ME 212: Dynamics
Homeyra Pourmohammadali
Estimated study time: 1 hr 22 min
Table of contents
Sources and References
Primary textbook — Russell C. Hibbeler, Engineering Mechanics: Dynamics, 14th ed., Prentice Hall, 2015.
Online resources — MIT OpenCourseWare 2.003SC Engineering Dynamics (Fall 2011); David Tong, Classical Dynamics, University of Cambridge lecture notes (2004); Richard Fitzpatrick, Newtonian Dynamics, University of Texas at Austin (2011).
Chapter 1: Kinematics of Particles
Kinematics is the branch of mechanics concerned exclusively with the geometry of motion — how positions, velocities, and accelerations are described in space and time, without reference to the forces that cause the motion. A particle is an idealization in which the body’s size and shape are irrelevant; its entire mass is concentrated at a single point. This idealization is valid whenever the dimensions of the body are negligible compared with the distances involved in the motion, or whenever rotational effects are unimportant to the analysis at hand.
The present chapter develops the full apparatus of particle kinematics in two and three dimensions, covering rectilinear (straight-line) motion, curvilinear motion in rectangular coordinates, normal-tangential coordinates, and polar coordinates, and then treating two important classes of constrained motion: absolute dependent motion and relative motion with translating axes.
Section 1.1: Rectilinear Kinematics
The Three Fundamental Kinematic Quantities
Consider a particle constrained to move along a straight line, which we take as the \(s\)-axis with a fixed reference origin \(O\). The position of the particle at time \(t\) is the signed scalar \(s(t)\), measured in metres. Three derived quantities are of central importance.
Let \(s(t)\) be the position coordinate of a particle on a straight line. Then:
- Displacement over an interval: \(\Delta s = s_2 - s_1\).
- Velocity: \(v = \dot{s} = \dfrac{ds}{dt}\), the instantaneous rate of change of position.
- Acceleration: \(a = \dot{v} = \ddot{s} = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\).
A useful derived relation eliminates time between velocity and acceleration:
\[ a \, ds = v \, dv \]This follows immediately from \(a = dv/dt\) and \(v = ds/dt\), so \(a = v \, dv/ds\).
The sign of velocity indicates direction of motion along the axis; the sign of acceleration indicates whether the velocity is increasing (same sign as \(v\)) or decreasing (opposite sign). The distance traveled is the total path length, distinct from the net displacement.
Constant Acceleration
When \(a = a_c = \text{const}\), the three kinematic equations integrate directly:
\[ v = v_0 + a_c \, t \]\[ s = s_0 + v_0 t + \tfrac{1}{2} a_c t^2 \]\[ v^2 = v_0^2 + 2 a_c (s - s_0) \]These are the classical constant-acceleration kinematic equations, valid only when \(a\) is truly constant throughout the interval.
Erratic Motion
When motion data are given graphically or numerically rather than as a smooth function, we use the integral definitions directly.
Given \(a(t)\):
\[ v(t) = v_0 + \int_0^t a(\tau)\,d\tau \]Given \(v(t)\):
\[ s(t) = s_0 + \int_0^t v(\tau)\,d\tau \]The geometric interpretation is that velocity is the area under the \(a\text{-}t\) graph and displacement is the area under the \(v\text{-}t\) graph. Similarly, using \(a = v\,dv/ds\):
\[ \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 = \int_{s_0}^s a\,ds' \]so the change in \(\tfrac{1}{2}v^2\) equals the area under the \(a\text{-}s\) graph.
A particle moves along the \(x\)-axis with position \(s(t) = 2t^3 - 9t^2 + 12t - 4\) metres, \(t\) in seconds. Find (a) the velocity and acceleration at \(t = 1\,\text{s}\); (b) the times at which the particle is momentarily at rest; (c) the total distance traveled from \(t = 0\) to \(t = 3\,\text{s}\).
Solution.
Differentiate: \(v(t) = \dot{s} = 6t^2 - 18t + 12 = 6(t-1)(t-2)\) m/s and \(a(t) = 12t - 18\) m/s².
(a) At \(t = 1\): \(v = 6(0)(−1) = 0\) m/s, \(a = 12(1) - 18 = -6\) m/s².
(b) The particle is momentarily at rest when \(v = 0\), i.e., \(t = 1\,\text{s}\) and \(t = 2\,\text{s}\).
(c) Evaluate positions: \(s(0) = -4\,\text{m}\), \(s(1) = 1\,\text{m}\), \(s(2) = 0\,\text{m}\), \(s(3) = 5\,\text{m}\). The particle moves right from \(-4\) to \(1\) (distance \(5\,\text{m}\)), then left from \(1\) to \(0\) (distance \(1\,\text{m}\)), then right from \(0\) to \(5\) (distance \(5\,\text{m}\)). Total distance = \(5 + 1 + 5 = 11\,\text{m}\). (Net displacement: \(5 - (-4) = 9\,\text{m}\).)
Section 1.2: Curvilinear Motion — Rectangular Coordinates
When a particle moves along a curved path in the \(xy\)-plane (or in three dimensions), we describe its state using vectors. Let \(\mathbf{r}(t) = x(t)\,\hat{\mathbf{i}} + y(t)\,\hat{\mathbf{j}}\) be the position vector from a fixed origin.
The velocity vector is always tangent to the path; the acceleration vector is generally not tangent, but points toward the concave side of the curve.
Projectile Motion
Projectile motion is rectilinear kinematics in the vertical direction combined with uniform (constant-velocity) motion in the horizontal direction, under the sole assumption that aerodynamic drag is negligible and \(g = 9.81\,\text{m/s}^2\) downward.
