Uniform Circular Motion

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Uniform Circular Motion Basic Principles & Formulas

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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 - PowerPoint PPT Presentation

Transcript of Uniform Circular Motion

Page 1: Uniform Circular Motion

Uniform Circular Motion

Basic Principles & Formulas

Page 2: Uniform Circular Motion

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

Page 3: Uniform Circular Motion

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)

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Uniform Circular Motion Diagram

Acceleration & Force = Parallel to Radius / Velocity = Perpendicular to Radius

r

v

F

a

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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 = InwardQ = In what direction is the acceleration?A = Inward

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Uniform Circular Motion Terms & Formulas

Velocity (v) = d or 2πr

t T

Acceleration (a) = v²

r

circumference

Time to travel around circumference once

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Uniform Circular Motion Terms & Formulas

Force (F) = ma

or

mv²

r

Remember, mass must be in Kg!

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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

= 14.3f = 1.00/14.3

= 0.0699 Hz

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UCM Example Problem (cont.)

V = (2)(π)(50.0)

14.3

= 22.0 m/sa = (22.0 m/s)²

50.0

= 9.68 m/s²

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UCM Example Problem (cont.)

F (on tires) = (615 kg)(22.0 m/s)²

50.0 m

= 5,950 N