Engineering Mechanics · Dynamics

Planar Kinematics of a Rigid Body

Particles have no size, so they cannot rotate. Rigid bodies can — and that single fact reshapes every kinematic tool from Chapter 12. This chapter builds the velocity and acceleration relationships for bodies that translate, spin about a fixed axis, or do both at once, the geometry behind gears, cams, linkages, and rolling wheels.

Learning objectives

Classify planar rigid-body motion from geometric evidence.
Calculate point velocities and accelerations with fixed-axis and relative-motion equations.
Select between absolute motion, instantaneous-center, translating-frame, and rotating-frame methods.
Apply rolling, guide, and rigid-link constraints with a declared sign convention.
Verify units, directions, compatibility, and applicability of a kinematic result.

Course Note Treat this whole chapter as one toolbox with two questions: what kind of motion is this body in? and which two points (or which frame) do I tie together? Nearly every problem reduces to picking the right relationship — fixed-axis, translating-frame, IC, or rotating-frame — before you ever write a number. Master the decision, not just the algebra. — added from teaching experience

16.1

Planar Rigid-Body Motion

A body is in planar motion when every particle moves in a plane parallel to a single fixed reference plane. Studying it is the foundation for designing gears, cams, and mechanisms; once the kinematics is in hand, the equations of motion can link the forces on a body to how it moves. Planar rigid-body motion comes in three types, listed in order of increasing complexity.

Translation

A line fixed in the body keeps its original orientation. If every point follows a straight path it is rectilinear; if the paths are congruent curves it is curvilinear.

Rotation about a Fixed Axis

Every particle except those on the axis travels a circular path centered on the axis.

General Plane Motion

A simultaneous combination of translation (in the reference plane) and rotation (about an axis perpendicular to that plane).

Physical meaning

Rigidity fixes every distance between body points. A velocity relation describes how that unchanging geometry moves now; it does not identify the forces causing the motion. The model is appropriate when deformation and joint clearance are negligible relative to the desired accuracy.

Planar Motion

Motion in which all particles of a rigid body travel along paths that stay equidistant from one fixed plane. The body's behavior can always be described as some mix of translation and rotation within that plane.

Fig. 16.1The three categories of planar motion: rectilinear and curvilinear translation, rotation about a fixed axis, and general plane motion. (Original redrawn schematic.)

16.2

Translation

Consider a rigid body translating in the x–y plane. Points A and B are located by position vectors measured from a fixed frame, while a translating frame is pinned to the body at the base point A. The position of B relative to A is the relative-position vector that ties them together.

Fig. 16.2A translating body: the relative-position vector from base point A to B has constant magnitude and direction. (Original redrawn schematic.)

Base Point

The point at which the translating coordinate frame is attached to the body. Its motion is taken as known, and the motion of every other point is referenced to it.

Position relation

\[ \mathbf{r}_B = \mathbf{r}_A + \mathbf{r}_{B/A} \]

Differentiating the position relation in time gives the velocity relation. Because the body is rigid and translating, the relative-position vector is constant in both magnitude and direction, so its time derivative vanishes. The acceleration relation follows by differentiating once more.

Velocity — pure translation

\[ \mathbf{v}_B = \mathbf{v}_A \]

Acceleration — pure translation

\[ \mathbf{a}_B = \mathbf{a}_A \]

Key Idea Every point on a translating body — whether the translation is rectilinear or curvilinear — shares the same velocity and the same acceleration. The particle kinematics of Chapter 12 therefore carry over directly to any point on a translating rigid body.

Course Note Curvilinear translation trips students up. A Ferris-wheel gondola travels a circular path yet still translates — it never changes orientation, so there's no angular velocity. "Curved path" does not mean "rotation." Check the orientation of a fixed line in the body, not the shape of the path. — added from teaching experience

16.3

Rotation about a Fixed Axis

When a body rotates about a fixed axis, every point traces a circular path about that axis. To describe it we first need the angular motion of the body, then the motion of a point on it.

Angular Motion

A point has no dimension and therefore no angular motion — only lines and bodies rotate. The angular motion of a radial line is captured by four quantities.

Angular Position & Displacement

The angular position \( \theta \) is the angle from a fixed reference line to a radial line in the body. A differential change \( d\theta \) is the angular displacement — a vector directed along the axis by the right-hand rule (1 rev = \(2\pi\) rad).

