7.4 Kinematics, Kinetics of Particles & Rigid Body Dynamics

6.4 Kinematics, Kinetics of Particles & Rigid Body Dynamics

Dynamic systems analysis is at the core of heavy mining operations—from skip hoists hoisting extracted coal at high acceleration out of vertical shafts, to haul trucks negotiating unpaved curves, to impact dynamics in crusher mills. This section synthesizes kinematic equations, Newton-Euler kinetics, energy methods, impact restitution, and rigid body plane motion for the CIL MT examination.

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1. Kinematics of Particles: Rectilinear Motion

Kinematics analyzes geometry of motion (displacement, velocity, acceleration, time) without considering the forces causing the motion.

Constant (Uniform) Acceleration Formulations

When acceleration $a = \text{constant}$:

$$v = u + a t$$ $$s = u t + \frac{1}{2} a t^2$$ $$v^2 = u^2 + 2 a s$$ $$s_n = u + \frac{a}{2}(2n - 1) \quad \text{(Distance traversed during the } n^{\text{th}} \text{ second)}$$

Variable Acceleration Formulations

When acceleration varies as a function of time $a(t)$, displacement $a(s)$, or velocity $a(v)$:

$$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = v \frac{dv}{ds}$$

Integration Paths for Variable Acceleration:
  1. If a = f(t) ───► v = ∫ f(t) dt ──────────────► s = ∫ v(t) dt
  2. If a = f(v) ───► t = ∫ dv / f(v) ────────────► s = ∫ v dv / f(v)
  3. If a = f(s) ───► ∫ v dv = ∫ f(s) ds ─────────► (1/2)(v² - u²) = ∫ f(s) ds

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2. Projectile Motion in a Gravitational Field

A projectile launched with initial velocity $u$ at elevation angle $\theta$ to the horizontal under constant downward gravity $g$ (neglecting aerodynamic drag):

                         Trajectory of a Projectile
               ▲ y
               │                  Peak (H_max)
               │                 . - - - .
         u     │             .               .
          ▲    │          .                     .
           \   │        .                         .
            \ θ│      .                             .
               └─────┴───────────────────────────────┴────► x
               O                                     Range (R)

Governing Mathematical Expressions

• Horizontal velocity (constant): $v_x = u \cos\theta$
• Vertical velocity (variable): $v_y = u \sin\theta - g t$
• Cartesian Trajectory Equation (Parabolic): $$y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta} = x \tan\theta \\left[1 - \frac{x}{R}\\right]$$
• Time to Reach Maximum Height ($t_h$): $t_h = \frac{u \sin\theta}{g}$
• Total Time of Flight ($T$): $T = 2 t_h = \frac{2 u \sin\theta}{g}$
• Maximum Vertical Height ($H_{\text{max}}$): $$H_{\text{max}} = \frac{u^2 \sin^2\theta}{2g}$$
• Horizontal Range ($R$): $$R = \frac{u^2 \sin(2\theta)}{g}$$
• Maximum Horizontal Range ($R_{\text{max}}$): Occurs at launch angle $\theta = 45^{\circ}$: $$R_{\text{max}} = \frac{u^2}{g}$$
• Complementary Angles Property: Launch angles $\theta$ and $(90^{\circ} - \theta)$ produce the exact same horizontal range $R$ for a given launch velocity $u$.

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3. Curvilinear Motion: Normal and Tangential Acceleration

When a particle traces a curved planar path, its velocity vector $\vec{v}$ is always tangent to the trajectory, while its acceleration decomposes into two orthogonal components:

                        Tangential Acceleration (a_t = dv/dt)
                                    ▲
                                    │   / Velocity Vector v
                                    │  /
                                    │ /
                     Particle (P) ──┼/────────► Normal Centripetal Accel (a_n = v²/ρ)
                                    │          (Directed toward Center of Curvature)
                                    │
                                    ▼

Acceleration Components

1. Tangential Acceleration ($a_t$): Changes the magnitude of velocity (speed): $$a_t = \frac{dv}{dt} = \dot{v} = v \frac{dv}{ds}$$
2. Normal (Centripetal) Acceleration ($a_n$): Changes the direction of velocity, directed perpendicular to the path toward the center of curvature: $$a_n = \frac{v^2}{\rho}$$ where $\rho$ is the local radius of curvature of the trajectory curve $y = f(x)$: $$\rho = \frac{\\left[1 + \\left(\frac{dy}{dx}\\right)^2\\right]^{3/2}}{\\left|\frac{d^2y}{dx^2}\\right|}$$
3. Total Resultant Acceleration ($a$): $$a = \sqrt{a_t^2 + a_n^2}, \qquad \beta = \tan^{-1}\\left(\frac{a_n}{a_t}\\right)$$

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4. Kinetics of Particles: Newton & D'Alembert's Principle

