AcademyRotational Dynamics
Academy
Rotation About Moving Axes
Level 1 - Physics topic page in Rotational Dynamics.
Principle
Moving-axis rotation couples center-of-mass translation with rotation about the center.
Notation
\(M\)
total mass
\(\mathrm{kg}\)
\(I_{\mathrm{cm}}\)
moment of inertia about the center of mass
\(\mathrm{kg\,m^{2}}\)
\(a_{\mathrm{cm}}\)
center-of-mass acceleration
\(\mathrm{m\,s^{-2}}\)
\(\alpha\)
angular acceleration
\(\mathrm{rad\,s^{-2}}\)
\(R\)
rolling radius
\(\mathrm{m}\)
\(f_s\)
static friction force
\(\mathrm{N}\)
Method
Derivation 1: Split translation and rotation
For a rolling body, the center of mass translates while the body rotates about that center.
Translate the center
\[\sum F_x=Ma_{\mathrm{cm}}\]
Rotate about the center
\[\sum\tau_{\mathrm{cm}}=I_{\mathrm{cm}}\alpha\]
Use the same force model
\[f_s\ \text{can appear in both equations}\]
A force can accelerate the center and produce a torque.
The free-body diagram shows the forces for a wheel on a horizontal surface. Translation uses the net force; rotation uses torques about the center.
Derivation 2: Add the rolling constraint
Rolling without slipping is a kinematic constraint. It is not guaranteed just because the body is round.
Rolling speed
\[v_{\mathrm{cm}}=R\omega\]
Rolling acceleration
\[a_{\mathrm{cm}}=R\alpha\]
Static-friction test
\[|f_s|\le\mu_sN\]
Derivation 3: Account for energy
A rolling body's kinetic energy has translational and rotational parts.
Translational part
\[K_{\mathrm{trans}}=\frac12Mv_{\mathrm{cm}}^2\]
Rotational part
\[K_{\mathrm{rot}}=\frac12I_{\mathrm{cm}}\omega^2\]
Total kinetic energy
\[K=\frac12Mv_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2\]
Rules
These are the compact results from the method above.
Translation
\[\sum F_x=Ma_{\mathrm{cm}}\]
Rotation
\[\sum\tau_{\mathrm{cm}}=I_{\mathrm{cm}}\alpha\]
Rolling speed
\[v_{\mathrm{cm}}=R\omega\]
Rolling acceleration
\[a_{\mathrm{cm}}=R\alpha\]
Total kinetic energy
\[K=\frac12Mv_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2\]
Examples
Question
A solid cylinder rolls without slipping at
\[v_{\mathrm{cm}}=4.0\,\mathrm{m\,s^{-1}}\]
with \[R=0.20\,\mathrm{m}\]
Find \(\omega\).Answer
Use
\[\omega=\frac{v_{\mathrm{cm}}}{R}=\frac{4.0}{0.20}=20\,\mathrm{rad\,s^{-1}}\]
Checks
- Rolling without slipping is a constraint.
- Static friction can point either way.
- Torques about the center avoid torque from weight and normal on level ground.
- Translational and rotational kinetic energies both matter.