Angular Velocity & Acceleration

The angular velocity \( \omega \) is the time rate of change of angular position; the angular acceleration \( \alpha \) is the time rate of change of \( \omega \). When \( \omega \) is decreasing, \( \alpha \) opposes it (angular deceleration).

Fig. 16.3Point P on a body rotating about a fixed axis through O; its velocity is tangent to the circular path. (Original redrawn schematic.)

Angular velocity

\[ \omega = \frac{d\theta}{dt} \]

(16–1)

Angular acceleration

\[ \alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2} \]

(16–2), (16–3)

Differential relation

\[ \alpha\,d\theta = \omega\,d\omega \]

(16–4)

These three differential relations mirror the rectilinear-particle relations \( v=ds/dt \), \( a=dv/dt \), and \( a\,ds=v\,dv \) exactly — angular position, velocity, and acceleration play the roles of \( s \), \( v \), and \( a \).

Constant Angular Acceleration

When \( \alpha = \alpha_c \) is constant, integrating the differential relations yields a kinematic set identical in form to the constant-acceleration particle equations.

Constant angular acceleration

\[ \omega = \omega_0 + \alpha_c t \] \[ \theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha_c t^2 \] \[ \omega^2 = \omega_0^2 + 2\alpha_c(\theta - \theta_0) \]

(16–5), (16–6), (16–7)

Caution The constant-\(\alpha\) equations apply only when \(\alpha\) is truly constant. If \(\alpha\) is given as a function of time, angle, or \(\omega\), you must integrate the differential relations (16–1, 16–2, or 16–4) instead.

Motion of Point P

As the body rotates by \( d\theta \), point P at radius r moves an arc length \( ds = r\,d\theta \). Dividing by \( dt \) gives its speed; the velocity is always tangent to the circular path.

Velocity of a point (scalar)

\[ v = \omega r \]

(16–8)

Velocity of a point (vector)

\[ \mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}_P = \boldsymbol{\omega}\times\mathbf{r} \]

(16–9), (16–10)

The acceleration of P splits into tangential and normal components — the tangential component tracks the change in speed, the normal component the change in direction (always pointing toward the center O).

Acceleration components (scalar)

\[ a_t = \alpha r \qquad a_n = \omega^2 r \]

(16–11), (16–12)

Acceleration of a point (vector)

\[ \mathbf{a} = \boldsymbol{\alpha}\times\mathbf{r} - \omega^2\mathbf{r} \]

(16–14)

Acceleration magnitude

\[ a = \sqrt{a_n^{\,2} + a_t^{\,2}} \]

Meshed / Belted Bodies in Contact

Where two rotating bodies contact (meshed gears, or a belt over pulleys), the contact points share the same velocity and the same tangential acceleration, giving \( \omega_B r_B = \omega_C r_C \) and \( \alpha_B r_B = \alpha_C r_C \). Their normal accelerations differ, however, because each point follows a different-radius circle.

Important Points

Points on a body rotating about a fixed axis follow circular paths.
Once \( \omega \) and \( \alpha \) are known, the velocity and acceleration of any point follow immediately.
Velocity is always tangent to the path; tangential acceleration \( a_t=\alpha r \) measures speed change, normal acceleration \( a_n=\omega^2 r \) measures direction change.

16.4

Absolute Motion Analysis

A body in general plane motion translates and rotates at once. One way to analyze it is to tie the linear position of a point to the angular position of a line in the body using the geometry of the problem, then differentiate.

Absolute Motion Analysis

A method that relates a point's rectilinear coordinate \( s \) to a body's angular coordinate \( \theta \) through a geometric equation \( s = f(\theta) \). Differentiating once relates \( v \) to \( \omega \); differentiating again relates \( a \) to \( \alpha \). The chain rule must be applied throughout.

Procedure

Locate the point with a position coordinate s along its straight-line path, and the body's orientation with an angle \( \theta \).
Write \( s = f(\theta) \) from geometry/trigonometry.
Take the first time derivative for the \( v\)–\(\omega \) relation, the second for the \( a\)–\(\alpha \) relation, using the chain rule.

16.5

Relative-Motion Analysis: Velocity

General plane motion can be split into a translation of a base point plus a rotation about it, viewed from a translating (non-rotating) frame attached to the base point A. The position of B is \( \mathbf{r}_B = \mathbf{r}_A + \mathbf{r}_{B/A} \); differentiating gives the velocity relation.