Newton's Second Law

$$\sum \vec{F} = m \vec{a}$$

D'Alembert's Principle (Dynamic Equilibrium)

D'Alembert's Principle states that any moving particle or body can be analyzed using static equilibrium equations by adding a fictitious reversed effective force (inertia force) $\vec{F}_{\text{inertia}} = -m \vec{a}$:

$$\sum \vec{F} + (-m \vec{a}) = \vec{0}$$

           Elevator Accelerating Upward (+a)       Elevator Accelerating Downward (-a)
                     ▲ Cable Tension T                     ▲ Cable Tension T
                     │                                     │
                ┌────┴────┐                           ┌────┴────┐
                │ Mass m  │                           │ Mass m  │
                └────┬────┘                           └────┬────┘
                     │                                     │
                     ▼ Weight W = mg                       ▼ Weight W = mg
                     ▼ Inertia Force = m*a                 ▲ Inertia Force = m*a

              T = m(g + a) (Feels heavier)          T = m(g - a) (Feels lighter)

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5. Work-Energy & Impulse-Momentum Principles

Work-Energy Theorem for Particles

The net work done on a particle by all external and frictional forces equals the net change in its kinetic energy:

$$W_{1 \to 2} = \Delta KE = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2$$

• Work done by constant force: $W = \vec{F} \cdot \vec{s} = F s \cos\theta$
• Work done by linear spring ($F = k x$): $W_{\text{spring}} = -\frac{1}{2} k (x_2^2 - x_1^2)$
• Work done by gravity: $W_{\text{gravity}} = -mg \Delta h = -mg (y_2 - y_1)$

Principle of Conservation of Mechanical Energy

For conservative force fields (gravity, ideal elastic springs):

$$KE_1 + PE_1 = KE_2 + PE_2 \implies \frac{1}{2} m v_1^2 + m g h_1 + \frac{1}{2} k x_1^2 = \frac{1}{2} m v_2^2 + m g h_2 + \frac{1}{2} k x_2^2$$

Impulse-Momentum Principle

The linear impulse imparted by resultant force $\vec{F}$ over time interval $\Delta t = t_2 - t_1$ equals the change in linear momentum:

$$\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \Delta \vec{p} = m \vec{v}_2 - m \vec{v}_1$$

• Conservation of Linear Momentum: If the net external force along a specific axis is zero ($\Sigma F_{\text{ext}, x} = 0$), linear momentum along that axis is conserved: $$m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$$

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6. Collision & Impact Mechanics

Impact is a sudden collision between two bodies occurring over an infinitesimal duration with exceptionally high contact forces.

                     Direct Central Impact along Line of Action
              Before Impact:      (m1, u1) ───►     ◄─── (m2, u2)  (u1 > u2)
              Line of Impact: ───────────────────────────────────────
              After Impact:       (m1, v1) ───►          (m2, v2) ───►

Coefficient of Restitution ($e$)

Defined as the ratio of relative velocity of separation to relative velocity of approach along the line of impact:

$$\ne = \frac{v_2 - v_1}{u_1 - u_2}$$

Loss of Kinetic Energy in Direct Central Impact

$$\Delta E_{\text{loss}} = KE_{\text{initial}} - KE_{\text{final}} = \frac{m_1 m_2}{2(m_1 + m_2)} (1 - e^2) (u_1 - u_2)^2$$

Rebound Dynamics of a Dropped Particle

For a ball dropped from initial height $h_0$ onto a massive rigid horizontal floor:

• Velocity after $n$ bounces: $v_n = e^n \sqrt{2 g h_0}$
• Height attained after $n$ bounces: $h_n = e^{2n} h_0$
• Total distance traveled before coming to rest: $S_{\text{total}} = h_0 \\left(\frac{1 + e^2}{1 - e^2}\\right)$
• Total time elapsed before coming to rest: $T_{\text{total}} = \sqrt{\frac{2 h_0}{g}} \\left(\frac{1 + e}{1 - e}\\right)$

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7. Planar Rigid Body Dynamics & Pure Rolling

Planar rigid body motion combines translation of the center of mass $G$ and rotation about $G$ (Chasles' Theorem).

