INGENIA

MEC-29

Flywheel fluctuation energy

ΔE = I ω² Cs with Cs = (ωmax−ωmin)/ω. Stored kinetic buffer.

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EnergyShigley

Governing equation

ΔE=Iω2Cs\Delta E=I\omega^2 C_s

ω∈[10,600]\omega \in \left[10, 600\right]

where

I
Inertia (kg·m²)
\omega
Mean speed (rad/s)
C_s
Fluctuation coeff. (—)
\Delta E
Energy fluctuation (kJ)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

MEC-29 · pendulum
00:0 / 00:08
①

Governing law — red

ΔE=Iω2Cs\Delta E=I\omega^2 C_s

ω∈[10,600]\omega \in \left[10, 600\right]

ω=10\omega = 10ΔE=0.02\Delta E = 0.02

②

Domain of the variable — blue

ωmin⁡=10,ωmax⁡=600\omega_{\min} = 10,\quad \omega_{\max} = 600

ω∈[10,600]\omega \in \left[10, 600\right]

ω=10\omega = 10ω=10\omega = 10

③

Response curve — green

ΔE=Iω2Cs\Delta E=I\omega^2 C_s

ω∈[10,600]\omega \in \left[10, 600\right]

ω=10\omega = 10ΔE=0.02\Delta E = 0.02

④

Rate of change — amber

dΔEdω\frac{\mathrm{d}\Delta E}{\mathrm{d}\omega}

ω∈[10,600]\omega \in \left[10, 600\right]

ω=13.71\omega = 13.71ΔE′=0.005\Delta E^{\prime} = 0.005

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-29 — Flywheel fluctuation energy) is the form associated with Shigley. Working symbols: II, ω\omega, CsC_s →\rightarrow ΔE\Delta E. A flywheel stores ½ I ω². The energy exchanged while speed fluctuates by Cs is I ω² Cs.

Purpose

Purpose: compute ΔE\Delta E from II, ω\omega, CsC_s in Mechanical via ΔE=Iω2Cs\Delta E=I\omega^2 C_s ΔE = I ω² Cs with Cs = (ωmax−ωmin)/ω. Stored kinetic buffer. Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given I=4.000 kg⋅m2I = 4.000\,\mathrm{kg·m^{2}}, ω=150.000 rad/s\omega = 150.000\,\mathrm{rad/s}, Cs=0.050 —C_s = 0.050\,\mathrm{—}, the governing relation ΔE=Iω2Cs\Delta E=I\omega^2 C_s yields ΔE=4.500 kJ\Delta E = 4.500\,\mathrm{kJ}. A disk, an ω band, an energy pulse. Move a slider: the numbers are this situation, not a canned story.

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Outputs

  • Energy fluctuation \Delta E4.500 kJ
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MEC-29 · pendulum
00:0 / 00:08

Narration of this film

A disk, an ω band, an energy pulse.

A flywheel stores ½ I ω². The energy exchanged while speed fluctuates by Cs is I ω² Cs.

Reading speed

Watch on YouTube