Relative Velocity

The velocity \( \mathbf{v}_{B/A} \) of B as seen by an observer translating with A. Because that observer sees only rotation, this term represents circular motion about A: it has magnitude \( \omega\, r_{B/A} \) and direction perpendicular to \( \mathbf{r}_{B/A} \).

Relative-velocity equation (scalar form)

\[ \mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A} \]

(16–15)

Relative-velocity equation (vector form)

\[ \mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}\times\mathbf{r}_{B/A} \]

(16–16)

Fig. 16.4The velocity of B equals a translation with \( \mathbf{v}_A \) plus a rotation \( \mathbf{v}_{B/A} \) about base point A. (Original redrawn schematic.)

Choosing Points Pick A and B as points whose motion direction you know — pin connections to other bodies, or contact points with known motion. For a wheel rolling without slipping, the ground-contact point momentarily has zero velocity, and the center moves horizontally.

Course Note The vector form rarely fails you; the scalar form is faster but only if you can sketch the directions correctly. My advice: always draw the kinematic diagram first with assumed senses for unknowns. A negative answer just flips the assumed direction — it is not a mistake. — added from teaching experience

16.6

Instantaneous Center of Zero Velocity

If the base point is chosen as the one point that has zero velocity at this instant, the velocity equation collapses to pure rotation. That point is the instantaneous center of zero velocity (IC), and the whole body momentarily appears to rotate about it.

Instantaneous Center of Zero Velocity (IC)

The point — on the body or its extension — whose velocity is zero at the instant considered. Every other point moves in a circle about it, so \( v = \omega\,r \) where \( r \) is the distance from the IC to the point, and each velocity is perpendicular to its radial line from the IC.

Velocity about the IC

\[ \mathbf{v}_B = \boldsymbol{\omega}\times\mathbf{r}_{B/IC},\qquad v_B = \omega\, r_{B/IC} \]

Locating the IC — three cases

Known \( \mathbf{v}_A \) and \( \omega \): the IC lies along the line perpendicular to \( \mathbf{v}_A \) at A, at distance \( r_{A/IC} = v_A/\omega \).
Two non-parallel velocity directions: draw perpendiculars to \( \mathbf{v}_A \) and \( \mathbf{v}_B \); their intersection is the IC.
Two parallel velocities (known magnitudes): use proportional triangles, with \( r_{A/IC}=v_A/\omega \) and \( r_{B/IC}=v_B/\omega \).

Fig. 16.5The IC sits at the intersection of the perpendiculars to two known velocity directions. (Original redrawn schematic.)

Two Cautions

The IC is valid only at the instant considered — its location traces a curve called the centrode as the body moves.
The IC generally has non-zero acceleration, so never use it to find accelerations.

16.7

Relative-Motion Analysis: Acceleration

Differentiating the relative-velocity equation in time gives the relative-acceleration equation. The relative term now splits into tangential and normal parts, because to the translating observer B moves on a circular arc about A.

Relative-Acceleration Components

The acceleration of B relative to A has a tangential part of magnitude \( \alpha\,r_{B/A} \) perpendicular to \( \mathbf{r}_{B/A} \) (tracking change in relative speed), and a normal part of magnitude \( \omega^2 r_{B/A} \) directed from B toward A (tracking change in relative direction).

Relative-acceleration equation (component form)

\[ \mathbf{a}_B = \mathbf{a}_A + (\mathbf{a}_{B/A})_t + (\mathbf{a}_{B/A})_n \]

(16–17)

Relative-acceleration equation (vector form)

\[ \mathbf{a}_B = \mathbf{a}_A + \boldsymbol{\alpha}\times\mathbf{r}_{B/A} - \omega^2\mathbf{r}_{B/A} \]

(16–18)

Rolling without slipping — center of mass

\[ v_G = \omega r \qquad a_G = \alpha r \]

(16–19), (16–20)

Procedure

First do a velocity analysis to get \( \omega \) (and \( \mathbf{v}_A,\mathbf{v}_B \) if their paths are curved).
Draw the kinematic diagram. Points on curved paths need both tangential and normal acceleration components.
Apply the vector equation (16–18) and equate i and j components, or build the scalar diagram directly.

