9.1 Kinematics of Particles and Rigid Bodies

9.1 Kinematics of Particles and Rigid Bodies

Kinematic Foundation: Kinematics is the study of motion without regard to the forces causing it. It establishes the mathematical relationships between position $\mathbf{r}$, displacement $\Delta \mathbf{s}$, velocity $\mathbf{v}$, and acceleration $\mathbf{a}$ for both single particles and rigid bodies undergoing planar motion.

Rectilinear Kinematics of Particles

Rectilinear motion occurs along a straight-line path defined by position coordinate $s(t)$. Velocity and acceleration are differential relationships:

$$v(t) = \frac{ds}{dt}, \quad a(t) = \frac{dv}{dt} = \frac{d^2 s}{dt^2} = v \frac{dv}{ds}$$

Constant Acceleration Equations

When acceleration is constant ($a = a_c$), direct integration yields the standard NCEES kinematic formulas:

$$v(t) = v_0 + a_c t$$ $$s(t) = s_0 + v_0 t + \frac{1}{2} a_c t^2$$ $$v^2 = v_0^2 + 2 a_c (s - s_0)$$ $$\text{Average Velocity } v_{\text{avg}} = \frac{v_0 + v}{2} = \frac{\Delta s}{\Delta t}$$

Variable Acceleration

When acceleration is a specified function of time $a(t)$, velocity $v(s)$, or position $a(s)$, integration limits must be established:

$$v(t) = v_0 + \int_{0}^{t} a(t) dt, \quad s(t) = s_0 + \int_{0}^{t} v(t) dt$$ $$\int_{v_0}^{v} v dv = \int_{s_0}^{s} a(s) ds \implies \frac{1}{2}(v^2 - v_0^2) = \int_{s_0}^{s} a(s) ds$$

Curvilinear Motion: Normal-Tangential (n-t) Components and Projectiles

When a particle moves along a curved path, its acceleration has components along orthogonal unit vectors tangent ($\mathbf{u}_t$) and normal ($\mathbf{u}_n$) to the path.

Normal-Tangential Acceleration

$$\mathbf{a} = a_t \mathbf{u}_t + a_n \mathbf{u}_n = \dot{v} \mathbf{u}_t + \left( \frac{v^2}{\rho} \right) \mathbf{u}_n$$

• Tangential Acceleration Component ($a_t = \frac{dv}{dt}$): Measures the rate of change of the magnitude of velocity (speed). If $a_t = 0$, speed is constant.
• Normal Acceleration Component ($a_n = \frac{v^2}{\rho}$): Measures the rate of change of the direction of velocity. It always points inward toward the center of curvature, where $\rho$ is the radius of curvature.
• Total Acceleration Magnitude ($a$):

$$a = \sqrt{a_t^2 + a_n^2}$$

Projectile Motion

Ignoring air resistance, a projectile experiences zero horizontal acceleration ($a_x = 0$) and constant vertical gravitational acceleration ($a_y = -g$). For launch speed $v_0$ at angle $\theta$:

$$x(t) = (v_0 \cos \theta) t, \quad y(t) = (v_0 \sin \theta) t - \frac{1}{2} g t^2$$ $$v_x = v_0 \cos \theta, \quad v_y = v_0 \sin \theta - g t$$ $$\text{Maximum Height } h_{\max} = \frac{(v_0 \sin \theta)^2}{2g}, \quad \text{Horizontal Range } R = \frac{v_0^2 \sin(2\theta)}{g}$$

Rigid Body Kinematics in Planar Motion

Rigid bodies undergo translation, fixed-axis rotation, or general planar motion (a combination of translation and rotation).

Fixed-Axis Rotation

For a body rotating about a fixed point $O$ with angular position $\theta$, angular velocity $\omega$, and angular acceleration $\alpha$:

$$\omega = \frac{d\theta}{dt}, \quad \alpha = \frac{d\omega}{dt} = \omega \frac{d\omega}{d\theta}$$

Linear velocity and acceleration of a point $P$ at distance $r$ from rotation center $O$:

$$v = \omega r, \quad a_t = \alpha r, \quad a_n = \omega^2 r$$

Relative Velocity Vector Equation

The velocity of point $B$ on a rigid body relative to point $A$ is expressed as:

$$\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A} = \mathbf{v}_A + (\boldsymbol{\omega} \times \mathbf{r}_{B/A})$$

In 2D planar motion, $\boldsymbol{\omega} = \omega \mathbf{k}$. The relative velocity $\mathbf{v}_{B/A}$ has magnitude $v_{B/A} = \omega r_{B/A}$ directed perpendicular to position vector $\mathbf{r}_{B/A}$.

