^2}{\rho}$.
  • Total Acceleration Magnitude: $$a = \sqrt{a_t^2 + a_n^2} = \sqrt{\left(\frac{dv}{dt}\right)^2 + \left(\frac{v^2}{\rho}\right)^2}$$
  • Radius of Curvature Formula: For a path expressed as $y = f(x)$: $$\rho(x) = \frac{[1 + (y')^2]^{3/2}}{|y''|}$$

§3.3 Projectile Motion in Two Dimensions (Flat and Inclined Planes)

Two-dimensional ballistic motion under constant gravitational acceleration provides the classical prototype of particle kinematics.

1. Kinematic Equations on Flat Terrain

For a particle launched from $(0, 0)$ with initial speed $v_0$ at angle $\theta_0$ to horizontal: $$a_x = 0, \quad a_y = -g$$ Integrating: $$v_x(t) = v_0 \cos \theta_0, \quad v_y(t) = v_0 \sin \theta_0 - gt$$ $$x(t) = (v_0 \cos \theta_0) t, \quad y(t) = (v_0 \sin \theta_0) t - \frac{1}{2}gt^2$$ Eliminating $t$ yields the parabolic trajectory: $$y(x) = x \tan \theta_0 - \frac{g}{2 v_0^2 \cos^2 \theta_0} x^2$$ Trajectory metrics:
  • Time of Flight ($T$): $T = \frac{2 v_0 \sin \theta_0}{g}$.
  • Maximum Height ($H$): $H = \frac{v_0^2 \sin^2 \theta_0}{2g}$.
  • Horizontal Range ($R$): $R = \frac{v_0^2 \sin 2\theta_0}{g}$, maximized at $\theta_0 = 45^\circ$.

2. Projectile on an Inclined Plane

When a projectile is launched at angle $\alpha$ relative to horizontal onto a hill inclined at angle $\beta$ ($alpha > \beta$): Rotating axes so that $x'$ is along the slope and $y'$ is normal to the slope: $$a_{x'} = -g \sin \beta, \quad a_{y'} = -g \cos \beta$$ $$v_{0x'} = v_0 \cos(\alpha - \beta), \quad v_{0y'} = v_0 \sin(\alpha - \beta)$$ Solving $y'(T) = 0$ gives time of flight: $$T = \frac{2 v_0 \sin(\alpha - \beta)}{g \cos \beta}$$ Range along the inclined slope: $$R_{\text{incline}} = x'(T) = \frac{2 v_0^2 \cos \alpha \sin(\alpha - \beta)}{g \cos^2 \beta}$$ Maximized when the launch angle bisects the remaining angle: $\alpha_{\text{opt}} = \frac{\pi}{4} + \frac{\beta}{2}$.

§3.4 Uniform and Non-Uniform Circular Dynamics

Circular motion occurs when a particle moves along a circular path of fixed radius $R$.

1. Uniform Circular Motion

If speed $v$ is constant: $$a_t = \frac{dv}{dt} = 0, \quad a_n = \frac{v^2}{R} = \omega^2 R$$ The acceleration is purely radial (centripetal), pointing toward the center. By Newton's second law: $$F_c = m \frac{v^2}{R} = m \omega^2 R$$

2. Motion in a Vertical Circle

When a particle of mass $m$ is attached to a string of length $R$ and swung in a vertical plane under gravity: By energy conservation between bottom (speed $v_0$) and angle $\theta$ from bottom: $$\frac{1}{2}m v_0^2 = \frac{1}{2}m v^2 + mgR(1 - \cos \theta) \implies v^2 = v_0^2 - 2gR(1 - \cos \theta)$$ Newton's second law along the radial direction gives string tension $T$: $$T - mg \cos \theta = \frac{mv^2}{R} \implies T = mg \cos \theta + \frac{m}{R}[v_0^2 - 2gR(1 - \cos \theta)]$$ $$T(\theta) = \frac{m v_0^2}{R} - mg(2 - 3\cos \theta)$$ Critical Conditions:
  • At top of loop ($\theta = \pi$): $T_{\text{top}} = \frac{m v_0^2}{R} - 5mg$. For the string not to go slack ($T_{\text{top}} \ge 0$): $$v_{\text{top}} \ge \sqrt{gR}, \quad v_0 \ge \sqrt{5gR}$$
  • Difference in tension between bottom and top is always independent of launch speed: $$T_{\text{bottom}} - T_{\text{top}} = 6mg$$

§3.5 Newton’s Laws of Motion and Coulomb-Amontons Dry Friction

Newtonian particle dynamics relates external forces to resulting particle trajectories.

1. Newton’s Three Laws

  1. First Law (Inertia): A body remains in rest or uniform straight-line motion unless acted upon by a net force: $\sum \mathbf{F} = \mathbf{0} \implies \mathbf{v} = \text{const}$.
  2. Second Law (Dynamical Evolution): Net force equals time rate of change of linear momentum $\mathbf{p} = m\mathbf{v}$: $$\mathbf{F} = \frac{d\mathbf{p}}{dt} = m \frac{d\mathbf{v}}{dt} + \mathbf{v} \frac{dm}{dt}$$ For constant mass: $\mathbf{F} = m \mathbf{a}$.
  3. Third Law (Reciprocity): Pairwise mutual interaction forces between bodies $A$ and $B$ are collinear, equal in magnitude, and opposite in direction: $\mathbf{F}_{AB} = -\mathbf{F}_{BA}$.