With \(x\) horizontal and \(y\) upward, initial position \((x_0, y_0)\), initial speed \(v_0\) at angle \(\theta\) from horizontal:
\[ x = x_0 + (v_0 \cos\theta)\,t \]\[ y = y_0 + (v_0 \sin\theta)\,t - \tfrac{1}{2}g t^2 \]\[ v_x = v_0 \cos\theta \quad \text{(constant)} \]\[ v_y = v_0 \sin\theta - g t \]The trajectory equation (eliminating \(t\)) is the parabola
\[ y - y_0 = (x - x_0)\tan\theta - \frac{g}{2(v_0\cos\theta)^2}(x-x_0)^2. \]A ball is launched from ground level at \(v_0 = 20\,\text{m/s}\) at \(\theta = 30°\) from the horizontal. Find the maximum height and the horizontal range.
Solution.
\(v_{0x} = 20\cos 30° = 17.32\,\text{m/s}\), \(v_{0y} = 20\sin 30° = 10\,\text{m/s}\).
Maximum height occurs when \(v_y = 0\): \(t^* = v_{0y}/g = 10/9.81 = 1.019\,\text{s}\).
\[ y_{\max} = 10(1.019) - \tfrac{1}{2}(9.81)(1.019)^2 = 10.19 - 5.09 = 5.10\,\text{m}. \]Total flight time (return to \(y = 0\)): \(t_{\text{flight}} = 2t^* = 2.039\,\text{s}\).
Range: \(R = v_{0x} \cdot t_{\text{flight}} = 17.32 \times 2.039 = 35.3\,\text{m}\).
Section 1.3: Normal-Tangential Coordinates
For a particle traveling along a curved path, it is often natural to express acceleration in terms of components along and perpendicular to the path itself. We introduce a moving reference frame attached to the particle.
At any instant, let \(\hat{\mathbf{e}}_t\) be the unit tangent to the path (in the direction of motion) and \(\hat{\mathbf{e}}_n\) be the unit normal pointing toward the center of curvature (inward). The binormal \(\hat{\mathbf{e}}_b = \hat{\mathbf{e}}_t \times \hat{\mathbf{e}}_n\) completes a right-handed triad.
The velocity is purely tangential:
\[ \mathbf{v} = v\,\hat{\mathbf{e}}_t, \quad v = \dot{s} \geq 0 \]The acceleration has two components:
\[ \mathbf{a} = \dot{v}\,\hat{\mathbf{e}}_t + \frac{v^2}{\rho}\,\hat{\mathbf{e}}_n \]where \(\rho\) is the radius of curvature of the path at that point.
We differentiate \(\mathbf{v} = v\,\hat{\mathbf{e}}_t\) with respect to time:
\[ \mathbf{a} = \dot{v}\,\hat{\mathbf{e}}_t + v\,\dot{\hat{\mathbf{e}}}_t. \]The key step is evaluating \(\dot{\hat{\mathbf{e}}}_t\). As the particle moves an arc length \(ds\) along the path, the tangent rotates through an angle \(d\theta\) (where \(d\theta = ds/\rho\) by the definition of radius of curvature). Thus:
\[ \left|\frac{d\hat{\mathbf{e}}_t}{ds}\right| = \frac{1}{\rho}, \quad \frac{d\hat{\mathbf{e}}_t}{ds} = \frac{1}{\rho}\hat{\mathbf{e}}_n. \]By the chain rule, \(\dot{\hat{\mathbf{e}}}_t = (d\hat{\mathbf{e}}_t/ds)(ds/dt) = (v/\rho)\hat{\mathbf{e}}_n\). Substituting:
\[ \mathbf{a} = \dot{v}\,\hat{\mathbf{e}}_t + \frac{v^2}{\rho}\,\hat{\mathbf{e}}_n. \]The tangential component \(a_t = \dot{v}\) changes the speed; the normal component \(a_n = v^2/\rho\) changes the direction of motion. The total magnitude is \(|\mathbf{a}| = \sqrt{a_t^2 + a_n^2}\).
Section 1.4: Polar Coordinates
Polar coordinates \((r, \theta)\) are most natural when the geometry of a problem involves rotation about a fixed point — for instance, a particle constrained to move along a radial arm or in a central-force field.
Let \(\hat{\mathbf{e}}_r\) be the unit vector pointing radially outward from the origin to the particle, and \(\hat{\mathbf{e}}_\theta\) be the unit vector perpendicular to \(\hat{\mathbf{e}}_r\) in the direction of increasing \(\theta\). Then:
\[ \dot{\hat{\mathbf{e}}}_r = \dot{\theta}\,\hat{\mathbf{e}}_\theta, \qquad \dot{\hat{\mathbf{e}}}_\theta = -\dot{\theta}\,\hat{\mathbf{e}}_r. \]Position: \(\mathbf{r} = r\,\hat{\mathbf{e}}_r\).
