3.11.1.2 - Rotational Kinetic Energy

3.11.1.2 - Rotational Kinetic Energy

A flywheel stores kinetic energy in its rotation. Follow the energy from moving mass elements to a complete wheel, calculate the energy available as it slows, and explain how flywheels support engines, vehicles and production machinery.

Deriving rotational kinetic energy

Every small mass element in a rotating rigid body is moving, so each has kinetic energy.

An element mim_i at radius rir_i has speed vi=ωriv_i=\omega r_i. All elements share the same angular speed ω\omega, but elements farther out have greater linear speeds. Thus

Ek=∑i12mi(ωri)2=12ω2∑imiri2.E_k=\sum_i\frac12m_i(\omega r_i)^2 =\frac12\omega^2\sum_i m_i r_i^2.

The sum is the moment of inertia about the rotation axis, giving

Rotational kinetic energy

Ek=12Iω2E_k=\frac12 I\omega^2

Use II in kg m2\mathrm{kg\,m^2}, ω\omega in rad s−1\mathrm{rad\,s^{-1}}, and EkE_k in joules.

The formula describes rotation about the specified axis. Reversing the direction does not change the energy at a given speed because ω\omega is squared. If the object's centre also travels through space, its translational kinetic energy must be included separately.

Stored and usable energy

A flywheel that slows down releases the difference between its initial and final rotational kinetic energies.

Convert revolutions per minute before substituting: ω=2πn/60\omega=2\pi n/60, where nn is the numerical speed in rpm. A flywheel with I=0.45 kg m2I=0.45\,\mathrm{kg\,m^2} at 3000 rpm3000\,\mathrm{rpm} has ω=100π rad s−1\omega=100\pi\,\mathrm{rad\,s^{-1}} and

Ek=12(0.45)(100π)2=2.22×104 J.E_k=\tfrac12(0.45)(100\pi)^2=2.22\times10^4\,\mathrm J.

Retain the unrounded angular speed in the calculation.

If this wheel slows to 2000 rpm2000\,\mathrm{rpm}, the available energy decrease is

ΔE=12I(ωhigh2−ωlow2)=1.23×104 J.\Delta E=\frac12I(\omega_{\rm high}^2-\omega_{\rm low}^2)=1.23\times10^4\,\mathrm J.

Do not use 12I(ωhigh−ωlow)2\tfrac12I(\omega_{\rm high}-\omega_{\rm low})^2. Energy below the minimum useful operating speed remains stored. If the transfer efficiency is η\eta, the load receives ηΔE\eta\Delta E.

Flywheel capacity and losses

A flywheel is a rotating energy store that can exchange energy with a machine as its angular speed changes.

At a fixed speed, increasing II increases stored energy in direct proportion. For a fixed mass and outer radius, concentrating mass nearer the rim increases II. At fixed II, doubling the angular speed gives four times the energy. These comparisons require the other quantity to stay unchanged.

Maximum usable capacity depends on the permitted speed range, 12I(ωmax⁡2−ωmin⁡2)\tfrac12I(\omega_{\max}^2-\omega_{\min}^2), as well as the wheel's mass and distribution. Higher speeds require larger inward forces on the material and create greater mechanical stress, so material strength limits the safe speed. A larger radius cannot be increased indefinitely at the same speed without consequences for stress and size.

Bearing friction and air resistance dissipate energy during storage and transfer. Suitable bearings, lubrication and a low-pressure enclosure can reduce these losses. Reducing friction does not increase EkE_k at the same II and ω\omega; it helps retain that energy for longer.

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