Lecture 6
Ch4. TWO- AND THREE-DIMENSIONAL MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2012
Uniform Circular Motion
 A particle is in uniform circular motion if
it travels around a circle or a circular arc
at constant (uniform) speed.
 Although the speed does not vary, the
particle is accelerating because the
velocity changes in direction.
 The velocity is always directed tangent to the circle in the
direction of motion.
 The acceleration is always directed radially inward.
 Because of this, the acceleration associated with uniform
circular motion is called a centripetal (“center seeking”)
acceleration.
Erwin Sitompul
University Physics: Mechanics
6/2
Uniform Circular Motion
→
 The magnitude of this centripetal acceleration a is:
a
v
2
(centripetal acceleration)
r
where r is the radius of the circle and v is the speed of the
particle.
 In addition, during this acceleration at constant speed, the
particle travels the circumference of the circle (a distance of
2πr) in time of:
T 
2 r
(period)
v
with T is called the period of revolution, or simply the period,
of the motion.
Erwin Sitompul
University Physics: Mechanics
6/3
Centripetal Acceleration
v

v
r
v
vt

v
a 
Erwin Sitompul
r
r
v
t

v
2
r
University Physics: Mechanics
6/4
Checkpoint
An object moves at constant speed along a circular path in a
horizontal xy plane, with the center at the origin. When the
object is at x=–2 m, its velocity is –(4 m/s) ^j.
Give the object’s (a) velocity and (b) acceleration at y=2 m.
a
→
v2
v
2
r
→
a

(4)
2
 8m s
2
2
2m
→
v1
→
v1 = –4 m/s ^j
→
v2 = –4 m/s ^i
→
a = –8 m/s2 ^j
Erwin Sitompul
University Physics: Mechanics
6/5
Example: Fighter Pilot
Fighter pilots have long worried about taking
a turn too tightly. As a pilot’s body
undergoes centripetal acceleration, with the
head toward the center of curvature, the
blood pressure in the brain decreases,
leading to unconsciousness.
What is the magnitude of the acceleration, in g units, of a pilot
whose aircraft enters a horizontal circular turn with a velocity of
vi = 400i^ + 500j^ m/s and 24 s later leaves the turn with a velocity
^
of vf = –400i
– 500j^ m/s?
a

v
2
r
2

2
T 
v
2 r
v
T
v
640.312
48
 83.818 m s
 8.553 g
Erwin Sitompul

2
v

2
r
(400)  (500)
2
 6 4 0 .3 1 2 m s
g  9.8 m s
T
2
1
2
T  24 s
2
University Physics: Mechanics
6/6
Example: Aston Martin
An Aston Martin V8 Vantage has a
“lateral acceleration” of 0.96g. This
represents the maximum centripetal
acceleration that the car can attain
without skidding out of the circular path.
If the car is traveling at a constant
speed of 144 km/h, what is the
minimum radius of curve it can
negotiate? (Assume that the curve is
unbanked.)
a
v
2
 r 
r
v
2
a

