Rotational Collisions & Angular Impulse
- Apply angular impulse–momentum, τΔt = ΔL, as the rotational analogue of FΔt = Δp
- Solve a ball-strikes-rod collision by conserving angular momentum about the pivot
- Combine a rotational collision with an energy analysis of the motion that follows
Angular impulse changes angular momentum
Just as a force applied over time changes momentum, a torque applied over time changes angular momentum: τΔt = ΔL. The same consequence follows too — spreading a given change in angular momentum over a longer time reduces the torque required. It is why a gymnast bends her knees on landing and why a long, gradual braking of a flywheel puts less stress on its shaft than a sudden stop.
The ball-and-rod problem
A putty ball traveling in a straight line strikes the end of a rod pivoted at its center and sticks. This is the signature Unit 6 free-response, and it is solved in three moves. One: the ball has angular momentum about the pivot before impact, L = mvr_⊥, even though it was not rotating. Two: angular momentum about the pivot is conserved through the collision, because the pivot force acts through the pivot and so exerts no torque about it. Three: after impact, the combined rotational inertia is I_rod + m r², and ω follows from L = Iω.
A 0.20 kg putty ball moving at 8.0 m/s strikes the end of a uniform rod (mass 1.0 kg, length 1.2 m) pivoted at its center, and sticks. Find the angular speed just after impact.
- 1.Ball's angular momentum about the pivot: L = mvr = (0.20)(8.0)(0.60) = 0.96 kg·m²/s.
- 2.Rod about its center: I_rod = ML²/12 = (1.0)(1.2)² ÷ 12 = 0.12 kg·m².
- 3.Ball stuck at the end: I_ball = mr² = (0.20)(0.60)² = 0.072 kg·m².
- 4.Total after: I = 0.12 + 0.072 = 0.192 kg·m².
- 5.ω = L/I = 0.96 ÷ 0.192 = 5.0 rad/s.
Then, and only then, switch to energy
If the question continues — "how high does the rod swing?" or "what is the maximum angle?" — the collision is over and mechanical energy takes back over. Use ½Iω² as the kinetic energy immediately after impact and set it equal to the gravitational potential energy gained, mgΔh_cm, where Δh is the rise of the center of mass. This mirrors the ballistic pendulum of Unit 4 exactly: momentum through the collision, energy for the swing, and never the other way around.
Do not use energy conservation through a sticking collision. Putty embedding in a rod dissipates a large fraction of the kinetic energy as heat and deformation, and an energy equation written across the impact gives an answer that is simply wrong — often by a factor of two or more.
A constant torque of 4.0 N·m acts on a wheel for 3.0 s. What is the change in the wheel's angular momentum?
A ball strikes and sticks to the end of a pivoted rod. Which is conserved during the collision?
Where should a ball strike a pivoted rod to produce the largest angular speed, for a given ball speed?
For a collision that then leads to a swing, write two clearly separated stages on your paper: "Stage 1, collision — angular momentum" and "Stage 2, swing — energy". Rubrics score the two stages independently, so a correct stage 2 still earns points even after a slip in stage 1.
Answer the 3 checkpoints as you read.
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