Course Note Students often forget the \( -\omega^2\mathbf{r}_{B/A} \) (normal) term because it's "invisible" in the velocity stage. Build the habit: the moment a point sits on a rotating body at finite \( \omega \), it has a normal relative-acceleration pointing inward. Velocity analysis must always come before acceleration analysis. — added from teaching experience

16.8

Relative-Motion Analysis Using Rotating Axes

When parts slide relative to one another, or when two points lie on different bodies, a frame that both translates and rotates is the cleanest tool. Attaching the moving frame to the body along which sliding occurs makes the relative motion easy to track.

Rotating Reference Frame

A frame with origin at base point A that translates with velocity \( \mathbf{v}_A \) and rotates with angular velocity \( \boldsymbol{\Omega} \) and angular acceleration \( \dot{\boldsymbol{\Omega}} \). Quantities measured by an observer riding this frame carry the subscript xyz.

Velocity — rotating frame

\[ \mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\Omega}\times\mathbf{r}_{B/A} + (\mathbf{v}_{B/A})_{xyz} \]

(16–26)

Acceleration — rotating frame

\[ \mathbf{a}_B = \mathbf{a}_A + \dot{\boldsymbol{\Omega}}\times\mathbf{r}_{B/A} + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r}_{B/A}) + 2\boldsymbol{\Omega}\times(\mathbf{v}_{B/A})_{xyz} + (\mathbf{a}_{B/A})_{xyz} \]

(16–29)

Coriolis Acceleration

The term \( 2\boldsymbol{\Omega}\times(\mathbf{v}_{B/A})_{xyz} \). It captures the interaction between a point's motion relative to the rotating frame and the frame's own rotation, and is always perpendicular to both \( \boldsymbol{\Omega} \) and the relative velocity. It is essential whenever rotating reference frames are used — for rockets, long-range projectiles, and earth-rotation effects.

Term-by-Term Meaning (Acceleration)

\( \mathbf{a}_A \): absolute acceleration of the moving-frame origin.
\( \dot{\boldsymbol{\Omega}}\times\mathbf{r}_{B/A} \): effect of the frame's angular acceleration.
\( \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r}_{B/A}) \): effect of the frame's angular velocity (centripetal-type).
\( 2\boldsymbol{\Omega}\times(\mathbf{v}_{B/A})_{xyz} \): the Coriolis acceleration (interaction term).
\( (\mathbf{a}_{B/A})_{xyz} \): acceleration of B as seen from the rotating frame.

Course Note Decision rule for the whole chapter: if the two points stay on the same rigid body, use translating axes (16.5–16.7). The moment there's sliding between members, or your two points live on different bodies, switch to rotating axes (16.8) and respect the Coriolis term. Getting this choice right is 90% of the battle. — added from teaching experience

Problem-solving strategy

Classify translation, fixed-axis rotation, general plane motion, or sliding in a rotating frame before selecting equations.
Declare fixed \(x\)-\(y\) axes and positive angular sense; mark every known path direction on a kinematic diagram.
Choose a base point with known motion, then write complete vector relations before resolving components.
Complete velocity analysis before acceleration analysis because the normal term requires the current \(\omega\).
Audit rigid-length, no-slip, slot, and contact compatibility; use the IC for velocity only.

Integrated applications

EXAMPLE 1Decelerating inspection turntable

An inspection turntable rotates counterclockwise at \(8.00 \mathrm{rad/s}\) while its angular acceleration is \(3.00 \mathrm{rad/s^2}\) clockwise. Find the velocity and acceleration of a marker at the rightmost edge, \(0.240 \mathrm m\) from the fixed axis.

Recognition strategy

The fixed center and requested instantaneous point motion identify fixed-axis kinematics, not a force problem; tangential and normal acceleration components are both required.

Given information

Take \(+x\) right, \(+y\) up, and counterclockwise \(+\mathbf k\); \(\mathbf r=0.240\mathbf i \mathrm m\), \(\omega=8.00 \mathrm{rad/s}\), and \(\alpha=-3.00 \mathrm{rad/s^2}\).

Model and interpretation

Model the table as rigid about a fixed axis. The marker velocity is tangent, while its acceleration combines angular-speed change with inward curvature.

Step-by-step solution

1. Select relations. For a point fixed in a body rotating about a fixed axis, use \(\mathbf v=\boldsymbol{\omega} \times\mathbf r\) and \(\mathbf a=\boldsymbol{\alpha} \times\mathbf r-\omega^2\mathbf r\).

