Transcript Document

Uniform Circular Motion (UCM)
• The object travels in a circular path with a
constant speed.
• Its velocity is tangent to the circle and is
changing due to the changing direction.
• The unbalanced force is always directed
toward the center of the circle.
• The acceleration is always directed toward
the center of the circle.
v
v
F
v
F
F
v
F
F
v
v
F
F
F
v
v
Without a force toward the center the object will fly off on
a line tangent to the circle.
Something you should already know:
The Circumference (C) of a circle:
A
C  2 r   d
By the way the area is given by:
Ar
2
New Definition:
T = Period = The time for one cycle. (measured in seconds)
In the case of circular motion the period is the time for
one cycle, which is the time for one complete revolution.
time
T
# of revolutions
What are the units for an angle:
1/4
π/2
90º
Revolutions
Degrees
Radians
π/3
θθ==1/6
60º = 3.14/3 ≈ 1 rad
θ
π
180º
1/2
3/4
270º
3π/2
000º 2π
Linear Velocity vs. Angular Velocity
V = Linear Velocity – The rate at which the object
travels a distance.
s 2 r
v

t
T
Units: m/s
ω = Angular Velocity – The rate at which the object
travels through an angle.


t
Units: deg/s, rad/s,
Revolutions per minute (rpm)
Two objects travel in a circular path. Object A has a radius
of 0.8 meters and object B has a radius of 2 meters. They
both complete a revolution in 5 seconds. Compare their
linear and angular velocities.
Object A
2 r 2 (.8)
m
v

 1.0
T
5
s
 2
rad


 1.26
.8m
t
A
5
s
Object B
2m
B
2 r 2 (2)
m
v

 2.5
T
5
s
 2
rad


 1.26
t
vA  vB
5
s
A  B
Let’s do some conversions:
?
3.35
80rpm = ______
m/s radius = 40cm
m
80rev 1 min 2 (.4) m
 3.35


min
1rev
60 s
s
2.1
? radians
120º = _____
240
2 radians

radians
120 deg 
360
360 deg
2

radians  2.1 radians
3
An object traveling in a circle with a
constant speed is still accelerating due
to its changing direction.
This acceleration is called the
centripetal acceleration and is given
by the following equation:
2
v
ac 
r
Units:
m s 
2
m
m

2
s2  m
2
m
s
The steps to solving a problem:
• Draw a force diagram
• Apply Newton’s 2nd Law
• Substitute the centripetal acceleration equation
for the acceleration.
 F  ma
2
v
F  m
r
Example 1
A 300g mass is swung on a string in a horizontal circle.
(Neglect gravity) The mass has an angular velocity of
120rpm. Find the tension in the string.
 F  ma
2
v
FT  m
r
(.3)(37.7) 2
FT 
3
FT  142N
3m
FT
m=300g
s 120(2 r )
v

t
t
120(2 (3))
m

 37.7
60
s
Example 2:
A car traveling with a constant speed rounds a circular
curve of radius 15 meters. The coefficient of friction
between the road and the tires is 0.8. What is the
maximum velocity the can have and still successfully
negotiate the curve?
 F  ma
2
v
f m
r
2
v
F  m
r2
v
mg  m
r
Top view
v2
g 
r
v  gr
v  gr
2
v  (.8)(10)(15)
m
v  11
s
15m
f
v
Example 3:
A 500kg satellite is in a geosynchronous orbit. (It takes 24
hours to orbit the earth one time.) The satellite has a orbital
radius of 42000km. What is the force of gravity on the
satellite?
F  ma2
v
Fg  m
r
 30542 

Fg  500
7 
 4.2 10 
Fg  111N
r = 42000km
Fg
2r 2 (42000000)
m
v

 3054
T
24(3600)
s