Unit 6 · Topic 6.3 Beta

Angular Momentum and Angular Impulse

Angular momentum is L = Iω for a rigid object turning about a fixed axis, and L = rmv sin θ for a point object about a chosen point.

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

A spinning bicycle wheel is hard to stop and hard to tilt, and a puck sliding past you in a straight line has, in a precise sense, some of the same turning quantity about the spot where you stand. Angular momentum measures "how much turning" an object carries, and torque acting over time is what changes it.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A 2.0 kg cart moves at 3.0 m/s. Its momentum is

  1. 6.0 kg·m/s
  2. 9.0 kg·m/s
  3. 1.5 kg·m/s
  4. 18 kg·m/s
Show the answer

p = mv.

  • Correct: 6.0 kg·m/s:
  • 9.0 kg·m/s:
  • 1.5 kg·m/s:
  • 18 kg·m/s:

2. The area under a force-time graph equals

  1. the impulse, the change in momentum
  2. the work done
  3. the change in kinetic energy
  4. the average force
Show the answer

J = FΔt = Δp.

  • Correct: the impulse, the change in momentum:
  • the work done:
  • the change in kinetic energy:
  • the average force:

Part 4 · See it

See it first

Top view of a level, frictionless table. A puck of mass m slides at constant velocity v along a straight dashed line. A point P lies off the line, a perpendicular distance d from it. The puck is shown at three positions; gray lines from P reach the first and third positions, and at the first one the angle θ between r and v is marked. The amber perpendicular from P to the line has length r⊥ = d. A panel gives L = rmv sin θ = mvd: r changes as the puck moves, but r sin θ always equals d, so the angular momentum about P is constant. About any point on the puck’s own line, L is zero.
A puck in straight-line motion has angular momentum mvd about a point a distance d from its path, the same at every position. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A rigid object turns at ω and has rotational inertia Iit has angular momentum L = Iω
  2. A point object moves with momentum mv past a chosen pointit has angular momentum L = rmv sin θ = mv × (perpendicular distance) about that point
  3. A net torque acts for a time Δtit delivers an angular impulse τΔt
  4. An angular impulse is deliveredthe angular momentum changes by exactly that much: ΔL = τΔt
  5. The net torque is zeroL stays the same, and the L-t graph is flat

Part 6 · Key ideas

Key ideas

  • L = Iω for a rigid object about a fixed axis; unit kg·m²/s.
  • A point object has L = rmv sin θ about a chosen point: momentum × perpendicular distance from the point to its line of motion. Change the point, change L.
  • A puck in straight-line motion keeps the same L about any fixed point, and has L = 0 about points on its path.
  • Angular impulse τΔt = ΔL; it is the area under a τ-t graph.
  • Net torque = slope of the L-t graph (and = Iα when I is constant).

Part 7 · Misconception

A common mistake

The wrong idea: Only spinning objects have angular momentum; a puck moving in a straight line has none.

What actually happens: About any point not on its line of motion, a moving point object has L = mv × (perpendicular distance). Whether something has angular momentum depends on the point you measure it about.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

A puck sliding past two points

v = 3.0 m/sPQ0.40 mTop view, level frictionless table. Puck mass 0.20 kg. Q lies on the puck’s path.

1. What is the size of the puck’s angular momentum about point P? Give the value and its unit.

Type a number and its unit.

Show the answer

L = rmv sin θ = mv(r sin θ) = mvd = (0.20 kg)(3.0 m/s)(0.40 m) = 0.24 kg·m²/s.

  • Answer: 0.24 kg·m^2/s

2. As the puck slides on past P, its distance r from P grows. Its angular momentum about P

  1. stays the same, because r sin θ stays 0.40 m
  2. grows, because r grows
  3. shrinks, because the puck moves away from P
  4. is zero, because the puck is not turning
Show the answer

L = rmv sin θ, and r sin θ is the perpendicular distance from P to the line of motion, 0.40 m at every position. With m and v constant, L about P stays 0.24 kg·m²/s.