2. Laws of Dry Friction (Coulomb-Amontons)

Contact forces parallel to surfaces arise from microscopic roughness and molecular bonding:
  • Static Friction ($f_s$): Self-adjusting force opposing applied force up to a maximum threshold: $$f_s \le \mu_s N$$
  • Kinetic Friction ($f_k$): Dynamic resistive force during relative sliding: $$f_k = \mu_k N, \quad \text{with } \mu_k < \mu_s$$
  • Angle of Friction ($\lambda$) and Angle of Repose ($\theta_r$): The maximum static friction angle satisfies $\tan \lambda = \mu_s$. On an inclined plane, a block begins sliding under gravity when slope exceeds the angle of repose $\theta_r = \arctan(\mu_s)$.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Medium Example 3.1: Projectile with High Elevation on Inclined Plane

A projectile is launched with speed $v_0$ at angle $\alpha$ above a planar hillside inclined at angle $\beta$ to the horizontal ($alpha > \beta$). Derive an exact analytical expression for the range $R$ of the projectile measured along the inclined surface.

Step 1: Set up rotated coordinates along the inclined plane
$$\text{Let } x' \text{ be parallel to incline, } y' \text{ perpendicular to incline}$$ $$a_{x'} = -g \sin \beta, \quad a_{y'} = -g \cos \beta$$ $$v_{0x'} = v_0 \cos(\alpha - \beta), \quad v_{0y'} = v_0 \sin(\alpha - \beta)$$

Decomposing gravity into components parallel and normal to the slope simplifies the impact boundary condition.

Step 2: Solve for time of flight and range along incline
$$y'(T) = v_{0y'} T - \frac{1}{2} g \cos \beta \, T^2 = 0 \implies T = \frac{2 v_0 \sin(\alpha - \beta)}{g \cos \beta}$$ $$R = x'(T) = v_{0x'} T - \frac{1}{2} g \sin \beta \, T^2$$ $$R = \frac{2 v_0^2}{g \cos^2 \beta} \sin(\alpha - \beta) \cos \alpha$$

The maximum range along the inclined plane is achieved when $\alpha = \frac{\pi}{4} + \frac{\beta}{2}$.

Hard Example 3.2: Particle Sliding Off a Frictionless Spherical Dome

A particle of mass $m$ rests at the top of a smooth frictionless sphere of radius $R$. Given an infinitesimal nudge, it slides down under gravity. Determine the exact angle $\theta_0$ measured from the vertical at which the particle loses contact with the sphere.

Step 1: Apply energy conservation from the top
$$E = mgR = \frac{1}{2}m v^2 + mgR \cos \theta \implies v^2 = 2gR(1 - \cos \theta)$$

Gravitational potential energy converted to kinetic energy depends solely on vertical drop.

Step 2: Apply Newton's Second Law in radial direction
$$mg \cos \theta - N = m \frac{v^2}{R} \implies N = mg \cos \theta - \frac{m}{R}(2gR(1 - \cos \theta)) = mg(3 \cos \theta - 2)$$

The normal contact force $N$ diminishes as speed increases.

Step 3: Condition for loss of contact (N = 0)
$$N = 0 \implies 3 \cos \theta_0 - 2 = 0 \implies \cos \theta_0 = \frac{2}{3} \implies \theta_0 = \arccos\left(\frac{2}{3}\right) \approx 48.19^\circ$$

The particle leaves the surface at $\cos\theta_0 = 2/3$, independent of particle mass and dome radius.

Medium Example 3.3: Minimum Horizontal Pushing Force on an Inclined Plane

A block of mass $m$ rests on an incline of angle $\theta$ with coefficient of static friction $\mu_s$. Determine the minimum horizontal force $F_h$ required to prevent the block from slipping down the incline.

Step 1: Resolve forces along and perpendicular to the incline
$$\Sigma F_\perp = N - mg \cos \theta - F_h \sin \theta = 0 \implies N = mg \cos \theta + F_h \sin \theta$$ $$\Sigma F_\parallel = F_h \cos \theta + f_s - mg \sin \theta = 0 \implies f_s = mg \sin \theta - F_h \cos \theta$$

Static friction $f_s$ acts up the incline to prevent downward sliding.

Step 2: Apply static friction threshold condition f_s <= μ_s N
$$mg \sin \theta - F_h \cos \theta \le \mu_s (mg \cos \theta + F_h \sin \theta)$$ $$mg(\sin \theta - \mu_s \cos \theta) \le F_h (\cos \theta + \mu_s \sin \theta)$$ $$F_{h,\min} = mg \left( \frac{\sin \theta - \mu_s \cos \theta}{\cos \theta + \mu_s \sin \theta} \right) = mg \tan(\theta - \lambda)$$

Here $\lambda = \arctan(\mu_s)$ is the angle of friction.