Tags: static equilibrium, support reactions, free body diagram, moment, torque, collisions, momentum conservation, energy conservation, coefficient of restitution, angular momentum, rotational inertia, moment of inertia, oscillations, simple harmonic motion, SHM, pendulum, damping, ENGR 216, PHYS 216, Texas A&M
The later modules cover classical mechanics: balancing forces and moments on static structures, analysing collisions with conservation laws, computing rotational quantities, and characterising oscillatory systems. The through-line is applying Newton's laws in various forms (translational, rotational, harmonic) and knowing which conservation principles hold under which conditions.
Static equilibrium
A body is in static equilibrium when both the net force and net moment (torque) about any point are zero. ΣF = 0 and ΣM = 0.
Free body diagram (FBD)
A sketch isolating a single body with all external forces and moments drawn. The starting point for every statics and dynamics problem.
Support reactions
Forces and/or moments exerted by supports (pins, rollers, fixed walls) on a structure. A roller provides one reaction (perpendicular to the surface). A pin provides two (horizontal and vertical). A fixed support provides two forces and one moment.
Moment (torque)
The tendency of a force to cause rotation about a point: M = r × F, or in 2D, M = F·d, where d is the perpendicular distance from the point to the line of action of the force.
Conservation of linear momentum
In the absence of external forces, the total momentum of a system is constant: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'.
Coefficient of restitution (e)
The ratio of relative speed after collision to relative speed before collision: e = (v₂' − v₁') / (v₁ − v₂). Perfectly elastic: e = 1. Perfectly inelastic: e = 0.
Kinetic energy lost in a collision
ΔKE = KE_before − KE_after. In an inelastic collision, some kinetic energy converts to heat, sound, or deformation.
Moment of inertia (rotational inertia)
Resistance to angular acceleration, analogous to mass for linear motion. Depends on mass distribution: I = Σ mᵢrᵢ² or I = ∫ r² dm.
Angular momentum
L = Iω for a rigid body. Conserved when no external torque acts on the system.
Simple harmonic motion (SHM)
Oscillatory motion where the restoring force is proportional to displacement: F = −kx. Period: T = 2π√(m/k).
Simple pendulum
A mass on a string swinging under gravity. For small angles: T = 2π√(L/g), where L is the string length.
Damped oscillations
Oscillations where energy is lost to friction or drag. Three regimes: underdamped (oscillates with decaying amplitude), critically damped (returns to equilibrium fastest without oscillating), overdamped (returns slowly without oscillating).
The method is always the same:
Draw the free body diagram with all forces and their points of application.
Choose a coordinate system.
Write ΣFₓ = 0, ΣFᵧ = 0, and ΣM_point = 0.
Choose the moment point strategically to eliminate unknowns (take moments about a point where an unknown force acts, so it drops out).
Beam with distributed weight:
A uniform beam of mass m and length L, supported at points A and B, has its weight W = mg acting at the centre of mass (L/2 from either end for a uniform beam).
For a simply supported beam (pin at A, roller at B):
Pin at A gives reactions Aₓ and Aᵧ.
Roller at B gives reaction Bᵧ only.
If only vertical loads are present, Aₓ = 0.
ΣM_A = 0 lets you solve for Bᵧ.
ΣFᵧ = 0 lets you solve for Aᵧ.
Cables and tension:
For a bar held by two cables at angles, resolve each tension into horizontal and vertical components, then apply equilibrium. If the bar is horizontal and the cables are at different angles, the tensions will generally be different.
The moment of a force F about a point O in 2D:
M_O = F × d (perpendicular distance)
Or using the cross product in component form:
M_O = (rₓ × Fᵧ) − (rᵧ × Fₓ)
where r is the position vector from O to the point of force application.
Angle for zero moment:
The moment about a point is zero when the line of action of the force passes through that point, i.e., when the perpendicular distance d = 0.
Elastic collisions conserve both momentum and kinetic energy.
For a 1D elastic collision between m₁ (velocity v₁) and m₂ (initially at rest):
v₁' = v₁(m₁ − m₂) / (m₁ + m₂)
v₂' = 2m₁v₁ / (m₁ + m₂)
Inelastic collisions conserve momentum but not kinetic energy. Use both conservation of momentum and the coefficient of restitution:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
e = (v₂' − v₁') / (v₁ − v₂)
Two equations, two unknowns (v₁' and v₂').
Superball problem (Problem 14):
When a small ball (soccer ball) sits atop a large ball (yoga ball) and both are dropped, the small ball bounces to a much greater height. This is a two-stage collision: the large ball hits the ground first and reverses, then the large ball hits the small ball. If both balls are elastic and the large ball is much heavier:
After ground collision, the large ball has velocity +v (upward).
The small ball still has velocity −v (downward).
Relative approach speed is 2v.
For m_large >> m_small, the small ball bounces off at approximately +3v.
Maximum height: h_bounce = (3v)² / (2g) = 9v² / (2g) = 9 × h_drop.
The small ball can bounce up to 9 times the drop height in the ideal case.
Newton's second law for rotation:
Στ = Iα
where τ is torque, I is moment of inertia, and α is angular acceleration.
