Transcript Circular Motion

Circular Motion
2D Forces and Motion
Which is faster? The horse on the outside
or the horse on the inside?
Merry-Go-Round
Same rotational speed for all animals on the
Merry-Go-Round because they are attached
rigidly.
Animals further out have a greater linear
speed.
Rotational Speed
 Also called
 Angular Speed
 Circular Speed
 ω (lower case
Omega Ω)
Linear Speed (v)
 Also called
 Tangential Speed
 v or vT
Describe Earth’s motion using the
words rotate, revolve, and axis.
Axis: Straight line around
which rotation takes place.
Days on Earth.
Rotation: Spin. Axis is
located within the object.
Days on Earth.
Revolution: Object turns
about an external axis.
Earth years.
Frequency vs. Period
 Period (T)- The time it takes for
one full rotation or revolution of
an object in seconds.
1
T
f
 Frequency (f)- The number or
rotations or revolutions per unit
time, measured in Hertz (Hz)
1
f
T
Revolution Lab
 QUESTIONS:
 1) HOW IS RADIUS RELATED TO REVOLUTION SPEED?
 2) WHAT HAPPENS WHEN YOU ARE SPINNING THE STOPPER AT A
CONSTANT RATE AND THEN SUDDENLY PULL DOWN ON THE STRING? WHY
DOES THIS HAPPEN?
 3) DOES A SPINNING OBJECT ACCELERATE? IF SO, WHAT IS THE DIRECTION
OF ACCELERATION?
 CHALLENGES:
 1) Spin a rubber stopper above your head at several different
lengths to answer the question above. Do multiple trials.
 2) Graph your data to help find the mathematical relationship
between radius and revolution speed.

Tangential Speed
C 2r
vT  
T
T
HOW IS RADIUS RELATED TO
REVOLUTION SPEED?
In Uniform Circular Motion (fixed tangential speed), a
larger radius will result in a smaller rotational speed.

vT

r
r

v
HOW IS RADIUS RELATED TO
REVOLUTION SPEED?
In Uniform Circular Motion (fixed tangential speed), a
larger radius will result in a smaller rotational speed.

r

v
 r
d
dr

dt dt
d
vT  r
dt
v T  r
Rotational Speed
2r
vT 
T
vT

r

2r 2


Tr
T
1
T
f
  2f
WHAT HAPPENS WHEN YOU ARE SPINNING THE STOPPER
AT A CONSTANT RATE AND THEN SUDDENLY PULL DOWN
ON THE STRING? WHY DOES THIS HAPPEN?
It spirals in because you apply a constant
force inward. You reduce the radius.
WHAT IS THE DIRECTION OF ACCELERATION OF
AN OBJECT IN UNIFORM CIRCULAR MOTION?
v
a
t

v r

v
r
r
v  v
r
2
vru v

ar 

rt
r
WHAT IS THE DIRECTION OF ACCELERATION OF
AN OBJECT IN UNIFORM CIRCULAR MOTION?
There is a radial component
of acceleration responsible
for the constant direction
change, and a tangential
component of acceleration
which results in an increase or
decrease in tangential speed.
2
dv
aT 
dt
2
v
ar 
r
a  a a
2
r
2
T
d vˆ v



ˆ
a
 r
a
1
T
  tan  
dt
r
a r 
Describe the path of the stopper IF you were to cut the
string between the tube and the bottom weight
Centripetal Force Fc
 A force of some kind is required to
maintain circular motion. Why?
 Any force that causes an object to
follow a circular path is called a
centripetal force.
 Centripetal means “center-seeking”
 Always acts inwards
Centripetal Acceleration ac
vT
ac 
r
2
Tangential
Velocity
Radius
Centripetal Force
F  ma
Fc  mac
2
v
ac 
r
2
mvT
Fc 
r
The banked ramp exit
 The goal is to design a banked ramp exit that drivers can
round safely even on ice.
 radius of curve is 50m
 speed of cars- 13.4m/s
 What should the angle of the bank be?
The banked ramp exit
FN
FNy
F
FNy  Fg  0
FNx  ma c
FNy  Fg
FN sin   ma c
y
FNx
Fg
F
FN cos  mg

mg
FN 

cos

 
1 a c
  tan   
 g 
2


1 v
  tan     20
rg 

x
mg
sin   ma c
cos
gtan  ac
Review Problems
Derive the expression (fully simplified) that
will determine the time it will take for a
projectile launched on flat ground to reach its
maximum height.
How long will it take to land?
v 0 sin 
t
g
2v 0 sin 
t
g
Review Problems
v0x=15m/s
What is the
range R?
a) 750m
b) 375m
c) 105m
d) 210m
e) 150m
h=250m
R
c) 105m
Review Problems
v0x=15m/s
h=250m
R
a) 72m/s
What is the
speed of the
object when it
hits the ground?
a) 72m/s
b) 15m/s
c) 150m/s
d) 70m/s
e) 21m/s
Review Problems
v0
A (peak)
What is the direction of the acceleration vector
and velocity vector at point A?
a) 0m/s2 and 0m/s
d) a v
b) a v
e) a
0m/s
c) a v
c)
Review Problems
A very agile physics student is standing on one
of those spinny things in a playground without
slipping. Which force provides the student’s
centripetal acceleration?
a) Normal Force
d) Centrifugal Force
b) Weight
e) None
c) Friction on shoes
f) Abnormal force
c)
Review Problems
a)
b)
c)
d)
e)
Two quarters are on a spinning turntable.
One head side up and one tail side up.
Heads is at a distance R/2 from the center.
Tails is a distance R from the center.
What is the ratio of accelerations (ah/at)?
b)
2
vh
R
2/1
2 
2 2 


a
v
r

r
r
rh 2 1
r
h
h
t
h
t
h
1/2
 2   2 
 2 2   
a t vt
rh v t  rh  rt  rt R 2
1
rt
2^(1/2)
2^-(1/2)
Review Problems
A
C
A
B
C
B (along rope)
C
E
D (tangent to curve)
For the pendulum on the left, which vector on the right
possibly shows the direction of acceleration at
point A?
Review Problems
A
B
C
A
A
B
C
E
D
For the pendulum on the left, which vector on the right
shows the direction of acceleration at point B?
Review Problems
L
θ
For the conical pendulum above, find a fully simplified
expression for the period in terms of theta, L, g,
and other constants.
Review Problems
We need an equation with T in it!
θ
L
2r
v
T
r
2r
T
v
What is r?
r  Lsin 

Ok, so what is v?
Take it easy, make an FBD.


FTx  ma c
FTy  Fg  0
Review Problems
2r
T
v
θ
L
r  Lsin 
r
v2
F
T sin   m
r
FTx  ma c

FTy  Fg  0
2
mg
v
sin  m
cos
r
FT cos  mg
v  rgtan 

Review Problems
r  Lsin 
θ
L
r
T
2r
T
v
2L sin  
L sin gtan 


L sin 
 L sin 
L sin 
sin  sin 

 cos
sin

tan 
cos
v  rgtan 
L cos
T  2
g