CELE Engineering Mechanics — Dynamics: KinematicsRevision Notes

Formulas

Exam Tips

• List knowns and unknowns first. Match them to the equation that contains exactly those variables — usually one equation is sufficient.
• The equation v² = v₀² + 2as is the fastest route whenever time is not given and not needed.
• For free-fall problems, draw a sketch with upward as positive; set v = 0 at the highest point.
• For a body decelerating to rest: v = 0 is your final velocity — plug it in and solve for a or s directly.
• Check your answer with a second equation (the 'check' step) — this catches arithmetic errors and earns confidence.

Key Points

• Rectilinear motion is one-dimensional; displacement s, velocity v, and acceleration a all act along the same straight line.
• The fundamental kinematic definitions are: v = ds/dt (velocity is the time-rate of change of position) and a = dv/dt = v(dv/ds) (acceleration is the time-rate of change of velocity).
• For CONSTANT acceleration, the three governing equations are: v = v₀ + at ; s = v₀t + ½at² ; v² = v₀² + 2as. Memorize all three — exam problems often give two knowns and ask for a third.
• FREE FALL is a special case of constant acceleration: a = g = 9.81 m/s² directed downward. Taking upward as positive, a = −9.81 m/s².
• For VARIABLE acceleration, integrate: v = ∫a dt and s = ∫v dt, or use a = v(dv/ds) to eliminate time.
• Always define a positive direction at the start of every problem and be consistent throughout. Reversal of direction occurs when v = 0.
• Average velocity = total displacement / total time; average speed = total distance / total time (not the same when the body reverses direction).

Definitions

Section Title

Rectilinear Motion — Motion Along a Straight Line

Common Mistakes

• Using g = 10 m/s² instead of 9.81 m/s² — use 9.81 unless the problem explicitly allows approximation.
• Forgetting to define a positive direction and mixing up signs mid-solution, leading to sign errors in the final answer.
• Confusing distance (scalar) with displacement (vector) when a body reverses direction — each segment must be analysed separately for distance.
• Applying the constant-acceleration equations when acceleration is variable — always check whether a is constant before using the trio.
• Dropping units during calculation and presenting an answer without units — always carry and check units.

Formulas

Exam Tips

• Draw the trajectory and label vₓ, vᵧ, H, R, and θ before writing any equation. This prevents formula mix-ups.
• For a ball thrown horizontally from a height (θ = 0°), use y = −½gt² for vertical and x = v₀t for horizontal — there is no initial vertical velocity.
• If the problem asks for the angle or speed at which a projectile hits the ground, compute vₓ and vᵧ at landing, then use the vector magnitude formula.
• When two angles are possible for a given range, the shallower angle gives a shorter time of flight and lower height; the steeper angle gives the opposite.
• Always double-check: at t = T, y should return to zero (or the specified landing height). Use this as a built-in check.

Key Points

• A projectile moves under gravity only (no air resistance). The horizontal and vertical motions are INDEPENDENT and are analyzed separately.
• Horizontal: constant velocity (aₓ = 0). x = v₀cosθ · t
• Vertical: constant downward acceleration g. y = v₀sinθ · t − ½gt²
• The two components share a single time variable t — this is the bridge between horizontal and vertical equations.
• For launch and landing at the SAME elevation, the simplified formulas apply: Range R = v₀²sin2θ/g ; Max height H = v₀²sin²θ/(2g) ; Time of flight T = 2v₀sinθ/g.
• Maximum range occurs at θ = 45°, giving R_max = v₀²/g.
• Complementary angles (e.g., 30° and 60°) produce equal ranges but different heights and times of flight.
• At the apex (maximum height), vertical velocity vᵧ = 0; horizontal velocity vₓ = v₀cosθ remains unchanged throughout.
• For problems where launch and landing heights differ, do NOT use the simplified range formula — instead write the full y(t) equation, set y equal to the landing height, and solve the resulting quadratic in t.

Definitions

Section Title

Projectile Motion — Two-Dimensional Kinematics

Common Mistakes

• Using sin2θ for height and sin²θ for range — they are REVERSED. Range uses sin2θ; height uses sin²θ.
• Applying R = v₀²sin2θ/g when the landing elevation differs from the launch elevation — this formula is only valid for equal heights.
• Forgetting that vₓ is CONSTANT throughout the flight — only the vertical component changes due to gravity.
• Using the wrong angle in the formula when the angle is given from the vertical instead of the horizontal — always clarify the reference direction.
• Incorrect quadratic solution when finding t from the y(t) equation — always identify the physically meaningful (positive) root.