Conservation of angular momentum:
When no external torque acts: I₁ω₁ = I₂ω₂.
Classic example: an ice skater pulling in their arms reduces I, so ω increases to conserve L.
Flywheels and rotating platforms:
For a person stepping onto a rotating platform, the system's angular momentum is conserved:
I_platform × ω_initial = (I_platform + I_person) × ω_final
Common moments of inertia to know:
Solid disc/cylinder (about central axis): I = ½mR²
Thin ring/hoop: I = mR²
Solid sphere: I = ⅖mR²
Thin rod (about centre): I = (1/12)mL²
Thin rod (about end): I = ⅓mL²
Point mass at distance r: I = mr²
Mass-spring system (SHM):
T = 2π√(m/k)
where m is the oscillating mass and k is the spring constant.
Simple pendulum (small angle):
T = 2π√(L/g)
Period depends only on length and gravitational acceleration, not on mass or amplitude (for small angles).
Physical (compound) pendulum:
T = 2π√(I / (mgh))
where I is the moment of inertia about the pivot and h is the distance from pivot to centre of mass.
Diving board as oscillator:
A diving board behaves like a cantilever beam. The effective spring constant depends on the board's material properties and geometry. The person's mass on the end of the board determines the oscillation period via T = 2π√(m/k_eff).
Damping classifications:
Damping is characterised by the damping ratio ζ = c / (2√(mk)):
ζ < 1: underdamped (oscillates with exponentially decaying amplitude)
ζ = 1: critically damped (fastest return to equilibrium, no oscillation)
ζ > 1: overdamped (slow exponential return, no oscillation)
Equilibrium conditions: ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0
Moment about a point (2D): M = r × F sin(θ) or M = F × d_perp
Linear momentum conservation: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Coefficient of restitution: e = (v₂' − v₁') / (v₁ − v₂)
Rotational Newton's second law: Στ = Iα
Angular momentum conservation: I₁ω₁ = I₂ω₂ (no external torque)
SHM period: T = 2π√(m/k)
Simple pendulum period: T = 2π√(L/g)
Damping ratio: ζ = c / (2√(mk))
⚠️ When summing moments, be consistent with sign convention (e.g., counterclockwise positive). A single sign error will propagate through the entire solution.
⚠️ For collision problems, always check whether the collision is elastic or inelastic before choosing your equations. Elastic means KE is conserved. Inelastic means it is not.
⚠️ The coefficient of restitution formula uses the speeds of separation over speeds of approach. Watch the sign convention: it should come out positive.
⚠️ The simple pendulum formula T = 2π√(L/g) is only valid for small angles (typically < 15°). If the problem states large angles, this approximation breaks down.
⚠️ Angular momentum is conserved only when there is no net external torque. Gravity acting at a pivot does not create torque about that pivot, but gravity acting on a point away from the pivot does.
⚠️ Know the common moments of inertia from memory. Exams rarely give them to you.
⚠️ In the stacked-ball drop problem, the key insight is that the relative velocity of approach doubles because the large ball has reversed direction before hitting the small ball.
Q: A 4 m uniform beam (mass 20 kg) is supported by a pin at A (left end) and a roller at B (3 m from A). What is the reaction at B?
A: Weight = 20 × 9.81 = 196.2 N acts at 2 m from A. Taking moments about A: B × 3 = 196.2 × 2, so B = 130.8 N upward.
Q: A 2 kg ball moving at 5 m/s collides head-on with a stationary 3 kg ball. If e = 0.8, what are the final velocities?
A: Momentum: 2(5) = 2v₁' + 3v₂', so 10 = 2v₁' + 3v₂'. Restitution: 0.8 = (v₂' − v₁')/(5 − 0), so v₂' − v₁' = 4. Solving: v₂' = 3.6 m/s, v₁' = −0.4 m/s (bounces back).
Q: A disc (I = 0.5 kg·m²) spins at 10 rad/s. A ring (I = 0.3 kg·m²) is dropped onto it. What is the final angular speed?
A: Angular momentum conserved: 0.5 × 10 = (0.5 + 0.3) × ω_f, so ω_f = 5/0.8 = 6.25 rad/s.
Q: What happens to the period of a simple pendulum if you double its length?
A: T = 2π√(L/g). Doubling L multiplies T by √2 ≈ 1.414. The period increases by about 41%.
Q: How do you classify damping if the system oscillates but the amplitude decreases over time?
A: Underdamped (ζ < 1). The system still oscillates, but each successive peak is smaller than the last.
static equilibrium, free body diagram, FBD, support reactions, pin support, roller support, fixed support, moment of a force, torque, sum of moments, elastic collision, inelastic collision, perfectly inelastic, coefficient of restitution, kinetic energy loss, stacked ball problem, superball, angular momentum, conservation of angular momentum, moment of inertia, rotational inertia, flywheel, angular velocity, angular acceleration, simple harmonic motion, SHM, spring-mass system, Hooke's law, simple pendulum, physical pendulum, compound pendulum, damped oscillation, underdamped, overdamped, critically damped, damping ratio, diving board oscillation, cantilever, ENGR 216, PHYS 216, Texas A&M, experimental physics lab