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Uniform Circular Motion

Uniform Circular Motion. Basic Principles & Formulas. Uniform Circular Motion Principles. Object moves at a constant speed (speed = scalar quantity) Object’s velocity is constantly changing (velocity = vector quantity) due to a continuous change in direction

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Uniform Circular Motion

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  1. Uniform Circular Motion Basic Principles & Formulas

  2. Uniform Circular Motion Principles • Object moves at a constant speed (speed = scalar quantity) • Object’s velocity is constantly changing (velocity = vector quantity) due to a continuous change in direction • Velocity is tangent to circular motion • While in motion, there is a force and an acceleration being applied to the object

  3. Uniform Circular Motion Terms & Formulas • Period (T) = time it takes to make 1 revolution (time to travel circumference). • Period may be given, obvious or determined by time/revolutions ex. 25 rev. in 10 sec. T = 10/25  .4 sec (T = 1/f) • Frequency (f) = number of revolutions made in 1 sec. (f = 1/T) units = Hz (Period & Frequency are Inversely Related)

  4. Uniform Circular Motion Diagram Acceleration & Force = Parallel to Radius / Velocity = Perpendicular to Radius v a r F

  5. Uniform Circular Motion Questions • Q = While moving in a circular path, if string breaks, in what direction will the object travel? • A = In a straight line, perp. to radius (Law of Inertia; resistance to change) • Q = In what direction is the force? • A = Inward • Q = In what direction is the acceleration? • A = Inward

  6. Uniform Circular Motion Terms & Formulas • Velocity (v) = d or 2πr t T • Acceleration (a) = v² r circumference Time to travel around circumference once

  7. Uniform Circular Motion Terms & Formulas • Force (F) = ma or mv² r Remember, mass must be in Kg!

  8. UCM Example Problem • A 615 kg race car completes 1.00 lap in 14.3 sec around a circular track with a radius of 50.0 m. What is the car’s T, f, v, a, F (on tires)? • T = 14.3/1.00 = 14.3 • f = 1.00/14.3 = 0.0699 Hz

  9. UCM Example Problem (cont.) • V = (2)(π)(50.0) 14.3 = 22.0 m/s • a = (22.0 m/s)² 50.0 = 9.68 m/s²

  10. UCM Example Problem (cont.) • F (on tires) = (615 kg)(22.0 m/s)² 50.0 m = 5,950 N

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