AcademyRigid-Body Rotation

Academy

Rotational Kinetic Energy

Level 1 - Physics topic page in Rigid-Body Rotation.

Principle

Rotational kinetic energy depends on angular speed and mass distribution about the axis.

Notation

\(K_{\mathrm{rot}}\)
rotational kinetic energy
\(\mathrm{J}\)
\(I\)
moment of inertia about the axis
\(\mathrm{kg\,m^{2}}\)
\(\omega\)
angular velocity
\(\mathrm{rad\,s^{-1}}\)
\(M\)
total mass
\(\mathrm{kg}\)
\(v_{\mathrm{cm}}\)
center-of-mass speed
\(\mathrm{m\,s^{-1}}\)
\(R\)
rolling radius
\(\mathrm{m}\)

Method

Derivation 1: Add the kinetic energy of many mass elements

A rigid body rotates with one angular speed, but each mass element has speed \(v_i=r_i\omega\).

Point-mass energy
\[K_i=\frac12m_iv_i^2\]
Use rotational speed
\[v_i=r_i\omega\]
Sum all elements
\[K_{\mathrm{rot}}=\frac12\left(\sum_i m_ir_i^2\right)\omega^2\]
Define moment of inertia
\[I=\sum_i m_ir_i^2\Rightarrow K_{\mathrm{rot}}=\frac12I\omega^2\]

Derivation 2: Include translation for rolling bodies

A rolling rigid body has center-of-mass translation and rotation about the center of mass.

Split kinetic energy
\[K=\frac12Mv_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2\]
No-slip link
\[\omega=\frac{v_{\mathrm{cm}}}{R}\]
Write in terms of speed
\[K=\frac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2\]

Rules

These are the compact results from the method above.

Rotational energy
\[K_{\mathrm{rot}}=\frac12I\omega^2\]
Point masses
\[I=\sum_i m_ir_i^2\]
Rolling total
\[K=\frac12Mv_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2\]
No-slip energy
\[K=\frac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2\]

Examples

Question
A flywheel has
\[I=0.80\,\mathrm{kg\,m^2}\]
and
\[\omega=12\,\mathrm{rad\,s^{-1}}\]
Find
\[K_{\mathrm{rot}}\]
Answer
\[K_{\mathrm{rot}}=\frac12(0.80)(12^2)=57.6\,\mathrm{J}\]

Checks

  • Moment of inertia depends on the chosen axis.
  • Rotational kinetic energy is nonnegative.
  • A rolling object's energy is not only translational.
  • For the same \(M\), \(R\), and \(\omega\), mass farther out means larger energy.