Formulas

Exam Tips

• Any time the problem states rpm or revolutions, your first step is always unit conversion: ω = 2πN/60 and θ = 2π(rev).
• The pattern 'from N₁ rpm to N₂ rpm in n revolutions, find α' maps directly to ω² = ω₀² + 2αθ — no time needed.
• For a point on the rim, always compute BOTH aₙ and aₜ separately, then find the total: a = √(aₙ² + aₜ²).
• Belt-and-pulley or gear problems: connected wheels have the same tangential velocity (v₁ = v₂) at the contact point — use v = rω to relate their angular velocities.
• Memorise the analogy table: s↔θ, v↔ω, a↔α. Any rectilinear formula you know applies to rotation by substitution.

Key Points

• Rotational kinematics describes motion about a fixed axis using angular quantities: angle θ (rad), angular velocity ω (rad/s), and angular acceleration α (rad/s²).
• The rotational equations are EXACT ANALOGUES of the rectilinear equations — replace s with θ, v with ω, and a with α.
• Constant-α equations: ω = ω₀ + αt ; θ = ω₀t + ½αt² ; ω² = ω₀² + 2αθ
• Conversion between rpm and rad/s: ω (rad/s) = 2πN/60, where N is in rpm. Always convert before substituting.
• Conversion: 1 revolution = 2π radians. So θ (rad) = 2π × (number of revolutions).
• For a point at radius r on a rotating body: tangential displacement s = rθ ; tangential velocity v = rω ; tangential acceleration aₜ = rα.
• Normal (centripetal) acceleration: aₙ = v²/r = rω². Directed toward the center of rotation. Not caused by α — it exists whenever ω ≠ 0.
• Total linear acceleration of a point: a_total = √(aₜ² + aₙ²). It is vector sum of tangential and normal components.
• For a wheel that decelerates to rest (ω = 0 is the final condition), the time-independent equation ω² = ω₀² + 2αθ is the fastest solution.

Definitions

Section Title

Rotational (Angular) Kinematics

Common Mistakes

• Failing to convert rpm to rad/s before substituting into kinematic equations — this is the single most common error in rotational problems.
• Using degrees instead of radians for θ in the kinematic equations — must use radians.
• Confusing aₙ (centripetal, directed inward) with aₜ (tangential, directed along the path) — they are perpendicular and must be combined as a vector sum.
• Treating revolutions directly as θ in the equations — multiply by 2π first: θ(rad) = 2π × rev.
• Assuming aₙ = 0 at constant speed — normal acceleration exists as long as ω ≠ 0, regardless of whether speed is changing.

Formulas

Exam Tips

• Variable acceleration problems on board exams are less frequent but more discriminating. Read the problem carefully: if a is given as an equation involving t, s, or v, it is variable.
• Check units of the given acceleration expression to ensure dimensional consistency before integrating.
• For a = f(t): straightforward double integration. For a = f(s): use energy-like relation v dv = a ds. For a = f(v): separate variables dt = dv/a.

Key Points

• When acceleration is NOT constant, the constant-acceleration trio (v = v₀ + at, etc.) CANNOT be used.
• Instead, use integration: given a(t), integrate to get v(t), then integrate v(t) to get s(t).
• Given a(s) — acceleration as a function of position — use a = v(dv/ds), rearrange to v dv = a(s) ds, then integrate both sides.
• Given a(v) — acceleration as a function of velocity — use dt = dv/a(v) or ds = v dv/a(v), then integrate.
• Initial conditions (v₀ at t = 0, s₀ at t = 0) are essential for evaluating constants of integration.
• Variable acceleration problems typically appear in the more challenging board exam items, but the technique — identify the form a = f(t), f(s), or f(v), then choose the correct differential equation — is systematic.

Definitions

Section Title

Variable Acceleration and Calculus-Based Kinematics

Common Mistakes

• Applying v = v₀ + at to a variable-acceleration problem — always check whether a is constant first.
• Forgetting to apply initial conditions when evaluating the constant of integration.
• Choosing the wrong differential relation — if a = f(s), use v dv = a ds, not dt = dv/a.