Instantaneous Center of Zero Velocity (ICZV)

The ICZV (denoted point $C$) is a point located on or off a body that has zero linear velocity at a specific instant ($\mathbf{v}_{C} = \mathbf{0}$).

1. If velocity vectors $\mathbf{v}_A$ and $\mathbf{v}_B$ of two points on a rigid body are known, the ICZV lies at the intersection of lines drawn perpendicular to $\mathbf{v}_A$ and $\mathbf{v}_B$.
2. Once the ICZV is located, the velocity of any point $P$ on the body is simply:

$$v_P = \omega r_{P/IC}$$

1. The angular velocity of the body is $\omega = \frac{v_A}{r_{A/IC}} = \frac{v_B}{r_{B/IC}}$.

Worked Engineering Problems

Problem 1: Curvilinear Normal-Tangential Acceleration on a Circular Curve

Scenario: An automobile travels along a horizontal circular curve of radius $\rho = 150\text{ m}$. Its speed at $t = 0$ is $v_0 = 12.0\text{ m/s}$, and its engine provides a constant tangential acceleration $a_t = 1.80\text{ m/s}^2$. Determine the vehicle's speed $v$, normal acceleration component $a_n$, and total acceleration magnitude $a$ at $t = 10.0\text{ s}$.

Solution:

1. Calculate velocity at $t = 10.0\text{ s}$ using constant $a_t$:

$$v(10) = v_0 + a_t t = 12.0\text{ m/s} + (1.80\text{ m/s}^2)(10.0\text{ s}) = 12.0 + 18.0 = 30.0\text{ m/s}$$

1. Calculate normal acceleration component $a_n$:

$$a_n = \frac{v^2}{\rho} = \frac{(30.0\text{ m/s})^2}{150\text{ m}} = \frac{900}{150} = 6.00\text{ m/s}^2$$

1. Calculate total acceleration magnitude $a$:

$$a = \sqrt{a_t^2 + a_n^2} = \sqrt{(1.80)^2 + (6.00)^2} = \sqrt{3.24 + 36.00} = \sqrt{39.24} \approx 6.264\text{ m/s}^2 \approx 6.26\text{ m/s}^2$$

1. Determine acceleration direction angle $\phi$ relative to path tangent:

$$\phi = \arctan\left(\frac{a_n}{a_t}\right) = \arctan\left(\frac{6.00}{1.80}\right) = \arctan(3.333) \approx 73.3^\circ$$

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Problem 2: Rigid Body Linkage Relative Velocity and ICZV Analysis

Scenario: In a planar four-bar mechanism, crank $AB$ of length $r_{AB} = 0.40\text{ m}$ rotates clockwise at constant angular velocity $\omega_{AB} = 5.0\text{ rad/s}$. At the instant shown, crank $AB$ is vertical, making velocity $\mathbf{v}_B$ horizontal to the right. Connecting link $BC$ ($L_{BC} = 0.80\text{ m}$) is inclined at $30^\circ$ to the horizontal, and slider $C$ is constrained to move vertically along a fixed slot. Locate the ICZV of link $BC$ and compute the angular velocity $\omega_{BC}$ and slider velocity $v_C$.

Solution:

1. Calculate velocity of point $B$:

$$v_B = \omega_{AB} r_{AB} = (5.0\text{ rad/s})(0.40\text{ m}) = 2.0\text{ m/s} \quad (\text{directed horizontally right, } +\mathbf{i})$$

1. Locate ICZV of link $BC$:

• Perpendicular to horizontal $\mathbf{v}_B$ is a vertical line passing through $B$.
• Perpendicular to vertical $\mathbf{v}_C$ is a horizontal line passing through $C$.
• The intersection of these perpendicular lines defines the ICZV location $N$.
• From geometry, right triangle $B C N$ gives $r_{B/IC} = L_{BC} \sin 30^\circ = 0.80 \cdot 0.5 = 0.40\text{ m}$, and $r_{C/IC} = L_{BC} \cos 30^\circ = 0.80 \cdot 0.866025 = 0.6928\text{ m}$.

1. Determine angular velocity $\omega_{BC}$:

$$\omega_{BC} = \frac{v_B}{r_{B/IC}} = \frac{2.0\text{ m/s}}{0.40\text{ m}} = 5.0\text{ rad/s} \quad (\text{counterclockwise})$$

1. Calculate velocity of slider $C$:

$$v_C = \omega_{BC} r_{C/IC} = (5.0\text{ rad/s})(0.6928\text{ m}) = 3.464\text{ m/s} \approx 3.46\text{ m/s}$$

This confirms slider $C$ moves upward along the vertical guide.