Uniform Circular Motion
A special form of 2D motion
is circular motion.
Examples: ball on a string,
cars exiting the highway,
planets orbiting the sun
WHY?
The direction of the
velocity changes at
every point
Circular motion is
accelerated motion, even
if the speed of the object
remains constant.
Uniform Circular Motion
• an object in uniform
circular motion experiences
a centripetal acceleration
of magnitude:
v2
ac 
r
• centripetal acceleration
always points toward the
center of the circle.
Period and Frequency
Uniform circular motion is periodic. Its
behavior repeats after one period, T.
• The period, T, is the amount of time it
takes for one complete revolution.
• The frequency, f, of motion is the rate
with which the motion repeats.
• The frequency is the inverse of the period.
per second
Hertz
1
f
T
seconds
Period and Speed
The period and frequency can be related to
the speed of an object in uniform circular
motion.
The object travels one circumference, 2πr,
every period.
Thus . . . . .
2r
v
T
The Dynamics of Circular Motion
v
To remain in uniform
circular motion, an
object must be constantly
accelerating toward the
center of the circle.
v
v
Fc  m
r
2
To constantly be
accelerating toward the
center of the circle, the
object must experience a
net centripetal force.
Horizontal circular motion
‘view from above’
Consider a box being whirled
in a horizontal circle on a
frictionless table. Draw the
forces acting on the box.
FT
Fg and FN are not drawn since
they point into and out of the
paper respectively. (They are not
 Fc = FT
v
Fc  m
r
2
important to the problem as they do not
act into or opposite the direction of the
acceleration)
v
FT  m
r
2
In what direction does this box
accelerate?
ALWAYS towards the center
Vertical circular motion
‘view from the side’
Note: At the top of
the circle, Fg
would be added to FT
FT
Fg
Consider a box being whirled
in a vertical circle. Draw the
forces acting on the box.
Fg is now drawn since it is
opposite to the direction the
box accelerates. (There is no FN as
it is not on a surface)
 F c = F T - Fg
 v2
Fc  m
r
v
FT  Fg  m
r
2
A person is flying a 0.088 kg model airplane
connected to a 10 m long string in a horizontal
circle. The string exerts a force of 2.65 N on the
plane. What is the plane’s speed?
Given:
m = 0.088 kg
r = 10 m
FT= 2.65 N
v = ? m/s
 Fc = FT
v
Fc  m
r
2
‘view from above’
v
FT  m
r
2
2.65 = 0.088
v2
10
v = 17.4 m/s
FT
A pendulum 80 cm long is lifted above its equilibrium
position and released. As the 0.050 kg pendulum bob
passes through its lowest position the tension in the
pendulum cord is 0.735 N, what is the pendulum's
speed at this point?
Given:
‘view from the side’
 Fc = FT - Fg
m = 0.050 kg
2
v
2
FT  Fg  m
r = 80 cm = 0.80 m
v
Fc  m
r
FT = 0.735N
r
v = ? m/s
2
v
0.735 – 0.49 = 0.050
0.80
FT
v = 2.0 m/s
ac
Fg
A special case – just completing a vertical loop
Unlike horizontal circular motion, in vertical circular motion the
speed, as well as the direction of the object, is constantly
changing. Gravity is constantly either speeding up the object
as it falls, or slowing the object down as it rises.
If we wanted to calculate the minimum or
critical velocity needed for the block to
just be able to pass through the top of the
circle without the rope sagging then
we would start by setting the tension in the
zero
rope to ____________.
* v  gr
ΣFc = Fg
mv 2
 Fc 
r
mv 2
Fg 
r
mv
mg 
r
2
*only valid for the
minimum velocity at the
top to just complete a
vertical loop.