  • Correct: stays the same, because r sin θ stays 0.40 m: Right: the perpendicular distance from P to the path never changes.
  • grows, because r grows: r grows, but sin θ shrinks so that r sin θ stays 0.40 m.
  • shrinks, because the puck moves away from P: Moving away along a straight line keeps the lever arm, and so L, the same.
  • is zero, because the puck is not turning: A point object moving past a point has angular momentum about it, turning or not.

3. What is the puck’s angular momentum about point Q, which lies on its path?

  1. Zero
  2. 0.24 kg·m²/s
  3. 0.60 kg·m²/s
  4. It depends on how far the puck is from Q
Show the answer

The puck’s velocity points along the line through Q, so the perpendicular distance from Q to its line of motion is zero: L = mv(0) = 0. Angular momentum depends on the point you choose.

  • Correct: Zero: Right: the puck moves straight along a line through Q, so θ = 0 or 180°.
  • 0.24 kg·m²/s: That is the value about P. About Q the lever arm is zero.
  • 0.60 kg·m²/s: That is the linear momentum, mv, in kg·m/s.
  • It depends on how far the puck is from Q: Along a line through Q, r sin θ = 0 at every distance.

4. A second puck of the same mass slides at 6.0 m/s along a parallel line 0.20 m from P. Compared with the first puck, its angular momentum about P is

  1. the same
  2. twice as large
  3. half as large
  4. four times as large
Show the answer

L = mvd: (0.20)(6.0)(0.20) = 0.24 kg·m²/s, the same as (0.20)(3.0)(0.40).

  • Correct: the same: Right: twice the speed × half the lever arm.
  • twice as large: The speed doubles, but the lever arm halves.
  • half as large: The lever arm halves, but the speed doubles.
  • four times as large: L = mvd is the product; the two factors cancel.

5. For a rigid object turning about a fixed axis, angular momentum is

  1. L = Iω, in kg·m²/s
  2. L = ½Iω², in joules
  3. L = Iα, in N·m
  4. L = mv, in kg·m/s
Show the answer

L = Iω pairs I with m and ω with v, as p = mv does. Its unit is kg·m²/s (the same as N·m·s and J·s).

  • Correct: L = Iω, in kg·m²/s: Right: the rotational partner of p = mv.
  • L = ½Iω², in joules: That is rotational kinetic energy.
  • L = Iα, in N·m: Iα is the net torque.
  • L = mv, in kg·m/s: That is linear momentum.

6. A torque of 0.60 N·m acts on a spinning top for 0.50 s in the direction it spins. By how much does its angular momentum change?

Type a number in kg·m^2/s.

Show the answer

ΔL = τΔt = (0.60 N·m)(0.50 s) = 0.30 kg·m²/s.

  • Answer: 0.30 kg·m^2/s

7. A graph of a wheel’s angular momentum against time is a horizontal line. What does that tell you?

  1. The net torque on the wheel is zero
  2. The wheel is not turning
  3. The wheel’s angular acceleration is constant and nonzero
  4. A constant torque is acting on the wheel
Show the answer

τ_net = ΔL/Δt, the slope of the L-t graph. A horizontal line has zero slope, so the net torque is zero and ω is constant.

  • Correct: The net torque on the wheel is zero: Right: the slope of L-t is the net torque.
  • The wheel is not turning: A horizontal line at a nonzero L means steady turning.
  • The wheel’s angular acceleration is constant and nonzero: Constant nonzero α would make L change steadily, a sloped line.
  • A constant torque is acting on the wheel: A constant nonzero torque gives a sloped straight line.

Part 9 · Summary

Summary

Angular momentum is L = Iω for a rigid object turning about a fixed axis, and L = rmv sin θ for a point object about a chosen point. A net torque acting over a time changes it: angular impulse τΔt = ΔL, the area under a τ-t graph, and the net torque is the slope of an L-t graph.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections