Lesson 5 · 40 min
Impulse, Momentum and Impact of Rigid Bodies
When forces act for a known time, or for an instant too short to measure, integrating the equations of motion over time is easier than solving them. For a rigid body that gives two principles, one for linear and one for angular momentum, and the second is often the one that survives an impact.
Learning objectives
- Apply the linear and angular impulse–momentum principles to a rigid body in plane motion.
- Identify when angular momentum about a point is conserved during an impact or coupling.
- Compute the energy lost in a plastic impact or a clutch engagement.
- Locate the center of percussion, where an impact produces no reaction at the pin.
The two impulse–momentum principles
Integrate \(\sum\Fvec = m\mathbf{a}_G\) and \(\sum M_G = I_G\alpha\) over an interval \(t_1\) to \(t_2\):
Impulse and momentum for a rigid body in plane motion
\[ m\vvec_{G1} + \sum\int_{t_1}^{t_2}\Fvec\,dt = m\vvec_{G2} \] \[ I_G\,\omega_1 + \sum\int_{t_1}^{t_2} M_G\,dt = I_G\,\omega_2 \]For rotation about a fixed axis through \(O\), the second becomes \(I_O\omega_1 + \sum\int M_O\,dt = I_O\omega_2\). In general, moments may be taken about any point \(P\) if the angular momenta are written as \(H_P = I_G\omega + (\rvec_{G/P} \times m\vvec_G)_z\) and \(P\) is fixed.
A good habit is to draw three pictures side by side: the momentum diagram at \(t_1\) (\(m\vvec_{G1}\) at \(G\) and the couple \(I_G\omega_1\)), the impulse diagram (every force times its duration), and the momentum diagram at \(t_2\). Then sum components and moments across the three.
Example 5.1 — Spinning up a flywheel against friction
A \(20\ \text{kg}\) disk flywheel of radius \(0.4\ \text{m}\) starts from rest. A motor applies \(10\ \text{N·m}\) for \(3\ \text{s}\) while bearing friction resists with \(2\ \text{N·m}\). Find the final angular velocity.
Show solution
\(I_O = \tfrac12(20)(0.4)^2 = 1.6\ \text{kg·m}^2\). About the fixed axis:
\[ 0 + (10 - 2)(3) = 1.6\,\omega_2 \quad\Rightarrow\quad \omega_2 = 15.0\ \text{rad/s} \]Conservation of angular momentum
If the external impulses have no net moment about a point over an interval, angular momentum about that point is conserved. Two situations make this especially useful:
- Couplings. Two shafts engaged by a clutch exert large friction moments on each other, but those are internal to the pair. The total angular momentum about the common axis is conserved.
- Impacts. During a collision lasting milliseconds, finite forces such as weights have negligible impulse. Only the impulsive forces matter. If the body is pinned, the pin exerts an unknown impulse, but it has no moment about the pin itself: angular momentum about the pin is conserved, while linear momentum is not.
Example 5.2 — Engaging a clutch
Disk \(A\) (\(I_A = 0.6\ \text{kg·m}^2\)) spins at \(120\ \text{rad/s}\). It is clutched to disk \(B\) (\(I_B = 0.2\ \text{kg·m}^2\)), initially at rest, on the same axis. Find the common final speed and the fraction of kinetic energy lost.
Show solution
\(1080\ \text{J}\), or \(25\%\), becomes heat in the clutch plates while they slip. In general the fraction lost is \(I_B/(I_A + I_B)\) when \(B\) starts at rest.
Example 5.3 — A bullet into a pinned rod
For the rod of Figure 5.1, the bullet strikes \(0.9\ \text{m}\) below the pin. Find the rod's angular velocity just after the impact, the horizontal impulse exerted by the pin, and how high the rod swings.
Show solution
Angular momentum about \(O\) (the pin impulse and the weights have no moment about \(O\) during the impact):
\[ m_b v_b h = \left(\tfrac13 m_r L^2 + m_b h^2\right)\omega \quad\Rightarrow\quad 0.02(400)(0.9) = (2.4 + 0.0162)\,\omega \quad\Rightarrow\quad \omega = 2.980\ \text{rad/s} \]Linear momentum (horizontal, positive in the bullet's direction), with \(P_x\) the pin impulse on the rod:
\[ m_b v_b + P_x = m_r(0.6\,\omega) + m_b(0.9\,\omega) \quad\Rightarrow\quad 8 + P_x = 8.993 \quad\Rightarrow\quad P_x = 0.993\ \text{N·s} \]The pin pushes in the bullet's direction: the impact is above the center of percussion.
Swing (energy after the impact): \(\tfrac12(2.416)(2.980)^2 = 10.73\ \text{J} = (5 \cdot 9.81 \cdot 0.6 + 0.02 \cdot 9.81 \cdot 0.9)(1 - \cos\theta)\), so \(\theta = 50.4^\circ\).
The bullet brought \(1600\ \text{J}\); \(99.3\%\) of it was lost in the impact. Angular momentum survives a plastic impact; kinetic energy does not.
The center of percussion
Where should a pinned body be struck so that the pin feels no impulse? Ignoring the bullet's own mass, the linear impulse \(F\,\Delta t\) must produce \(m v_G = m\omega r_G\) and the angular impulse \(F\,\Delta t\,h\) must produce \(I_O\omega\). Dividing:
Center of percussion
\[ h_P = \frac{I_O}{m\,r_G} = \frac{k_O^2}{r_G} \]For a slender rod pinned at one end, \(h_P = \dfrac{\tfrac13 L^2}{L/2} = \tfrac23 L\).
Check your understanding
Key takeaways
- \(m\vvec_{G1} + \sum\int\Fvec\,dt = m\vvec_{G2}\) and \(I_G\omega_1 + \sum\int M_G\,dt = I_G\omega_2\) (or about a fixed axis, \(I_O\)).
- Angular momentum about a point is conserved when the external impulses have no moment about it: clutches, and impacts on pinned bodies (about the pin).
- Plastic impacts and slipping clutches lose kinetic energy even though \(H\) is conserved.
- Center of percussion: \(h_P = k_O^2/r_G\), \(\tfrac23 L\) for a rod pinned at its end.
- Next: Lesson 6 takes angular momentum into three dimensions.