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6.1 Rotational Kinetic Energy

Syllabus
2024
Topic
6.1
Level
—

Learning objectives

6.1A—Describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity…Describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system.• The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system and is given by the equation- i. The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy.- ii. The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass.• A rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.• Rotational kinetic energy is a scalar quantity. TOPIC 6.1 Rotational Kinetic Energy TOPIC 6.2 Torque and Work

Calculate rotational kinetic energy

Connect rotation to stored motion energy

Rotational kinetic energy is the energy a rigid system has because it rotates. It depends on the system's rotational inertia and angular speed about the same axis.

Krot=12Iω2K_{\mathrm{rot}}=\frac12 I\omega^2

Match quantities to one axis

Symbol Meaning SI unit
II Rotational inertia about the chosen axis kg m2\text{kg}\,\text{m}^2
ω\omega Angular velocity; its magnitude is angular speed rad s−1\text{rad}\,\text{s}^{-1}
KrotK_{\mathrm{rot}} Rotational kinetic energy J\text{J}

Recover particle kinetic energy

For one object of mass mm rotating a distance rr from a fixed axis, I=mr2I=mr^2 and v=rωv=r\omega. Therefore

Krot=12(mr2)ω2=12m(rω)2=12mv2K_{\mathrm{rot}}=\tfrac12(mr^2)\omega^2=\tfrac12m(r\omega)^2=\tfrac12mv^2.

The rotational expression adds the translational kinetic energies of the moving parts.

Add center-of-mass and rotational motion

Ktotal=12Mvcm2+12Icmω2K_{\mathrm{total}}=\frac12 Mv_{\mathrm{cm}}^2+\frac12 I_{\mathrm{cm}}\omega^2

Calculate and scale the energy

Rotating system: if I=0.80 kg m2I=0.80\,\text{kg}\,\text{m}^2 and ω=5.0 rad s−1\omega=5.0\,\text{rad}\,\text{s}^{-1},

Krot=12(0.80)(5.0)2=10 JK_{\mathrm{rot}}=\tfrac12(0.80)(5.0)^2=10\,\text{J}.

Doubling angular speed would make the energy four times as large because Krot∝ω2K_{\mathrm{rot}}\propto\omega^2.

Keep energy scalar

Rotational kinetic energy is a scalar: reversing the rotation changes the sign of ω\omega but not KrotK_{\mathrm{rot}}. A stationary center of mass means the translational term is zero; the system can still have rotational kinetic energy because its parts are moving.