Velocity:
\[ \mathbf{v} = \dot{r}\,\hat{\mathbf{e}}_r + r\dot{\theta}\,\hat{\mathbf{e}}_\theta \]Acceleration:
\[ \mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\,\hat{\mathbf{e}}_r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\,\hat{\mathbf{e}}_\theta \]Differentiate \(\mathbf{v} = \dot{r}\,\hat{\mathbf{e}}_r + r\dot{\theta}\,\hat{\mathbf{e}}_\theta\):
\[ \mathbf{a} = \ddot{r}\,\hat{\mathbf{e}}_r + \dot{r}\dot{\hat{\mathbf{e}}}_r + (\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\mathbf{e}}_\theta + r\dot{\theta}\dot{\hat{\mathbf{e}}}_\theta \]\[ = \ddot{r}\,\hat{\mathbf{e}}_r + \dot{r}\dot{\theta}\,\hat{\mathbf{e}}_\theta + (\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\mathbf{e}}_\theta - r\dot{\theta}^2\,\hat{\mathbf{e}}_r \]\[ = (\ddot{r} - r\dot{\theta}^2)\hat{\mathbf{e}}_r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\mathbf{e}}_\theta. \]The term \(-r\dot{\theta}^2\) is the familiar centripetal (inward radial) acceleration. The term \(2\dot{r}\dot{\theta}\) is the Coriolis acceleration, which appears whenever the particle has both a radial velocity component and an angular velocity.
Section 1.5: Absolute Dependent Motion
In many mechanisms, two or more particles are connected by inextensible strings or cables over pulleys, so that one particle’s position uniquely determines the other’s. The analysis proceeds by expressing all positions relative to a fixed datum, writing the constraint (the cable length is constant), and differentiating.
- Define positive coordinate directions from fixed datums to each particle, measuring cable segments that change in length.
- Write the constraint equation: the sum of all active cable segments equals the total inextensible length: \(l = \text{const}\).
- Differentiate once with respect to time to obtain a velocity constraint; differentiate again for the acceleration constraint.
Block \(A\) moves downward with velocity \(v_A = 2\,\text{m/s}\) in a two-movable-pulley system where one end of the cable is fixed and the cable wraps around two pulleys attached to block \(B\). Express the velocity of block \(B\).
Solution.
Let \(s_A\) be the position of block \(A\) (positive downward) and \(s_B\) the position of block \(B\) (positive downward). With one fixed end and two cable segments supporting \(B\)’s pulley, the constraint is:
\[ s_A + 2s_B = \ell \]Differentiating: \(v_A + 2v_B = 0\), so \(v_B = -v_A/2 = -1\,\text{m/s}\). The negative sign means \(B\) moves upward at \(1\,\text{m/s}\). Differentiating again: \(a_A + 2a_B = 0\).
Section 1.6: Relative Motion with Translating Axes
When we want to describe the motion of particle \(B\) as seen by an observer who is themselves moving (but not rotating) along with particle \(A\), we use translating axes attached to \(A\).
Let \(xyz\) be a frame with origin at \(A\) that translates (does not rotate) relative to the fixed frame \(XYZ\). Then:
\[ \mathbf{r}_B = \mathbf{r}_A + \mathbf{r}_{B/A} \]\[ \mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A} \]\[ \mathbf{a}_B = \mathbf{a}_A + \mathbf{a}_{B/A} \]where \(\mathbf{r}_{B/A}\), \(\mathbf{v}_{B/A}\), \(\mathbf{a}_{B/A}\) are the position, velocity, and acceleration of \(B\) relative to \(A\), measured in the translating frame.
Aircraft \(A\) flies east at \(v_A = 300\,\text{km/h}\) and aircraft \(B\) flies north at \(v_B = 400\,\text{km/h}\). Find the velocity of \(B\) relative to \(A\).
Solution.
\(\mathbf{v}_A = 300\,\hat{\mathbf{i}}\,\text{km/h}\), \(\mathbf{v}_B = 400\,\hat{\mathbf{j}}\,\text{km/h}\).
\[ \mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A = -300\,\hat{\mathbf{i}} + 400\,\hat{\mathbf{j}}\,\text{km/h} \]\[ |\mathbf{v}_{B/A}| = \sqrt{300^2 + 400^2} = 500\,\text{km/h} \]The direction is \(\theta = \arctan(400/300) = 53.1°\) west of north as seen from \(A\).
Chapter 2: Kinetics of Particles — Force, Work, and Momentum
Kinetics brings forces into the picture. For a particle of mass \(m\), Newton’s second law \(\mathbf{F} = m\mathbf{a}\) is the single foundational equation, but it can be expressed in several coordinate systems and integrated in different ways to yield the work-energy and impulse-momentum principles — each form suited to a different class of problems.
Section 2.1: Newton’s Second Law and Equations of Motion
The resultant force \(\mathbf{F}\) acting on a particle of mass \(m\) produces an acceleration \(\mathbf{a}\) of the particle in an inertial reference frame:
\[ \sum \mathbf{F} = m\mathbf{a} \]In scalar component form (rectangular):
\[ \sum F_x = m\ddot{x}, \quad \sum F_y = m\ddot{y}, \quad \sum F_z = m\ddot{z} \]The free-body diagram (FBD) is indispensable: every force acting on the particle must appear, and only forces acting on the particle (not forces that the particle exerts on other bodies) should be included.
Equations of Motion in Normal-Tangential Coordinates
\[ \sum F_t = ma_t = m\dot{v} \]\[ \sum F_n = ma_n = \frac{mv^2}{\rho} \]The normal equation is particularly useful for circular motion problems, where the net inward force provides the centripetal acceleration.
Equations of Motion in Polar Coordinates
\[ \sum F_r = m(\ddot{r} - r\dot{\theta}^2) \]\[ \sum F_\theta = m(r\ddot{\theta} + 2\dot{r}\dot{\theta}) \]A block of mass \(m = 5\,\text{kg}\) rests on a rough inclined plane (\(\mu_k = 0.3\)) inclined at \(\alpha = 30°\) from horizontal. It is released from rest. Find the acceleration and the distance traveled in \(t = 2\,\text{s}\).