2. Compute velocity. The cross product gives \(\mathbf v=(8.00\mathbf k) \times(0.240\mathbf i)=1.92\mathbf j \mathrm{m/s}\), so the right-edge marker moves upward.

3. Resolve acceleration. Tangential acceleration is \(-0.720\mathbf j \mathrm{m/s^2}\), and normal acceleration is \(-15.36\mathbf i \mathrm{m/s^2}\), directed toward the axis.

4. Form the resultant. Thus \(\mathbf a=(-15.36\mathbf i-0.720\mathbf j) \mathrm{m/s^2}\) and \(|\mathbf a|=15.38 \mathrm{m/s^2}\).

Verification

Units: Multiplying radians per second by metres gives metres per second, and each acceleration term has metres per second squared.

Signs and direction: Counterclockwise motion makes the rightmost point move upward; clockwise angular acceleration makes its tangential acceleration downward, while normal acceleration points left toward the axis.

Plausibility: The normal component \(\omega^2r=15.36 \mathrm{m/s^2}\) greatly exceeds \(|\alpha|r=0.720 \mathrm{m/s^2}\), consistent with rapid rotation and modest deceleration.

Final result: \(\boxed{\mathbf v=1.92\mathbf j \mathrm{m/s},\quad \mathbf a=(-15.36\mathbf i-0.720\mathbf j) \mathrm{m/s^2}}\)

Engineering interpretation

The nearly radial acceleration shows why turntable fixtures are often governed by centripetal loading even during controlled slowdown; attachment capacity still requires a separate force and safety assessment.

EXAMPLE 2Guided packaging-machine link

A rigid link has \(\mathbf r_{B/A}=(0.600\mathbf i+0.800\mathbf j) \mathrm m\). End A moves right at \(1.50 \mathrm{m/s}\) with acceleration \(0.400\mathbf i \mathrm{m/s^2}\), while end B is constrained to a vertical guide. Determine \(\omega\), \(\alpha\), and B's velocity and acceleration.

Recognition strategy

Two points on one rigid link have known path directions, so translating-frame relative velocity followed by relative acceleration supplies the needed compatibility equations.

Given information

Use fixed \(+x\) right, \(+y\) up, counterclockwise positive; \(\mathbf v_A=1.50\mathbf i \mathrm{m/s}\), \(\mathbf a_A=0.400\mathbf i \mathrm{m/s^2}\), with \(\mathbf v_B=v_B\mathbf j\) and \(\mathbf a_B=a_B\mathbf j\).

Model and interpretation

The link is rigid and the guides are ideal, so B has no horizontal velocity or acceleration. Kinematics determines motion without using mass or force.

Step-by-step solution

1. Write velocity compatibility. Substitute the guide directions into \(\mathbf v_B=\mathbf v_A+\omega\mathbf k \times\mathbf r_{B/A}\), keeping both scalar components visible.

2. Solve angular velocity. The components are \(0=1.50-0.800\omega\) and \(v_B=0.600\omega\), giving \(\omega=1.875 \mathrm{rad/s}\) and \(v_B=1.125 \mathrm{m/s}\) upward.

3. Write acceleration compatibility. Use \(a_B\mathbf j=0.400\mathbf i+\alpha\mathbf k \times\mathbf r_{B/A}-(1.875)^2\mathbf r_{B/A}\); the last term is mandatory normal relative acceleration.

4. Solve acceleration components. From x, \(0=0.400-0.800\alpha-2.109\), so \(\alpha=-2.137 \mathrm{rad/s^2}\); from y, \(a_B=0.600\alpha-2.813=-4.095 \mathrm{m/s^2}\).

Verification

Units: Each velocity term is angular rate times length, while both \(\alpha r\) and \(\omega^2r\) correctly reduce to metres per second squared.

Signs and direction: Positive \(\omega\) is counterclockwise, negative \(\alpha\) is clockwise, and negative \(a_B\) means B accelerates downward despite currently moving upward.

Plausibility: Substituting \(\omega=1.875\) makes the horizontal relative velocity \(-1.50 \mathrm{m/s}\), exactly cancelling A's motion as the vertical guide requires.

Final result: \(\boxed{\omega=1.875 \mathrm{rad/s}, \alpha=-2.137 \mathrm{rad/s^2}, \mathbf v_B=1.125\mathbf j \mathrm{m/s}, \mathbf a_B=-4.095\mathbf j \mathrm{m/s^2}}\)

Engineering interpretation

B is moving upward but slowing at this instant. That distinction matters when sizing a guide actuator, although joint clearance and compliance would need inclusion in a production mechanism model.

