Lesson 4 · 35 min
Rigid Bodies in Plane Motion
For a rigid body the sum over particles collapses into something short: the moment of inertia times the angular velocity, plus the moment of the body's momentum. This lesson builds those formulas, shows why the point still matters, and connects them to the moment equations of plane kinetics.
Learning objectives
- Show that a rigid body in plane motion has \(H_G = I_G\omega\).
- Find the angular momentum about any point, \(H_P = I_G\omega + m v_G d\), with the correct sign.
- Use \(H_O = I_O\omega\) for rotation about a fixed axis and for the contact point of a rolling wheel.
- Derive \(\sum M_G = I_G\alpha\) from \(\sum M_G = \dot H_G\).
Angular momentum about the center of mass
Consider a rigid slab in the \(xy\)-plane (or a body symmetric about that plane) turning at \(\omega\khat\). Relative to \(G\), every particle moves on a circle about \(G\), with speed \(\omega\rho_i\) perpendicular to \(\boldsymbol\rho_i\). Its contribution to \(H_G\) is \(m_i\rho_i(\omega\rho_i)\), so
\[ H_G = \sum m_i\rho_i^2\,\omega = \left(\int \rho^2\,dm\right)\omega \]Rigid body in plane motion
\[ H_G = I_G\,\omega \] \[ H_P = I_G\,\omega + (\rvec_{G/P} \times m\vvec_G)_z = I_G\,\omega \pm m v_G d \]\(d\) is the perpendicular distance from \(P\) to the line of action of \(m\vvec_G\) through \(G\). Take the sign of each term from its sense of rotation about \(P\) (counterclockwise positive).
The second line is Lesson 3's split, \(\Hvec_P = \rvec_{G/P} \times m\vvec_G + \Hvec_G\), with \(\Hvec_G = I_G\omega\,\khat\). A useful picture: the momentum of a rigid body is a vector \(m\vvec_G\) acting at \(G\), together with a "momentum couple" \(I_G\omega\). Take moments of that pair about any point, exactly as you would a force and a couple.
Example 4.1 — A rolling disk about three points
A \(10\ \text{kg}\) uniform disk of radius \(0.3\ \text{m}\) rolls to the right without slipping at \(v_G = 6\ \text{m/s}\). Find its angular momentum about its center \(G\), about its contact point \(C\), and about the top of the disk \(A\).
Show solution
Rolling to the right means turning clockwise: \(\omega = -v_G/r = -20\ \text{rad/s}\). \(I_G = \tfrac12(10)(0.3)^2 = 0.45\ \text{kg·m}^2\).
\[ H_G = I_G\omega = 0.45(-20) = -9\ \text{kg·m}^2/\text{s}\ \ (\text{clockwise}) \]About \(C\), \(m\vvec_G = 60\,\ihat\ \text{kg·m/s}\) acts \(0.3\ \text{m}\) above \(C\), turning clockwise about it:
\[ H_C = -9 - 60(0.3) = -27\ \text{kg·m}^2/\text{s} \qquad \left(= I_C\omega = \tfrac32 (10)(0.09)(-20)\right) \]About \(A\), the same momentum acts \(0.3\ \text{m}\) below \(A\), turning counterclockwise about it:
\[ H_A = -9 + 18 = +9\ \text{kg·m}^2/\text{s} \]Same body, same instant, three different angular momenta: \(-9\), \(-27\) and \(+9\).
Rotation about a fixed axis
If the body turns about a fixed axis through \(O\), then \(v_G = \omega r_G\) with the moment arm \(d = r_G\), and the two terms combine through the parallel-axis theorem:
\[ H_O = I_G\omega + m(\omega r_G)r_G = (I_G + m r_G^2)\,\omega = I_O\,\omega \]Fixed-axis rotation
\[ H_O = I_O\,\omega \]Also true about the contact point \(C\) of a wheel rolling without slipping at that instant, because \(C\) is the instantaneous center and \(v_G = \omega r\).
Example 4.2 — A pinned rod, two ways
A \(4\ \text{kg}\) slender rod, \(1.2\ \text{m}\) long, turns about a pin at one end at \(5\ \text{rad/s}\). Find \(H_O\) about the pin.
Show solution
Fixed axis: \(I_O = \tfrac13(4)(1.2)^2 = 1.92\ \text{kg·m}^2\), so \(H_O = 1.92(5) = 9.60\ \text{kg·m}^2/\text{s}\).
Split: \(I_G\omega = \tfrac1{12}(4)(1.44)(5) = 2.40\); \(v_G = 5(0.6) = 3\ \text{m/s}\) with arm \(0.6\ \text{m}\), so \(m v_G d = 4(3)(0.6) = 7.20\). Total \(9.60\ \text{kg·m}^2/\text{s}\).
Example 4.3 — A translating plate
A \(6\ \text{kg}\) plate translates (no rotation) with \(\vvec_G = 2\,\ihat\ \text{m/s}\) while its center is \(0.5\ \text{m}\) above a point \(O\) on the floor. Find \(H_G\) and \(H_O\).
Show solution
\(\omega = 0\), so \(H_G = 0\). About \(O\), the momentum \(12\ \text{kg·m/s}\) acts \(0.5\ \text{m}\) above it, clockwise: \(H_O = -12(0.5) = -6\ \text{kg·m}^2/\text{s}\). A body that does not rotate can still have angular momentum about a point.
Back to the moment equations
Put \(H_G = I_G\omega\) into Lesson 3's \(\sum M_G = \dot H_G\). For a rigid body \(I_G\) is constant, so
\[ \sum M_G = I_G\,\alpha, \qquad \text{and for a fixed axis } \sum M_O = I_O\,\alpha \]These are the moment equations of plane rigid-body kinetics. They are not new laws: they are the angular-momentum principle written for a rigid body.
Check your understanding
Key takeaways
- \(H_G = I_G\omega\); about any point, \(H_P = I_G\omega + (\rvec_{G/P} \times m\vvec_G)_z\).
- The body's momentum is \(m\vvec_G\) at \(G\) plus the couple \(I_G\omega\): take moments like a force and a couple.
- Fixed axis (and rolling contact point): \(H_O = I_O\omega\).
- \(\sum M_G = I_G\alpha\) and \(\sum M_O = I_O\alpha\) follow from \(\sum M = \dot H\).
- Next: Lesson 5 integrates over time: impulse, momentum and impact.