Solution.
Set \(x\) along the incline (positive down the slope) and \(y\) perpendicular (positive away from surface).
Normal: \(\sum F_y = 0\): \(N - mg\cos\alpha = 0 \Rightarrow N = mg\cos\alpha = 5(9.81)(0.866) = 42.5\,\text{N}\).
Friction: \(f_k = \mu_k N = 0.3(42.5) = 12.74\,\text{N}\) (up the slope).
Along slope: \(\sum F_x = ma\): \(mg\sin\alpha - f_k = ma\).
\[ a = g\sin\alpha - \mu_k g\cos\alpha = 9.81(0.5 - 0.3 \times 0.866) = 9.81(0.5 - 0.260) = 2.35\,\text{m/s}^2 \]Distance: \(s = \tfrac{1}{2}at^2 = \tfrac{1}{2}(2.35)(4) = 4.70\,\text{m}\).
Section 2.2: Work and Energy
The work-energy method is powerful for problems where forces are known as functions of position (not time), and we seek relationships between speed and position.
Work of a Force
The work done by a force \(\mathbf{F}\) as the particle moves along a path from position 1 to position 2 is:
\[ U_{1 \to 2} = \int_{\mathbf{r}_1}^{\mathbf{r}_2} \mathbf{F} \cdot d\mathbf{r} \]For constant force: \(U = \mathbf{F} \cdot \Delta\mathbf{r} = F \cos\theta \cdot \Delta s\).
For a spring with stiffness \(k\) compressed or stretched from \(s_1\) to \(s_2\) from its natural length:
\[ U_{\text{spring}} = -\left(\tfrac{1}{2}ks_2^2 - \tfrac{1}{2}ks_1^2\right) \](negative because the spring force opposes displacement from natural length).
Work of gravity (upward \(y\) positive):
\[ U_g = -mg(y_2 - y_1) = -W\Delta y \]Principle of Work and Energy
where \(T = \tfrac{1}{2}mv^2\) is the kinetic energy. The net work done on the particle equals the change in kinetic energy.
Start from \(\sum F_t = ma_t = mv(dv/ds)\). Multiply both sides by \(ds\) and integrate:
\[ \int_{s_1}^{s_2} \sum F_t \, ds = \int_{v_1}^{v_2} mv\,dv = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2. \]The left side is precisely the total work done on the particle (since \(F_n\) is perpendicular to motion and does no work). Hence \(T_1 + U_{1\to 2} = T_2\).
Conservative Forces and Potential Energy
A force is conservative if the work it does depends only on the initial and final positions, not the path taken. Gravity and spring forces are conservative. Friction is non-conservative (path-dependent).
For conservative forces, define potential energy \(V\) such that \(U = -(V_2 - V_1) = V_1 - V_2\). Then:
\[ T_1 + V_1 = T_2 + V_2 \quad \Longleftrightarrow \quad E = T + V = \text{const} \]Gravitational PE: \(V_g = mgy\) (with datum at \(y = 0\)).
Elastic PE: \(V_e = \tfrac{1}{2}ks^2\) (where \(s\) is stretch/compression from natural length).
If non-conservative forces (friction, applied forces) are present, the general form is:
\[ T_1 + V_1 + U_{1\to 2}^{\text{nc}} = T_2 + V_2 \]A particle of mass \(m = 0.5\,\text{kg}\) is pushed against a spring of stiffness \(k = 500\,\text{N/m}\) compressing it by \(x_0 = 0.1\,\text{m}\). The particle is released on a frictionless horizontal surface, leaves the spring, and then travels up a smooth ramp of height \(h = 0.3\,\text{m}\). Find the speed at the top of the ramp.
Solution.
Take position 1 as the compressed-spring state (at rest, \(v_1 = 0\)) and position 2 as the top of the ramp. No non-conservative forces do work.
\[ T_1 + V_{e1} + V_{g1} = T_2 + V_{e2} + V_{g2} \]At position 1: \(T_1 = 0\), \(V_{e1} = \tfrac{1}{2}(500)(0.1)^2 = 2.5\,\text{J}\), \(V_{g1} = 0\) (datum here).
At position 2: \(V_{e2} = 0\) (spring fully released), \(V_{g2} = mgh = 0.5(9.81)(0.3) = 1.472\,\text{J}\).
\[ 0 + 2.5 + 0 = \tfrac{1}{2}(0.5)v_2^2 + 0 + 1.472 \]\[ v_2 = \sqrt{\frac{2(2.5 - 1.472)}{0.5}} = \sqrt{4.112} = 2.03\,\text{m/s}. \]Section 2.3: Linear Impulse and Momentum
When forces vary with time and we want to relate velocities at two instants, the impulse-momentum approach is most direct.
The linear momentum of a particle is \(\mathbf{L} = m\mathbf{v}\).
The linear impulse of a force \(\mathbf{F}\) over a time interval \([t_1, t_2]\) is:
\[ \mathbf{J} = \int_{t_1}^{t_2} \mathbf{F}\,dt \]Newton’s second law in impulse-momentum form:
\[ m\mathbf{v}_1 + \sum \int_{t_1}^{t_2} \mathbf{F}\,dt = m\mathbf{v}_2 \]If the net external force on a particle (or system of particles) is zero in a given direction, the component of linear momentum in that direction is conserved.
For a system of \(n\) particles with no external forces:
\[ \sum_{i=1}^n m_i \mathbf{v}_i = \text{const} \]Impact
Impact is a collision occurring over a very short time interval, during which the impulsive contact forces are much larger than any other external forces (which can be neglected during the collision).
For two particles colliding along the line of impact:
\[ e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}} = \frac{(v_B')_n - (v_A')_n}{(v_A)_n - (v_B)_n} \]where subscript \(n\) denotes the component along the line of impact, primes denote post-impact velocities.
- \(e = 1\): perfectly elastic impact (kinetic energy conserved).
- \(e = 0\): perfectly plastic (particles stick together, maximum energy loss).
- \(0 < e < 1\): real (partially elastic) impact.
Section 2.4: Angular Impulse and Momentum
The angular momentum of a particle about a fixed point \(O\) is:
\[ \mathbf{H}_O = \mathbf{r} \times m\mathbf{v} \]where \(\mathbf{r}\) is the position vector from \(O\) to the particle. In 2D, only the \(z\)-component (scalar \(H_O\)) is relevant:
\[ H_O = m(x v_y - y v_x) = m r v_\perp \]where \(v_\perp\) is the velocity component perpendicular to \(\mathbf{r}\).
The net moment (torque) of all external forces about fixed point \(O\) equals the rate of change of angular momentum:
\[ \sum \mathbf{M}_O = \dot{\mathbf{H}}_O \]In integrated form (angular impulse-momentum principle):
\[ \mathbf{H}_{O,1} + \int_{t_1}^{t_2} \sum \mathbf{M}_O \, dt = \mathbf{H}_{O,2} \]If \(\sum \mathbf{M}_O = 0\), angular momentum about \(O\) is conserved: \(\mathbf{H}_{O,1} = \mathbf{H}_{O,2}\).
Chapter 3: Planar Kinematics of Rigid Bodies
A rigid body is an idealization in which the distance between any two material points in the body remains constant at all times. In reality all bodies deform under load, but the rigid-body approximation is excellent when deformations are negligibly small compared with the overall motion. The kinematic description of a rigid body is richer than that of a particle: a body in three dimensions has six degrees of freedom (three translational, three rotational), and in planar motion it has three (two translational, one rotational).
Section 3.1: Types of Planar Rigid-Body Motion
- Pure translation: every line in the body remains parallel to its original orientation. All points have the same velocity and the same acceleration at any instant. The body moves like a single particle.
- Rotation about a fixed axis: all points in the body move in circular arcs about a single fixed axis perpendicular to the plane of motion.
- General plane motion: a combination of translation and rotation. The instantaneous motion can always be decomposed into a translation of the center of mass plus a rotation about an axis through the center of mass.
Section 3.2: Rotation About a Fixed Axis
Let \(\theta\) denote the angular position of the body, measured from a fixed reference line. Define angular velocity \(\omega = \dot{\theta}\) and angular acceleration \(\alpha = \dot{\omega} = \ddot{\theta}\).
The formal analogies with rectilinear particle kinematics are exact:
\[ \omega = \dot{\theta}, \quad \alpha = \dot{\omega} = \ddot{\theta}, \quad \alpha\,d\theta = \omega\,d\omega \]For constant angular acceleration \(\alpha = \alpha_c\):
\[ \omega = \omega_0 + \alpha_c t \]\[ \theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha_c t^2 \]\[ \omega^2 = \omega_0^2 + 2\alpha_c(\theta - \theta_0) \]For a point \(P\) at radial distance \(r\) from the axis of rotation:
\[ v = r\omega, \quad a_t = r\alpha, \quad a_n = r\omega^2 = v^2/r \]Section 3.3: Absolute Motion Analysis
The absolute motion analysis method relates the angular motion of a body to the linear motion of a specific point by writing a geometric constraint between position coordinates and then differentiating.
A crank \(OA\) of length \(r\) rotates with constant angular velocity \(\omega\). Point \(A\) is connected to slider \(B\) on a horizontal track by connecting rod \(AB\) of length \(L\). Express the velocity of slider \(B\) in terms of \(\theta\) (the crank angle).
Solution.
From the geometry: \(x_B = r\cos\theta + \sqrt{L^2 - r^2\sin^2\theta}\).
Differentiate with respect to time:
\[ v_B = \dot{x}_B = -r\omega\sin\theta - \frac{r^2\omega\sin\theta\cos\theta}{\sqrt{L^2 - r^2\sin^2\theta}} = -r\omega\sin\theta\left(1 + \frac{r\cos\theta}{\sqrt{L^2 - r^2\sin^2\theta}}\right). \]This exact result is used in engine design; for \(r \ll L\) (long connecting rod), the approximation \(v_B \approx -r\omega\sin\theta\) holds — purely sinusoidal motion.
Section 3.4: Relative Velocity Analysis
The fundamental equation for planar rigid-body velocity analysis decomposes the motion of any point \(B\) into a translation with point \(A\) plus a rotation about \(A\).
For any two points \(A\) and \(B\) on (or rigidly attached to) the same body rotating with angular velocity \(\boldsymbol{\omega}\):
\[ \mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega} \times \mathbf{r}_{B/A} \]In planar (2D) motion with \(\boldsymbol{\omega} = \omega\,\hat{\mathbf{k}}\):
\[ \mathbf{v}_B = \mathbf{v}_A + \omega\,\hat{\mathbf{k}} \times \mathbf{r}_{B/A} \]The term \(\boldsymbol{\omega} \times \mathbf{r}_{B/A}\) represents the velocity of \(B\) relative to \(A\) due to rotation; its magnitude is \(\omega r_{B/A}\) and its direction is perpendicular to \(\mathbf{r}_{B/A}\).
Section 3.5: Instantaneous Center of Zero Velocity
At any instant in planar motion, there exists a unique point \(C\) (possibly at infinity) called the instantaneous center of zero velocity, with respect to which the body appears to be in pure rotation.
If the velocity directions of two points \(A\) and \(B\) on the body are known, the IC is located at the intersection of the lines perpendicular to \(\mathbf{v}_A\) and \(\mathbf{v}_B\).
For any point \(P\) on the body:
\[ v_P = \omega \cdot \overline{CP} \]where \(\overline{CP}\) is the distance from \(P\) to the IC and \(\omega\) is the angular velocity of the body.
A wheel of radius \(R\) rolls without slipping on a flat surface at forward speed \(v_C\) (center). Find the velocity of the top of the wheel and confirm using the IC.
Solution.
No-slip condition: the contact point \(P\) has zero velocity, so \(P\) is the IC. The center \(C\) is at distance \(R\) from \(P\), and the top point \(T\) is at distance \(2R\).
\[ v_C = \omega R \Rightarrow \omega = v_C / R. \]\[ v_T = \omega \cdot 2R = 2v_C. \]The top of a rolling wheel moves at twice the speed of the center — a result with important implications for tire wear and vehicle dynamics.
Section 3.6: Relative Acceleration Analysis
In planar motion:
\[ \mathbf{a}_B = \mathbf{a}_A + \underbrace{\alpha\,\hat{\mathbf{k}} \times \mathbf{r}_{B/A}}_{\text{tangential}} - \underbrace{\omega^2 \mathbf{r}_{B/A}}_{\text{centripetal (inward)}} \]The tangential relative acceleration \(|\boldsymbol{\alpha} \times \mathbf{r}_{B/A}| = \alpha r_{B/A}\) is perpendicular to \(AB\); the centripetal relative acceleration \(\omega^2 r_{B/A}\) is directed from \(B\) toward \(A\).
Rod \(AB\) has length \(0.5\,\text{m}\). At the instant shown, end \(A\) moves horizontally to the right at \(v_A = 3\,\text{m/s}\) with zero acceleration. The rod makes angle \(\theta = 60°\) with the horizontal, and its angular velocity is \(\omega = 2\,\text{rad/s}\) clockwise with \(\alpha = 4\,\text{rad/s}^2\) counterclockwise. Find the acceleration of end \(B\).
Solution.
Let \(\mathbf{r}_{B/A} = 0.5(\cos 60°\,\hat{\mathbf{i}} + \sin 60°\,\hat{\mathbf{j}}) = 0.25\,\hat{\mathbf{i}} + 0.433\,\hat{\mathbf{j}}\,\text{m}\).
With \(\boldsymbol{\omega} = -2\hat{\mathbf{k}}\) and \(\boldsymbol{\alpha} = +4\hat{\mathbf{k}}\):
Tangential: \(\boldsymbol{\alpha} \times \mathbf{r}_{B/A} = 4\hat{\mathbf{k}} \times (0.25\hat{\mathbf{i}} + 0.433\hat{\mathbf{j}}) = 1.0\hat{\mathbf{j}} - 1.732\hat{\mathbf{i}}\,\text{m/s}^2\).
Centripetal: \(-\omega^2 \mathbf{r}_{B/A} = -4(0.25\hat{\mathbf{i}} + 0.433\hat{\mathbf{j}}) = -1.0\hat{\mathbf{i}} - 1.732\hat{\mathbf{j}}\,\text{m/s}^2\).
\[ \mathbf{a}_B = \mathbf{a}_A + (-1.732 - 1.0)\hat{\mathbf{i}} + (1.0 - 1.732)\hat{\mathbf{j}} = 0 - 2.732\hat{\mathbf{i}} - 0.732\hat{\mathbf{j}}\,\text{m/s}^2 \]\[ |\mathbf{a}_B| = \sqrt{2.732^2 + 0.732^2} = 2.83\,\text{m/s}^2 \]Section 3.7: Relative Motion Using Rotating Axes
When the reference frame is both translating and rotating — for instance, when analyzing a body moving within a rotating mechanism — the velocity and acceleration expressions acquire additional terms.
Let \(xyz\) be a frame attached to a body rotating with angular velocity \(\boldsymbol{\Omega}\) relative to the fixed frame. For a point \(B\) whose position relative to the moving origin \(A\) is \(\mathbf{r}_{B/A}\):
\[ \mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\Omega} \times \mathbf{r}_{B/A} + (\mathbf{v}_{B/A})_{xyz} \]\[ \mathbf{a}_B = \mathbf{a}_A + \dot{\boldsymbol{\Omega}} \times \mathbf{r}_{B/A} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{r}_{B/A}) + 2\boldsymbol{\Omega} \times (\mathbf{v}_{B/A})_{xyz} + (\mathbf{a}_{B/A})_{xyz} \]The term \(2\boldsymbol{\Omega} \times (\mathbf{v}_{B/A})_{xyz}\) is the Coriolis acceleration, arising from the interaction between the rotation of the frame and the velocity of the point as seen in the rotating frame.
Chapter 4: Planar Kinetics of Rigid Bodies
The kinetics of a rigid body extends Newton’s second law to account for the body’s mass distribution. The translational motion of the center of mass is governed by \(\sum \mathbf{F} = m\mathbf{a}_G\), exactly as for a particle, while the rotational motion about the center of mass is governed by the moment equation \(\sum M_G = I_G \alpha\). The work-energy and impulse-momentum methods then provide integrated forms of these laws.
