Unit 6.3 | Conservation of Angular Momentum

Problem Type 2: Landing on a turntable

How to spot it: Something with no spin lands on, or is placed on, a rotating platform or disk and moves with it: clay dropped on a turntable, a child jumping onto a merry-go-round, a disk dropped onto a spinning disk.

Setup: It is a perfectly inelastic collision, rotated: [katex] I_1\omega_1 = (I_1 + I_2)\omega_f [/katex]. A small object at radius [katex] r [/katex] adds [katex] mr^2 [/katex]. Kinetic energy is lost.

A turntable with [katex] I = 0.020 \text{ kg·m}^2 [/katex] spins freely at [katex] 3.0\,\text{rad/s} [/katex]. A [katex] 0.10\,\text{kg} [/katex] lump of clay is dropped straight down onto it and sticks [katex] 0.20\,\text{m} [/katex] from the center. Find the new angular speed and the kinetic energy lost.

The clay adds [katex] mr^2 = (0.10)(0.20)^2 = 0.0040 \text{ kg·m}^2 [/katex]. [katex] (0.020)(3.0) = (0.024)\omega_f [/katex], so [katex] \omega_f = 2.5 \text{ rad/s} [/katex].

Energy: [katex] K_i = \tfrac{1}{2}(0.020)(3.0)^2 = 0.090 \text{ J} [/katex] and [katex] K_f = \tfrac{1}{2}(0.024)(2.5)^2 = 0.075 \text{ J} [/katex]. Friction between the clay and the disk turned [katex] 0.015\,\text{J} [/katex] into thermal energy.

AP trap: Saying the turntable alone keeps its angular momentum. The clay exerts a friction torque on the turntable while it speeds up, so the turntable loses angular momentum to the clay. Only the total, turntable plus clay, is conserved.

Problem Type 3: Hitting a pivoted object

How to spot it: A ball, bullet or lump of clay moving in a straight line hits a rod, door or disk that can rotate about a fixed pivot, and sticks or bounces.

Setup: Use angular momentum about the pivot, so the pivot force drops out. Before: [katex] mvr_\perp [/katex] for the moving object (plus [katex] I\omega [/katex] if the target already spins). After: [katex] I_{total}\,\omega_f [/katex]. Linear momentum is usually not conserved, because the pivot pushes during the hit.

A uniform [katex] 0.60\,\text{kg} [/katex] rod, [katex] 1.2\,\text{m} [/katex] long, hangs from a frictionless pivot at its top. A [katex] 0.10\,\text{kg} [/katex] lump of clay moving horizontally at [katex] 4.0\,\text{m/s} [/katex] hits the bottom end and sticks. How fast does the rod turn just after the hit?

Before: [katex] L = mvr_\perp = (0.10)(4.0)(1.2) = 0.48 \text{ kg·m}^2\text{/s} [/katex]. After: [katex] I = \tfrac{1}{3}(0.60)(1.2)^2 + (0.10)(1.2)^2 = 0.288 + 0.144 = 0.432 \text{ kg·m}^2 [/katex]. So [katex] \omega_f = \frac{0.48}{0.432} \approx 1.1 \text{ rad/s} [/katex].

Linear momentum is not conserved here. Before the hit it is [katex] (0.10)(4.0) = 0.40 \text{ kg·m/s} [/katex]. Just after, the rod’s center of mass and the clay move at [katex] 0.60\omega [/katex] and [katex] 1.2\omega [/katex], for a total of [katex] (0.60)(0.60)(1.11) + (0.10)(1.2)(1.11) \approx 0.53 \text{ kg·m/s} [/katex]. The pivot pushed forward on the rod during the hit.

AP trap: Using conservation of linear momentum, or [katex] v = r\omega [/katex] with the rod’s center. When something is pinned at a pivot, conserve angular momentum about that pivot.

Problem Type 4: Collide, then swing

How to spot it: After the hit, the rod swings up to some angle or height. Or before it, the rod swings down and then hits something.

Setup: Split it at the collision, exactly like Unit 4.4. Angular momentum through the hit. Energy before and after it, with the rotational kinetic energy [katex] \tfrac{1}{2}I\omega^2 [/katex] and the rise of each center of mass.

Continue the problem above. How high does the rod swing?

Just after the hit: [katex] K = \tfrac{1}{2}(0.432)(1.11)^2 \approx 0.267 \text{ J} [/katex]. If the rod swings to an angle [katex] \theta [/katex], its center rises [katex] 0.60(1 – \cos\theta) [/katex] and the clay rises [katex] 1.2(1 – \cos\theta) [/katex]:

[katex] 0.267 = (9.8)(1 – \cos\theta)\left[(0.60)(0.60) + (0.10)(1.2)\right] [/katex], so [katex] 1 – \cos\theta \approx 0.057 [/katex] and [katex] \theta \approx 19^\circ [/katex].

The clay brought [katex] 0.80\,\text{J} [/katex] of kinetic energy, and only [katex] 0.267\,\text{J} [/katex] survived the sticky collision. Using energy across the hit would give a much bigger swing.

▶ Watch: a worked collision and angular momentum problem (starts at 14:17).

AP trap: Using energy conservation across the collision. A sticky hit loses kinetic energy. Use angular momentum for the hit, then energy for the swing.

Lesson 6.4 Preview

In the last lesson of the unit, objects move and spin at once. A ball rolling down a ramp carries both kinds of kinetic energy, and a planet orbiting a star keeps both its energy and its angular momentum. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-6-3-conservation-of-angular-momentum/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-6-3-conservation-of-angular-momentum/