Download uniform circular motion - Erwin Sitompul

Lecture 6
Ch4. TWO- AND THREE-DIMENSIONAL MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
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:
v2
a
r
(centripetal acceleration)
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:
2 r
(period)
T
v
with T is called the period of revolution, or simply the period,
of the motion.
Erwin Sitompul
University Physics: Mechanics
6/3
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.
v 2 (4)2
 8 m s2
a 
2
r
→
v
2
→
a
2m
→
v1
→
v1 = –4 m/s ^j
→
^i
v
=
–4
m/s
2
→
a = –8 m/s2 ^j
Erwin Sitompul
University Physics: Mechanics
6/4
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?
v2
2
a

v
r
T
2

640.312
48
 83.818 m s2
 8.553g
Erwin Sitompul
T
2 r
v 2
 
v
r T
v  (400)2  (500)2
 640.312 m s
1
2
T  24 s
g  9.8m s2
University Physics: Mechanics
6/5
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.)
v2
v2
a
r
r
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
6/6