Ch 14: Angular MomentumWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Solution: A particle of mass m near the surface of Earth is launched with an initial velocity v0 at an angle θ above the horizontal. Using the origin as the pivot, find the angular momentum when the particle is

Problem

A particle of mass m near the surface of Earth is launched with an initial velocity v0 at an angle θ above the horizontal. Using the origin as the pivot, find the angular momentum when the particle is at the highest point of its trajectory. The z-axis points out of the page.

1. = mv03/2g sinθ cos2 θ k

2. = mv03/2g sin2 θ cosθ k

3. = mv02/2g sin2 θ cosθ k

4. = - mv02/2g sinθ cos2 θ k

5. = - mv03/2g sin2 θ cosθ k

6. = mv02/2g sinθ cosθ k

7. = - mv02/2g sin2 θ cosθ k

8. = - mv02/2g tan3 θ k

9. = mv02/2g tan3 θ k

10. = - mv02/2g sin θ cosθ k