Section 4.1: Mass Moment of Inertia
The mass moment of inertia is the rotational analog of mass: it quantifies a body’s resistance to angular acceleration.
For a rigid body with mass \(m\) and a chosen axis, the mass moment of inertia about that axis is:
\[ I = \int r^2 \, dm \]where \(r\) is the perpendicular distance from the mass element \(dm\) to the axis. The SI unit is kg·m².
Parallel-Axis Theorem: if \(I_G\) is the moment of inertia about an axis through the center of mass \(G\), then the moment of inertia about a parallel axis at distance \(d\) from \(G\) is:
\[ I = I_G + md^2 \]- Slender rod of length \(L\), mass \(m\): \(I_G = \tfrac{1}{12}mL^2\) (about centroidal axis perpendicular to rod); \(I_{\text{end}} = \tfrac{1}{3}mL^2\).
- Solid disk/cylinder, radius \(R\): \(I_G = \tfrac{1}{2}mR^2\).
- Thin ring of radius \(R\): \(I_G = mR^2\).
- Solid sphere, radius \(R\): \(I_G = \tfrac{2}{5}mR^2\).
- Rectangular plate, width \(b\), height \(h\): \(I_G = \tfrac{1}{12}m(b^2 + h^2)\).
Consider a uniform disk of radius \(R\), mass \(m\), density \(\sigma = m/(\pi R^2)\) per unit area. Using annular rings of radius \(r\), width \(dr\):
\[ I = \int_0^R r^2 \, dm = \int_0^R r^2 \sigma (2\pi r\,dr) = 2\pi\sigma \int_0^R r^3\,dr = 2\pi\sigma \frac{R^4}{4} = \frac{\pi \sigma R^4}{2}. \]Since \(m = \sigma \pi R^2\): \(I = \dfrac{m R^2}{2}\). \(\square\)
Section 4.2: Planar Equations of Motion for a Rigid Body
For a rigid body undergoing planar motion:
\[ \sum F_x = m(\mathbf{a}_G)_x \]\[ \sum F_y = m(\mathbf{a}_G)_y \]\[ \sum M_G = I_G \alpha \]where \(\mathbf{a}_G\) is the acceleration of the center of mass \(G\) and \(\alpha\) is the angular acceleration of the body. Alternatively, moments may be taken about any fixed point \(O\) or the IC, with appropriate care:
\[ \sum M_O = I_O \alpha \quad \text{(only when } O \text{ is a fixed point or the IC)} \]Equations of Motion: Pure Translation
For a body in pure translation, \(\alpha = 0\), and \(\sum M_G = 0\). All points have the same acceleration.
Equations of Motion: Rotation About a Fixed Axis
When a body rotates about a fixed axis through point \(O\):
\[ \sum M_O = I_O \alpha \]The reaction forces at the pin \(O\) are found from the \(F_x\) and \(F_y\) equations.
Equations of Motion: General Plane Motion
General plane motion requires all three equations simultaneously, combined with kinematic constraints.
A uniform solid cylinder of mass \(m\) and radius \(R\) rolls without slipping down an incline of angle \(\theta\). Find the acceleration of the center and the friction force.
Solution.
Let \(x\) be along the incline (positive down). For rolling without slip: \(a_G = R\alpha\).
Force equations:
\[ \sum F_x = ma_G: \quad mg\sin\theta - f = ma_G \]\[ \sum F_y = 0: \quad N - mg\cos\theta = 0 \Rightarrow N = mg\cos\theta \]Moment about \(G\):
\[ \sum M_G = I_G\alpha: \quad fR = \tfrac{1}{2}mR^2 \cdot \alpha = \tfrac{1}{2}mR^2 \cdot \frac{a_G}{R} = \tfrac{1}{2}mRa_G \]So \(f = \tfrac{1}{2}ma_G\). Substituting into the \(x\)-equation:
\[ mg\sin\theta - \tfrac{1}{2}ma_G = ma_G \Rightarrow a_G = \tfrac{2}{3}g\sin\theta. \]\[ f = \tfrac{1}{2}m \cdot \tfrac{2}{3}g\sin\theta = \tfrac{1}{3}mg\sin\theta. \]Compare with a block sliding without friction: \(a = g\sin\theta\). The rolling cylinder is slower by a factor of \(2/3\) because some of the gravitational PE goes into rotational KE.
Section 4.3: Work and Energy for Rigid Bodies
For a rigid body in planar motion:
\[ T = \tfrac{1}{2}mv_G^2 + \tfrac{1}{2}I_G\omega^2 \]The first term is the translational kinetic energy of the center of mass; the second is the rotational kinetic energy about the center of mass.