Kinetic Energy of a Planar Rigid Body

$$KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} m v_G^2 + \frac{1}{2} I_G \omega^2$$

For pure rotation about a fixed pivot $O$:

$$KE = \frac{1}{2} I_O \omega^2 = \frac{1}{2} (I_G + m d^2) \omega^2$$

Pure Rolling of a Body Down an Incline

For a symmetrical circular body of mass $m$, radius $R$, and centroidal radius of gyration $k$ rolling without slipping down an incline $\theta$:

                                  Rolling Cylinder Down Incline
                                         /\
                                        /  \
                                       /  O ──► a = (g*sin θ) / (1 + k²/R²)
                                      /   / \
                                     /   /   \
                                    /   /     \
                                   /   /       \
                                  /   /  θ      \
                                 /───┴───────────\

• Linear Acceleration down the incline: $$a = \frac{g \sin\theta}{1 + \frac{I_G}{m R^2}} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}$$

Acceleration Hierarchy for Standard Rolling Shapes

• Minimum Friction Coefficient for Pure Rolling (No Slipping): $$\mu_{\text{min}} = \frac{\tan\theta}{1 + \frac{m R^2}{I_G}} = \frac{\tan\theta}{1 + \frac{R^2}{k^2}}$$

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8. Step-by-Step Worked Engineering Calculations

Worked Example 6.4.1: Curvilinear Acceleration of a Haul Truck

Problem: A heavy mining haul truck travels along a curved haul road section defined by $y = \frac{x^2}{800}$ (where $x$ and $y$ are in meters). When the truck is at $x = 200\text{ m}$, its speed is $v = 20\text{ m/s}$ and its speed is increasing at a rate of $2.5\text{ m/s}^2$. Determine the total magnitude of acceleration acting on the truck.

Step-by-step Solution:

1. Tangential Acceleration:

   $$a_t = \frac{dv}{dt} = 2.5\text{ m/s}^2$$

2. Radius of Curvature $\rho$ at $x = 200\text{ m}$:

   • First derivative: $\frac{dy}{dx} = \frac{2x}{800} = \frac{x}{400}$. At $x = 200\text{ m} \implies \frac{dy}{dx} = \frac{200}{400} = 0.5$.
   • Second derivative: $\frac{d^2y}{dx^2} = \frac{1}{400} = 0.0025\text{ m}^{-1}$.
   • Formula for $\rho$: $$\rho = \frac{[1 + (dy/dx)^2]^{3/2}}{\lvert d^2y/dx^2 \rvert} = \frac{[1 + (0.5)^2]^{3/2}}{0.0025} = \frac{[1.25]^{1.5}}{0.0025} = \frac{1.39754}{0.0025} \approx 559.02\text{ m}$$

3. Normal Centripetal Acceleration:

   $$a_n = \frac{v^2}{\rho} = \frac{20^2}{559.02} = \frac{400}{559.02} \approx 0.7155\text{ m/s}^2$$

4. Total Resultant Acceleration:

   $$a = \sqrt{a_t^2 + a_n^2} = \sqrt{(2.5)^2 + (0.7155)^2} = \sqrt{6.25 + 0.5120} = \sqrt{6.7620} \approx 2.60\text{ m/s}^2$$

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Worked Example 6.4.2: Inelastic Collision and Kinetic Energy Dissipation

Problem: A $2000\text{ kg}$ loaded coal tub traveling at $u_1 = 6\text{ m/s}$ along a mine rail track collides directly with an empty stationary tub of mass $1000\text{ kg}$ ($u_2 = 0$). If the coefficient of restitution is $e = 0.5$, calculate (a) the post-collision velocities $v_1$ and $v_2$, and (b) the kinetic energy lost during the impact.

Step-by-step Solution:

1. Apply Conservation of Linear Momentum:

   $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$(2000 \times 6) + (1000 \times 0) = 2000 v_1 + 1000 v_2$$ $$12000 = 2000 v_1 + 1000 v_2 \implies 2 v_1 + v_2 = 12 \quad \text{--- (Eq. 1)}$$

2. Apply Coefficient of Restitution Formula:

   $$e = \frac{v_2 - v_1}{u_1 - u_2} \implies 0.5 = \frac{v_2 - v_1}{6 - 0} \implies v_2 - v_1 = 3 \implies v_2 = v_1 + 3 \quad \text{--- (Eq. 2)}$$

3. Solve for Final Velocities: Substitute Eq. 2 into Eq. 1:

   $$2 v_1 + (v_1 + 3) = 12 \implies 3 v_1 + 3 = 12 \implies 3 v_1 = 9 \implies v_1 = 3\text{ m/s}$$ $$v_2 = 3 + 3 = 6\text{ m/s}$$

4. Calculate Energy Loss Directly via Formula:

   $$\Delta E_{\text{loss}} = \frac{m_1 m_2}{2(m_1 + m_2)} (1 - e^2) (u_1 - u_2)^2$$ $$\Delta E_{\text{loss}} = \frac{2000 \times 1000}{2(2000 + 1000)} (1 - 0.5^2) (6 - 0)^2$$ $$\Delta E_{\text{loss}} = \frac{2{,}000{,}000}{6000} \times (1 - 0.25) \times 36 = \frac{1000}{3} \times 0.75 \times 36 = 1000 \times 0.75 \times 12 = 9000\text{ J} = 9.0\text{ kJ}$$