Circular Motion
Uniform Circular Motion
Speed is constant, VELOCITY is NOT.
Direction of the velocity is
ALWAYS changing.
Period (T) = time to travel around circular
path once. (C = 2πr).
v
d 2r

T
T
We call this velocity, TANGENTIAL velocity as its
direction is TANGENT to the circle.
Centripetal Acceleration
v
v
θ
s
q
r
s  v t
vt v

r
v
v
v
v

r
t
2
v
q
v
2
v
ac 
r
ac 
centripetal
acceleration
Centripetal means “center
seeking” so that means
that the acceleration
points towards the
CENTER of the circle.
Drawing the Directions correctly
So for an object traveling in a
counter-clockwise path. The
velocity would be drawn
TANGENT to the circle and the
acceleration would be drawn
TOWARDS the CENTER.
To find the MAGNITUDES of
each we have:
2r
vt 
T
v2
ac 
r
Circular Motion and N.S.L
2
Recall that according to
Newton’s Second Law,
the acceleration is
directly proportional to
the Force. If this is true:
v
FNET  m ac  m
r
2
mv
FNET  Fc 
r
Fc  Centripetal Force
NOTE: The centripetal force is a NET FORCE. It
could be represented by one or more forces. So
NEVER draw it in an FBD.
Examples
2r
vt 
T
The blade of a windshield wiper moves
through an angle of 90 degrees in 0.28
seconds. The tip of the blade moves on
the arc of a circle that has a radius of
0.76m. What is the magnitude of the
centripetal acceleration of the tip of the
blade?
2 (.76m)
vt 
 4.26 m/s
(.28s * 4)
v 2 (4.26m/s2 ) 2
ac  
 23.92 m/s2
r
0.76m
Examples
Top view
What is the minimum coefficient of static friction
necessary to allow a penny to rotate along a 33
1/3 rpm record (diameter= 0.300 m), when
the penny is placed at the outer edge of the
record?
F f  Fc
FN
mg
Side view
Ff
m v2
FN 
r
m v2
m g 
r
v2

rg
rev 1 min
33.3
*
 0.555rev
sec
min 60 sec
1sec
 1.80 sec
T
rev
0.555 rev
2r 2 (0.15)

 0.524 m / s
T
1.80
v2
(0.524) 2


 0.187
rg (0.15)(9.8)
vc 
Satellites in Circular Orbits
Consider a satellite travelling in a circular orbit around Earth. There is
only one force acting on the satellite: gravity. Hence,
 F  ma
c
Fg
m ME m v2
G 2 
r
r
mM E
Fg  G 2
r
ME
v G
r
There is only one speed a satellite may have
if the satellite is to remain in an orbit
with a fixed radius.
Examples
Venus rotates slowly about its
axis, the period being 243
days. The mass of Venus is
4.87 x 1024 kg. Determine the
radius for a synchronous
satellite in orbit around
Venus. (assume circular
orbit)
Mm m v2
Fg  Fc
G 2 
r
r
GM
2r
 v 2 vt 
r
T
Fg
2
GM 4 2 r 2
GMT 2
GMT
3

r 
r3
2
2
r
T
4
4 2
11
24
7 2
(
6
.
67
x
10
)(
4
.
87
x
10
)(
2
.
1
x
10
)
r3

2
4
1.54x109 m
Welcome


Today, Mr. Souza will give some notes about
banked curves and vertical circular motion.
You may
Sit in the first several rows in order to join the
discussion about the above two topics
OR
Sit towards the back part of the room and
begin work on tonight’s homework:

C&J p.150 # 21, 24, 26, 35, 36, 37, 40, 42
Banked Curves
A car of mass, m, travels around
a banked curve of radius r.
m v2
 Fx  r
m v2
FN sin q 
r
FN
FN cos(θ)
θ
FN sin(θ)
 Fy  0
θ
W=mg
FN cosq  m g
2
mv
FN sin q
 r
FN cosq
mg
mg
FN  cos
q
2
v
tanθ 
rg
mg
cosq
sin q  mvr
2
Banked Curves Example
Design an exit ramp so that cars travelling at 13.4 ms-1 (30.0 mph) will not
have to rely on friction to round the curve (r = 50.0 m) without skidding.
FN
2
v
tanθ 
rg
θ
FN cos(θ)
FN sin(θ)
θ
2
1 2




(13.4m  s )
1 v
1

q  tan    tan 
2 
 rg 
 (50.0m)(9.80m s ) 
W=mg
q  20.1

Vertical Circular Motion
Gratuitous Hart
Attack
F 
m v2
NB  m g 
r
m v2
NL  NR 
r
NT
NL
mg
NR
mg
NB
mv 2
r
mg
m v2
NT  m g 
r
What minimum speed must
she have to not fall off at
the top?
mg
Vertical Circular Motion
What minimum speed must she have
to not fall off at the top?
2
mv
NT  m g 
r
2
mv
mg 
r
NT
mg
v  rg
2
v  rg
Examples
The maximum tension that a 0.50 m
string can tolerate is 14 N.
A 0.25-kg ball attached to this string
is being whirled in a vertical circle.
What is the maximum speed the ball
can have
(a) at the top of the circle, and
(b)at the bottom of the circle?
(a) FNET  Fc
m v2
T  mg 
r
r
(T  mg )  v 2
m
r
v
(T  m g)
m
v
T
mg
v  5.74 m  s-1
0.50m
(14N  (0.25kg)(9.8m  s-2 ))
0.25kg
Examples
(b) At the bottom?
FNET  Fc
m v2
T  mg 
r
r
(T  mg )  v 2
m
v
r
(T  m g)
m
v
0.50m
(14N  (0.25kg)(9.8m  s-2 ))
0.25kg
T
mg
v  4.81 m  s
-1
Homework

C&J p.150 # 21, 24, 26, 35, 36, 37, 40, 42

Please watch these two video clips (they
relate to problem #37).


http://www.youtube.com/watch?v=v1VrkWb0l2M
http://www.youtube.com/watch?v=2V9h42yspbo
Homework problems and links to video clips
will be posted on Mr. Souza’s ASD site.