For a body rotating about a fixed axis through \(O\):
\[ T = \tfrac{1}{2}I_O\omega^2 \]Express the velocity of any mass element \(dm\) at position \(\mathbf{r}'\) from \(G\) as \(\mathbf{v} = \mathbf{v}_G + \boldsymbol{\omega} \times \mathbf{r}'\). Then:
\[ T = \tfrac{1}{2}\int |\mathbf{v}|^2\,dm = \tfrac{1}{2}\int |\mathbf{v}_G + \boldsymbol{\omega}\times\mathbf{r}'|^2\,dm \]\[ = \tfrac{1}{2}v_G^2 \int dm + \mathbf{v}_G \cdot \left(\boldsymbol{\omega}\times \int \mathbf{r}'\,dm\right) + \tfrac{1}{2}\int |\boldsymbol{\omega}\times\mathbf{r}'|^2\,dm. \]Since \(G\) is the center of mass, \(\int \mathbf{r}'\,dm = 0\), so the cross term vanishes. The last integral gives \(\tfrac{1}{2}I_G\omega^2\). Hence:
\[ T = \tfrac{1}{2}mv_G^2 + \tfrac{1}{2}I_G\omega^2. \quad \square \]Work of Forces and Couples
A couple of moment \(M\) (constant) acting on a body that rotates through angle \(\Delta\theta\):
\[ U_{\text{couple}} = M\,\Delta\theta = \int_{\theta_1}^{\theta_2} M\,d\theta \](sign: positive if the couple is in the same direction as the rotation).
where \(T = \tfrac{1}{2}mv_G^2 + \tfrac{1}{2}I_G\omega^2\) and \(U_{1\to 2}\) includes the work of all external forces and couples acting on the body.
For conservative systems (elastic springs, gravity), define potential energy \(V\) as before:
\[ T_1 + V_1 = T_2 + V_2 \]A uniform slender rod of mass \(m\) and length \(L\) is pinned at one end and released from rest in the horizontal position. Find the angular velocity when it reaches the vertical position.
Solution.
Use conservation of energy. The center of mass drops \(h = L/2\) as the rod swings from horizontal to vertical.
\[ T_1 + V_1 = T_2 + V_2 \]\[ 0 + 0 = \tfrac{1}{2}I_O\omega^2 - mg\frac{L}{2} \]\[ I_O = \tfrac{1}{3}mL^2 \quad \text{(parallel axis theorem: } I_G + m(L/2)^2 = \tfrac{1}{12}mL^2 + \tfrac{1}{4}mL^2 = \tfrac{1}{3}mL^2\text{)}. \]\[ \omega = \sqrt{\frac{mgL}{\tfrac{1}{3}mL^2 \cdot 1}} = \sqrt{\frac{3g}{L}}. \]At \(L = 1\,\text{m}\): \(\omega = \sqrt{3 \times 9.81} = 5.42\,\text{rad/s}\).
Section 4.4: Impulse and Momentum for Rigid Bodies
Linear momentum: \(\mathbf{G} = m\mathbf{v}_G\) (same as for a particle of mass \(m\) at the center of mass).
Angular momentum about the center of mass: \(\mathbf{H}_G = I_G\boldsymbol{\omega}\).
Angular momentum about a fixed point \(O\): \(\mathbf{H}_O = I_G\boldsymbol{\omega} + \mathbf{r}_G \times m\mathbf{v}_G\). For rotation about a fixed axis through \(O\): \(\mathbf{H}_O = I_O\boldsymbol{\omega}\).
If no external forces act (or their sum is zero): linear momentum is conserved.
If no external moments act about \(G\) (or their sum is zero): angular momentum about \(G\) is conserved.
A uniform bar of mass \(m = 3\,\text{kg}\), length \(L = 1.2\,\text{m}\), is at rest on a frictionless surface when a horizontal impulse \(\hat{J} = 6\,\text{N·s}\) is applied at one end (perpendicular to the bar). Find the velocity of the center of mass and the angular velocity immediately after.
Solution.
Linear impulse-momentum:
\[ \hat{J} = mv_G \Rightarrow v_G = \frac{6}{3} = 2\,\text{m/s}. \]Angular impulse-momentum about \(G\) (moment arm = \(L/2\)):
\[ \hat{J} \cdot \frac{L}{2} = I_G \omega = \tfrac{1}{12}mL^2\,\omega \]\[ 6 \times 0.6 = \tfrac{1}{12}(3)(1.44)\,\omega = 0.36\,\omega \Rightarrow \omega = 10\,\text{rad/s}. \]The end where the impulse is applied has velocity \(v_G + \omega(L/2) = 2 + 10(0.6) = 8\,\text{m/s}\). The other end has velocity \(v_G - \omega(L/2) = 2 - 6 = -4\,\text{m/s}\) (moves in the opposite direction).
Summary and Conceptual Review
The four chapters of ME 212 build a hierarchy of mechanics:
Kinematics (Chapters 1 and 3) describes motion in purely geometric terms. For particles, the key tool is differentiation and integration of the position vector, with the choice of coordinate system (rectangular, normal-tangential, or polar) dictated by the geometry of the problem. For rigid bodies, the constraint that the distance between any two points is fixed leads to the relative velocity equation \(\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}\times\mathbf{r}_{B/A}\) and its acceleration counterpart, plus the powerful concept of the instantaneous center.
Kinetics (Chapters 2 and 4) introduces forces and moments. Three complementary approaches are available:
| Method | Best used when |
|---|---|
| Newton’s 2nd Law (\(\mathbf{F} = m\mathbf{a}\), \(M = I\alpha\)) | Forces known as functions of position or time; need acceleration or internal forces |
| Work-Energy (\(T_1 + U = T_2\)) | Forces known as functions of position; need speed/position relationship; no need for time |
| Impulse-Momentum (\(m\mathbf{v}_1 + \int \mathbf{F}\,dt = m\mathbf{v}_2\)) | Forces known as functions of time, or impulsive forces; need velocity change over a time interval |
The rigid-body versions of work-energy and impulse-momentum are direct extensions of the particle equations, with the kinetic energy gaining a rotational term \(\tfrac{1}{2}I_G\omega^2\) and the angular impulse-momentum equation adding the rotational degree of freedom.
Forgetting this correction is one of the most frequent sources of error in rigid-body kinetics.
