Transcript Lecture 16 Friday October 10

Lecture 16
Chapter 6: Circular Motion
Please sit in the first six rows
Mid Grades
9
8
7
6
5
4
3
2
1
0
A
B
C
D
F
Exam 2
9
8
7
6
5
4
3
2
1
0
A
B
C
D
F
Quick Review of Last Friday
•
Last Friday
•
The kinematics of uniform circular motion
•
TODAY
– The dynamics of uniform circular motion
– Circular orbits of satellites
– Newton’s law of gravity
Uniform Circular Motion
• Uniform
is constant
magnitude of velocity (speed)
 ( t )  angular position
  angular velocity 

t
  angular acceleration 

d  (t )
dt

t

d  (t )
dt
• Going from angular velocity to angular
displacement:
 f  i   t
T  period 
1
w here f is frequency (rad/s)
f
 
2  rad
T
for uniform circular m otion
Uniform Circular Motion
• Uniform
magnitude of velocity (speed)
ω, is constant
• But α is not zero because direction of velocity
is changing.
 
v
2
r
 r
2
Centrifugal Force
• NOT A REAL FORCE
• What supplies the force to make your acceleration
such as to cause you to move in a circle?
• Centripetal FORCE—towards the center.
This is the real force involved.
a
v
Vertical Circle
• Ferris wheel—free body diagram when at top
FNET
nTOP
FNET
w=mg
F N E T , y   m g  nT O P  
nT O P 
mv
r
2
 mg 
mv
r
mv
2
r
2
w
Ferris Wheel
At the bottom
nBOTTOM
FN ET , y  n BO TTO M  w 
n BOTTOM  w 
w=mg
mv
mv
2
r
2
r
“Apparent weight” = n is greater at bottom of
cycle than at top
Centrifuges
•
•
•
•
Diameter = 18 cm
Centripetal acceleration = 250,000 g
What is the frequency? Rev/sec and rpm
Apparent weight of a 3 mg mass in this
machine?
Workbook due today
• Go to board and explain
– 1a and c
– 1b and d
–2
–3
–4
–5
– 6a,b and c
A cyclist goes around a circular track at
constant speed. Since her speed is
constant, her acceleration is zero
33%
33%
33%
io
n
se
m
or
e
in
fo
rm
at
Fa
l
Ne
ed
Tr
ue
1. True
2. False
3. Need more
information
6.15
• A 1500 kg car drives around a flat 200m
diameter circular track. What are the
magnitude and direction of the net force on
the car? What causes this force?
When a ball on the end of a string
is swung in a vertical circle: What is
the direction of the acceleration
of
50%
50%
the ball?
d
ar
To
w
Ta
ng
en
tt
o
th
e
th
e
ce
n
te
ro
c ir
c le
,i
ft
n
...
he
.
..
1. Tangent to the
circle, in the
direction of the
ball’s motion
2. Toward the center
of the circle