Exploring Mechanics / 10. Capstones

10. Real-World Capstone Simulations

Larger multi-topic projects, each combining several earlier sections into one working model.

Bowling Physics

Make a prediction first: a bowling ball is thrown down the lane without any spin at all, sliding at first before it starts to roll normally. Does it slide down the whole lane, or does something change partway?

Rolling without slippingFrictionMoment of inertiaMomentumCollisions

While skidding: a = −μg (linear), α = 5μg/(2R) (angular)   ·   rolling begins at v = 5v₀/7

v₀ — the ball's initial thrown speed with no spin; μ — kinetic friction between ball and lane; R — ball radius
Friction acts backward on the ball's motion but forward as a torque, simultaneously slowing its slide and spinning it up, until the surface speed at the contact point reaches zero — the rolling condition v = ωR

Widget — Skid-to-Roll Transition and Pin Impact

Initial speed, v₀6.0 m/s
Friction, μ0.10
Time to start rolling—
Distance skidded first—
Speed once rolling, 5v₀/7—
Pin speed after impact—
For math enthusiasts: where 5v₀/7 comes from

With v(t) = v₀ − μgt and ω(t) = (5μg/2R)t (starting from zero spin), rolling begins when v = ωR: v₀ − μgt = (5μg/2)t, giving t = v₀/(3.5μg). Substituting back, vroll = v₀ − μg·v₀/(3.5μg) = v₀(1 − 1/3.5) = 5v₀/7 — independent of μ and R entirely, though the time and distance to reach it are not.

Widget — The Hook: Skid → Roll (Top-Down)

Real bowling shots almost always have side spin. Watch how the ball travels arrow-straight while skidding on the oiled part of the lane, then breaks sharply toward the pocket the moment it grips and starts true rolling.

Bicycle Mechanics

Make a prediction first: you're climbing a hill and shift to a lower gear at the same pedaling cadence. Your wheel spins slower — but does that also mean you're now pushing the pedals with less force?

GearsTorquePowerHill climbingRolling resistance

Wheel speed = cadence × (front teeth / rear teeth) × wheel radius   ·   Power = Force × speed

Gear ratio = front chainring teeth / rear cog teeth — a lower gear (smaller ratio) trades wheel speed for more torque at the same rider power and cadence
On a hill, the rider's fixed power output caps top speed once gravity's pull along the slope grows large enough

Widget — Gearing, Power, and Hill-Climbing Speed

Cadence80 rpm
Front chainring teeth34
Rear cog teeth18
Hill slope4°
Rider power200 W
Wheel speed from cadence/gearing—
Power-limited max speed on this hill—
Actual speed (the smaller of the two)—
Rear-wheel torque—

This is the same gear-ratio tradeoff from Section 5's Gears tab, now driving a full hill-climb speed calculation.

Roller-Coaster Design

Make a prediction first: you're designing a coaster with one hill leading into one vertical loop. If the hill is exactly twice as tall as the loop, is that automatically enough height for the cart to safely complete the loop?

Energy conservationLoopsNormal forceFrictional lossSafety constraints

Minimum loop-top speed: vmin = √(gr)   ·   Available energy after losses: (1 − loss%)·mgH

H — hill height; r — loop radius (so the loop's top sits at height 2r)
"Loss%" bundles track friction and air drag as a simple fraction of the hill's potential energy that never makes it to kinetic energy

Widget — Hill Height vs. Loop Safety

Hill height, H25 m
Loop radius, r8 m
Friction + drag loss10%
Speed reaching loop top—
Minimum required speed—

—

This combines the loop condition from Section 6's Vertical Circles tab with the energy-loss ideas from Section 3's Roller Coaster Physics tab.

Widget — Ride the Coaster: Velocity, Forces, and G's at Every Point

This runs the full track profile at once — a lift hill into a valley into a smaller hill — and tracks apparent weight continuously, color-coding the track itself by G-force as the cart passes over it.

Bridge Resonance

Make a prediction first: a pedestrian bridge is jostled at a rate far from its natural sway frequency, then at a rate very close to it, with the same-sized push each time. Does the bridge respond the same way both times?

Mass-spring modelNatural frequencyDampingDriven oscillationResonance

Natural frequency: ωn = √(k/m)   ·   Amplitude ratio: 1/√((1−r²)² + (2ζr)²),  r = ω/ωn

k — effective structural stiffness; m — effective oscillating mass; ζ — damping ratio (structural damping, dampers, etc.)
r — ratio of driving frequency to natural frequency; the closer r is to 1, the larger the amplification, especially for small ζ

Widget — Resonance Curve

r = ω/ωₙ amplitude (relative)
Stiffness, k500000 N/m
Oscillating mass, m20000 kg
Damping ratio, ζ0.05
Driving frequency (footsteps/wind)0.80 Hz
Natural frequency, ωn/2π—
Amplification factor—

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Real-world case: the Millennium Bridge (London, 2000)

  1. Pedestrians' side-to-side sway synchronized with the bridge's lateral natural frequency, producing exactly this kind of amplitude growth, and the bridge was closed within days to add dampers — increasing ζ directly, the same lever this widget exposes.

Same resonance mechanism as Section 8's Resonance tab, applied here to a structure instead of a single mass on a spring.

Widget — Watch the Deck Resonate

The deck's actual swing amplitude here follows the resonance-response formula continuously, so as you slide the driving frequency toward the natural frequency, watch the bridge visibly whip up in real time.

Bow Energy Model

Make a prediction first: drawing a bow twice as far back, with the same bow, stores how much more elastic energy — twice as much, or more?

Elastic energyDraw lengthProjectile massEfficiency

Stored energy (simplified linear-draw model): PE = ½kd²   ·   Launch speed: v = √(2·PE·η/m)

k — effective draw stiffness; d — draw length; η — energy-transfer efficiency (some energy stays in the limbs and string rather than reaching the arrow)
m — arrow mass

Widget — Force-Draw Curve and Arrow Speed

draw (m) force (N)
Draw stiffness, k300 N/m
Draw length, d0.60 m
Arrow mass25 g
Transfer efficiency, η80%
Stored elastic energy—
Arrow launch speed—

This is an educational model of elastic-energy transfer, not an instructional guide for building or drawing a real bow — real bow force-draw curves are noticeably non-linear.

Catapult Motion Model

Make a prediction first: a longer catapult arm, given the same driving torque, releases the projectile at a higher speed. True or false?

TorqueRotational inertiaEnergy transferProjectile motionNumerical methods

Angular acceleration: α = τ/I,  I = mL²   ·   Launch speed: v = ωL, with ω from ω² = 2αθrelease

τ — driving torque; L — arm length; m — projectile mass (dominating the arm's own inertia in this simplified model)
θrelease — the swing angle at which the projectile is released, as an abstract parameter rather than a mechanical latch design

Widget — Launch Speed and Trajectory

This is a numerical model of abstract mechanics, not a guide for constructing or optimizing a physical device.

Arm length, L2.0 m
Driving torque, τ400 N·m
Projectile mass, m2.0 kg
Release angle (swing traveled)60°
Launch speed—
Range—
For math enthusiasts: why the prediction above is false

A longer arm increases moment of inertia I = mL² quadratically, while it only increases the lever-arm advantage (and the tip's speed for a given ω) linearly. Since angular acceleration α = τ/I ∝ 1/L², and tip speed v = ωL with ω ∝ √α ∝ 1/L, the net launch speed actually scales as v ∝ 1/√L for a fixed torque — a longer arm gives a slower launch unless the driving torque also increases to compensate.