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
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
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)
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
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
1. Newton’s Three Laws
- First Law (Inertia): A body remains in rest or uniform straight-line motion unless acted upon by a net force: $\sum \mathbf{F} = \mathbf{0} \implies \mathbf{v} = \text{const}$.
- 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: $\mathbf{F} = m \mathbf{a}$.
- 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 proofsA 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.
Decomposing gravity into components parallel and normal to the slope simplifies the impact boundary condition.
The maximum range along the inclined plane is achieved when $\alpha = \frac{\pi}{4} + \frac{\beta}{2}$.
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.
Gravitational potential energy converted to kinetic energy depends solely on vertical drop.
The normal contact force $N$ diminishes as speed increases.
The particle leaves the surface at $\cos\theta_0 = 2/3$, independent of particle mass and dome radius.
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.
Static friction $f_s$ acts up the incline to prevent downward sliding.
Here $\lambda = \arctan(\mu_s)$ is the angle of friction.