(40 m s)
2
2
(0.96)(9.8 m s )
 170 m
Erwin Sitompul
• The required turning radius r is
proportional to the square of
the speed v
• Reducing v by small amount
can make r substantially
smaller
University Physics: Mechanics
6/7
Relative Motion in One Dimension
 The velocity of a particle depends on the reference frame of
whoever is observing or measuring the velocity.
 For our purposes, a reference frame is the physical object to
which we attach our coordinate system.
 In every day life, that object is the ground.
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University Physics: Mechanics
6/8
Thom(p)son Encounters Relative Velocity
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University Physics: Mechanics
6/9
Relative Motion in One Dimension
 Suppose that Alex (at the origin of frame A) is parked by the
side of a highway, watching car P (the ”particle”) speed past.
Barbara (at the origin of frame B) is driving along the highway
at constant speed and is also watching car P.
 Suppose that both Alex and Barbara measure the position of
the car at a given moment. From the figure we see that
x PA  x PB  x BA
“The coordinate of P as measured by A is equal
to the coordinate of P as measured by B plus
the coordinate of B as measured by A”
Erwin Sitompul
University Physics: Mechanics
6/10
Relative Motion in One Dimension
 Taking the time derivative of the previous equation, we obtain
d
dt
( x PA ) 
d
dt
( x PB ) 
d
dt
( x BA )
v PA  v PB  v BA
“The velocity of P as measured by A is equal to
the velocity of P as measured by B plus the
velocity of B as measured by A”
Erwin Sitompul
University Physics: Mechanics
6/11
Relative Motion in One Dimension
 Here we consider only frames that move at constant velocity
relative to each other.
 In our example, this means that Barbara drives always at
constant velocity vBA relative to Alex.
 Car P (the moving particle), however, can accelerate.
d
dt
(v PA ) 
d
dt
(v PB ) 
a PA  a PB
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d
dt
(v BA )
• Constant
University Physics: Mechanics
6/12
Example: Relative Velocity
Suppose that Barbara’s velocity relative
to Alex is a constant vBA = 52 km/h and
car P is moving in the negative direction
of the x axis.
(a) If Alex measures a constant vPA = –78 km/h for car P, what
velocity vPB will Barbara measure?
v BA  52 km h
v PA   78 km h
P m o v in g in th e n eg ativ e d irectio n
v PA  v PB  v BA
v PB  v PA  v BA  (  78)  (52)   130 km h
Erwin Sitompul
University Physics: Mechanics
6/13
Example: Relative Velocity
Suppose that Barbara’s velocity relative
to Alex is a constant vBA = 52 km/h and
car P is moving in the negative direction
of the x axis.
(b) If car P brakes to a stop relative to Alex (and thus relative to
the ground) in time t = 10 s at constant acceleration, what is
its acceleration aPA relative to Alex?
v 0 , PA   78 km h , t  10 s
a PA 
v PA  v0 ,PA
t

0  (  78)
10
 7.8 km h s
 2.167 m s
2
(c) What is the acceleration aPB of car P relative to Barbara
during the braking?
a PB 
Erwin Sitompul
v PB  v0,PB
t

v AB  v0 ,PB
t

 52  (  130)
10
 7.8 km h s
2
 2.167 m s
University Physics: Mechanics
6/14
Relative Motion in Two Dimensions
 In this case, our two observers are again watching a moving
particle P from the origins of reference
frames A and B, while
→
B moves at a constant velocity vBA relative to A.
 The corresponding axes of these two frames remain parallel,
as shown, for a certain instant during the motion, in the next
figure.
 The following equations
describe the position, velocity,
and acceleration vectors:
rPA  rPB  rBA
v PA  v PB  v BA
a PA  a PB
Erwin Sitompul
University Physics: Mechanics
6/15
Example: Relative Velocity
A plane moves due east while the pilot points the plane
somewhat south of east, toward a steady wind that blows to the
northeast.
→
The plane has velocity vPW relative to the wind, with an airspeed
(speed relative to wind) of 215.0 km/h, directed at angle θ south
of east.
→
The wind has velocity vWG relative to the ground with speed
65.0 km/h, directed 20.0° east of north.
→
What is the magnitude of the velocity vPG of the plane relative to
the ground, and what is θ.
Erwin Sitompul
University Physics: Mechanics
6/16
Example: Relative Velocity
v PG  v PW  vW G
v PW  215 km h  
vW G  65 km h  70 
v PG  v PG km h  0 
v PW sin    vW G sin 70 
2 1 5  sin    6 5  sin 7 0 
   16.50 
Erwin Sitompul
v PG  v PW cos   vW G cos 70 
 215  cos(  16.50  )  65  cos 70 
 228.38 km h
University Physics: Mechanics
6/17
Exercise Problems
1. A cat rides a mini merry-go-round turning with
uniform circular motion. At time t1 = 2 s, the
^
^
cat’s velocity is v1 = 3i + 4j m/s, measured on
a horizontal xy coordinate system. At t2 = 5 s,
^
^
its velocity is v2 = –3i – 4j m/s.
What are (a) the magnitude of the cat’s
centripetal acceleration and (b) the cats
average acceleration during the time interval
t2 – t1?
Answer: (a) 5.236 m/s2; (b) –2i – 2.667 j m/s2.
2. A suspicious-looking man runs as fast as he can along a moving sidewalk
from one end to the other, taking 2.50 s. Then security agents appear, and
the man runs as fast as he can back along the sidewalk to his starting
point, taking 10.0 s. What is the ratio of the man’s running speed to the
sidewalk’s speed?
Answer: 1.67
Erwin Sitompul
University Physics: Mechanics
6/18