Transcript Uniform Circular Motion

Uniform Circular Motion
Basic Principles & Formulas
Uniform Circular Motion Principles
Objects move at a constant speed (speed
= scalar quantity)
Objects 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
Uniform Circular Motion Terms &
Formulas
Period (T) = time it takes to make 1
revolution
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)
Uniform Circular Motion Diagram
Acceleration & Force = Parallel to Radius / Velocity = Perpendicular to Radius
v
a
r
F
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)
Q = In what direction is the force?
A = Inward
Q = In what direction is the acceleration?
A = Inward
Uniform Circular Motion Terms &
Formulas
Velocity (v) =
Acceleration (a) =
or
2πr
T
4π²r
T²
v²
r
Uniform Circular Motion Terms &
Formulas
Force (F) =
or
mv²
r
m 4π²r
T²
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
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
UCM Example Problem (cont.)
F (on tires) = (615 kg)(22.0 m/s)²
50.0 m
= 5,950 N