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Lecture 7
Purdue University, Physics 220
1
PHYSICS 220
Lecture 07
Circular Motion
Lecture 7
Purdue University, Physics 220
2
Examples of Circular Motion
Lecture 7
Purdue University, Physics 220
3
Uniform Circular Motion
• Assume constant
speed
• The direction of the
velocity is continually
changing
– The vector is always
tangent to the circle
• Uniform circular
motion assumes
constant speed
period of the motion: T = 2r/v
[s]
Angular Variables
• The motion of objects moving in circular (or nearly
circular) paths, is often described by angles measured in
radians rather than degrees.
• The angle  in radians, is defined as:
s

r
• If s = r the angle is 1 rad
• If s = 2r (the circumference of the circle) the angle is
2 rad. (In other words, 360° = 2 rad.)
Lecture 7
Purdue University, Physics 220
5
Circular Motion
• Period = 1/frequency
T = 1/f
Time to complete 1 revolution
• Angular displacement D = 2-1
How far it has rotated
• Angular velocity wav = D/Dt
How fast it is rotating
D
Units: radians/second w  lim
Dt 0 Dt
(2 = 1 revolution)
w 2/T
Lecture 7
Purdue University, Physics 220
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Circular to Linear
• Displacement Ds = r D(in radians)
• Speed |v| = |Ds/Dt| = r |D/Dt| = r|w
|v| = 2rf
|v| = 2r/T
• Direction of v is tangent to circle
Lecture 7
Purdue University, Physics 220
7
Examples
• The wheel of a car has a radius of 0.29 m and is being
rotated at 830 revolution per minute (rpm) on a tirebalancing machine. Determine the speed (in m/s) at which
the outer edge of the wheel is moving:
The speed could be obtained by |v| = 2r/T
T
1
 1.2  103 min/rev=0.072 s
830rev/min
2 r 2 (0.29m)
v

 25m / s
T
0.072s
• A CD spins with an angular velocity 20 radians/second.
What is the linear speed 6 cm from the center of the CD?
v = r w = 0.06  20 = 1.2 m/s
Lecture 7
Purdue University, Physics 220
8
iClicker
Bonnie sits on the outer rim of a merry-go-round with
radius 3 meters, and Klyde sits midway between the center
and the rim. The merry-go-round makes one complete
revolution every two seconds.
Klyde
Bonnie
Klyde’s speed is:
A) the same as Bonnie’s
B) twice Bonnie’s
C) half Bonnie’s
VKlyde
Bonnie travels 2  R in 2 seconds
1
 VBonnie
2
vB = 2  R / 2 = 9.42 m/s
Klyde travels 2  (R/2) in 2 seconds vK = 2  (R/2) / 2 = 4.71 m/s
Lecture 7
Purdue University, Physics 220
9
Centripetal Acceleration
•Magnitude of the velocity vector is constant, but direction is
constantly changing
•At any instant of time, the direction of the instantaneous
velocity is tangent to the path
•Therefore: nonzero acceleration
Lecture 7
Purdue University, Physics 220
10
Uniform Circular Motion
Circular motion with constant speed
R
2
a
v
Recall:
v=wR
v
2
ar 
w R
R
centripetal
acceleration
• Instantaneous velocity is tangent to circle
• Instantaneous acceleration is radially inward
• There must be a force to provide the acceleration
Lecture 7
Purdue University, Physics 220
11
Circular Motion
A ball is going around in a circle attached to a string. If the string
breaks at the instant shown, which path will the ball follow?
2
1
3
4
v
5
Answer: 2
Lecture 7
Purdue University, Physics 220
12
Artificial Gravity
Lecture 7
Purdue University, Physics 220
13
Roller Coaster Example
What is the minimum speed you must have at the
top of a 20 meter diameter roller coaster loop, to
keep the wheels on the track.
y-direction: F = ma
N + mg = m a
N
y
Let N = 0, just touching
mg = m a
mg = m v2/R
g = v2 / R
v = sqrt(g*R) = 10 m/s
Lecture 7
mg
Purdue University, Physics 220
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Unbanked Curve
What force accelerates a car around a turn on a level
road at constant speed?
A) it is not accelerating
B) the road on the tires
C) the tires on the road
D) the engine on the tires
Lecture 7
Purdue University, Physics 220
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Unbanked Curve
What is the maximum velocity a car can go around an
unbanked curve in a circle without slipping?
y : N  mg  0 N  mg
mv
x : Fc  fs   s N   s mg 
r
2
v   s gr
fs  s N
The maximum
velocity to go
around an unbanked curve
depends only on s
(for a given r)
Dry road: s=0.9
Icy road: s=0.1
Lecture 7
Purdue University, Physics 220
16
Banked Curve
A car drives around a curve with radius 410 m at a
speed of 32 m/s. The road is banked at 5.0°. The
mass of the car is 1400 kg.
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
Lecture 7
Purdue University, Physics 220
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Banked Curve
y

x
  50
y-direction
F y  may  0
N cos  mg  f sin   0
(1)
r  410m
v  32m / s
N
x-direction
F x  max  ma
2
v
N sin   f cos  ma  m
r
Lecture 7
Purdue University, Physics 220
(2)
W
f
18
Banked Curve
2 equations and 2 unknown we can solve for N in
(1) and substitute in (2)
f sin   mg
v2
N 
N sin   f cos  m
r
cos 
2
f
sin


mg
mv



 sin   f cos 
cos
r
2
mv
f (sin 2   cos2  ) 
cos  mg sin 
r
 v2

f  m  cos  g sin    2300N
 r

Lecture 7
Purdue University, Physics 220
19
Banked Curve
A car drives around a curve with radius 410 m at a
speed of 32 m/s. The road is banked at 5.0°. The
mass of the car is 1400 kg.
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
f 0
2
v
cos  gsin 
r
v  gr tan   19m / s
Lecture 7
Purdue University, Physics 220
20
iClicker
Suppose you are driving through a valley whose bottom
has a circular shape. If your mass is m, what is the
magnitude of the normal force N exerted on you by the
car seat as you drive past the bottom of the hill.
A) N < mg
B) N = mg
C) N > mg
a=v2/R
R
correct
Since there is centripetal acceleration, the
normal force is greater than simply mg
N
v
F = ma
mg
N - mg = mv2/R
N = mg + mv2/R
Lecture 7
Purdue University, Physics 220
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