EXAMPLE 3Sensor carriage in a rotating radial slot

A sensor carriage slides outward in a radial slot at \(r=0.450 \mathrm m\), \(\dot r=0.700 \mathrm{m/s}\), and \(\ddot r=-0.300 \mathrm{m/s^2}\). The arm rotates counterclockwise with \(\Omega=2.50 \mathrm{rad/s}\) and \(\dot\Omega=-0.800 \mathrm{rad/s^2}\). Find the carriage velocity and acceleration in radial-transverse components.

Recognition strategy

Sliding relative to a rotating arm is the cue for rotating axes; the acceleration must include frame-normal, frame-tangential, relative, and Coriolis terms.

Given information

Use \(\mathbf e_r\) outward and \(\mathbf e_\theta\) counterclockwise; the pivot is fixed, and all supplied values describe the same instant.

Model and interpretation

Treat the carriage as a point constrained to an ideal radial slot. The arm-fixed frame rotates, while the carriage has nonzero relative radial motion.

Step-by-step solution

1. Select rotating-frame relations. For a fixed pivot and radial sliding, \(\mathbf v=\dot r\mathbf e_r+\Omega r\mathbf e_\theta\) and \(\mathbf a=(\ddot r-\Omega^2r)\mathbf e_r+(\dot\Omega r+2\Omega\dot r)\mathbf e_\theta\).

2. Compute velocity. The transverse contribution is \((2.50)(0.450)=1.125 \mathrm{m/s}\), so \(\mathbf v=0.700\mathbf e_r+1.125\mathbf e_\theta \mathrm{m/s}\).

3. Compute radial acceleration. Relative and frame-normal parts combine as \(a_r=-0.300-(2.50)^2(0.450)=-3.1125 \mathrm{m/s^2}\).

4. Compute transverse acceleration. The arm term is \(-0.800(0.450)=-0.360\), while Coriolis contributes \(2(2.50)(0.700)=3.50\), giving \(a_\theta=3.140 \mathrm{m/s^2}\) and \(|\mathbf a|=4.42 \mathrm{m/s^2}\).

Verification

Units: Both velocity components have metres per second, and every acceleration contribution has metres per second squared after angular units are treated as dimensionless.

Signs and direction: Negative radial acceleration points inward; positive transverse acceleration is counterclockwise because outward sliding creates a large positive Coriolis contribution.

Plausibility: If relative sliding vanished, both \(\dot r\) and \(\ddot r\) terms would disappear; here the independently computed Coriolis term of \(3.50 \mathrm{m/s^2}\) explains the transverse dominance.

Final result: \(\boxed{\mathbf v=(0.700\mathbf e_r+1.125\mathbf e_\theta) \mathrm{m/s},\quad \mathbf a=(-3.113\mathbf e_r+3.140\mathbf e_\theta) \mathrm{m/s^2}}\)

Engineering interpretation

The Coriolis term is larger than the arm's tangential term and reverses the net transverse tendency; guide reactions and allowable bearing speed must be evaluated in a subsequent kinetics and component-design analysis.

★

Chapter at a Glance

Method Selector

Translation: all points share velocity and acceleration — use particle kinematics.
Fixed-axis rotation: \( v=\omega r \), \( a_t=\alpha r \), \( a_n=\omega^2 r \); constant-\(\alpha\) set when applicable.
Absolute motion: write \( s=f(\theta) \) and differentiate.
Relative velocity (translating): \( \mathbf{v}_B=\mathbf{v}_A+\boldsymbol{\omega}\times\mathbf{r}_{B/A} \).
IC: velocities only — find the zero-velocity point, then \( v=\omega r \).
Relative acceleration (translating): \( \mathbf{a}_B=\mathbf{a}_A+\boldsymbol{\alpha}\times\mathbf{r}_{B/A}-\omega^2\mathbf{r}_{B/A} \).
Rotating axes: sliding contacts / points on different bodies — include the Coriolis term \( 2\boldsymbol{\Omega}\times(\mathbf{v}_{B/A})_{xyz} \).

Before moving on

Can you declare axes and distinguish velocity constraints from acceleration constraints?
Can you identify when an IC or Coriolis term is valid?
Can you test no-slip and rigid-link compatibility from your computed vectors?