Transcript circular motion

CIRCULAR
MOTION
REMEMBER :
The
curved path of a
projectile was due to a force
(gravity) acting on a body in
a direction NOT parallel to
its line of motion.
CIRCULAR MOTION is due
to a net force acting
perpendicular to its line of
motion (radially inward)
F
V
Uniform Circular Motion (UCM)
Path is circular and the speed is constant
Such as the motion of the moon
Variable Circular Motion
Path is circular and the speed changes
Such as vertical circular motion
NOTE: For a body traveling
in a circular path, the velocity
is not constant because the
direction is always changing.
Thus it must be accelerating.
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
A
V0
R
O
B
R
V1
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
v
a
t
v  v1  v0  v1  ( v0 )
A
V0
R
O
B
R
V1
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
v  v1  v0  v1  ( v0 )
A
ΔV
V1
V0
-V0
R
O
B
R
NOTE : If θ is very small, ΔV points radially inward
(centripetally)
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
Y
A
ΔV
V
Z
R
V
X
O
B
R
Since in UCM, |V0| = |V1| = V , ΔXYZ is
isosceles --- as is ΔOAB.
Isosceles triangles with congruent vertex angles
are similar.
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
Y
A
ΔV
V
Z
R
V
X
O
B
R
ΔOAB ~ ΔXYZ
So corresponding parts are in proportion
YZ AB
V chord _ AB



XY OA
V
R
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
V chord _ AB

V
R
As θ → 00 , chord AB → arc AB ≡ Δd
V d

V
R
From kinematics, Δd = VΔt
DERIVING CENTRIPETAL
ACCELERATION FOR A BODY IN UCM
V
t
( )
V Vdt

V
R
V V

t
R
2
2
V

ac R
V
t
( )
By Newton’s Law of Acceleration
(a=F/m)
Fc mac
Centripetal force
mv
Fc 
R
2
Horizontal Circles - UCM
A 2.5 kg ball is spun in a horizontal circle
at 5.0 m/s at the end of a rope 0.75 m long.
Find (a) the centripetal acceleration and
(b) the tension in the rope.
m = 2.5 kg
R = 0.75 m
V = 5.0 m/s
T
R
V
Horizontal Circles - UCM
a)
2
2
v
(50
. m/ s)
2

ac 
=
33
m/s
075
. m
R
b)
2
. kg(50
. m/ s)
mv  25
T  Fc 
075
.
m
R
T = 83 N
2
VERTICAL CIRCLES –
VARIABLE CIRCULAR MOTION
At the top
Vmin
Fnet=Fc=Fw+Ttop
Fw
Ttop
R
Tbot
Vmax
Fw
At the bottom
Fnet=Fc=Tbot-Fw
VERTICAL CIRCLES –
VARIABLE CIRCULAR MOTION
Critical velocity, Vcrit
The lowest possible speed for
a body at the top of a
vertical circle to maintain
that circular path.
VERTICAL CIRCLES –
VARIABLE CIRCULAR MOTION
Vmin
At the top : if Vmin = Vcrit
Ttop
then Ttop = 0
Fw
R
Fc = Fw = mg
Tbot
Vmax
Fw
2
crit
mV
R
 mg
V crit  Rg
NOTE THAT THE GREATER THE RADIUS AT THE TOP,
VERTICAL CIRCLES –
VARIABLE CIRCULAR MOTION
At the bottom
Fc = Tbot - Fw
Tbot = Fc + Fw
Vmin
Fw
Ttop
2
max
mv
T bot  R
R
Tbot
Vmax
Fw
T
bot
m
(
v
 mg
2
max
R
g
)
the greater the radius at the bottom, the small
VERTICAL CIRCLES –
VARIABLE CIRCULAR MOTION
You may have noticed that
modern looping
roller coasters are never
designed with circular
loops. They are designed
with “clothoid” loops.
These have small radii at
the top and large radii at
the bottom. WHY?
r
R
CENTRIFUGAL FORCE ?
The apparent outward force acting on a
body rounding a curve.
The inertial effect of a centripetal force
acting on a body.
NOT A TRUE FORCE !!!
The body is in a non-inertial (accelerating)
frame of reference.
PROBLEM
The moon has a mass of 7.3 x 1022 kg
and orbits the Earth at a radius of
3.8 x 108 m once every 27.4 days. Find
a) The orbital speed of the moon,
b) The acceleration of the moon
towards the Earth, and
c) The gravitational force the Earth
exerts on the moon.
PROBLEM
= 2.37 x 106 s
T = 27.4 days
m
R = 3.8 x 108 m
m = 7.3 x
1022
Fgrav
kg
V
R
a) For 1 revolution,
Δd = C = 2πR and Δt = T
8
d 2R 2 (38
. x10 m)
3m/s
v


=1.0x10
6
2.37 x10 s
t
T
PROBLEM
b) ac = ?
v
1 2R
ac 

R
R T
[ ]
2
2
4 R

2
T
2
4 (38
. x10 m)
-3 m/s2
ac 
=
2.7
x
10
6
2
(2.37 x10 s)
2
8
PROBLEM
c) Fgrav = ?
2
Fgrav = Fc
Mv
M


R
R
4 MR

2
T
2
Fgrav
2R
T
[ ]
4 (7.3x10 kg)(3.8x10 m)
Fgrav 
6 2
(2.37 x10 s)
2
22
Fgrav = 2.0 x 1020 N
2
8
PROBLEM
The track at Talladega Superspeedway
has curves that are banked at 330 with
a radius of 450 m.
At what speed must
a 750 kg car make
these turns if there
is no friction due to
oil on the track?
PROBLEM
N
R
Fc
Fw
V
R
The only two forces acting on the car are
gravity and the normal force.
Of these the only one having a component
that runs radially inward is the normal force.
PROBLEM
o
N
Ny
Fw
Nx
Given : θ = 33
R = 450 m
m = 750 kg
Since Nx acts radially inward
N x = Fc
N sin θ = mv2/R
Vertically, the car is at rest
At equilibrium, Σ Fup = Σ Fdown
N y = Fw
N cos θ = mg
VERT
PROBLEM
N cos θ = mg
mg
N
cos
N sin θ = mv2/R
mg
cos
2
v
g tan  
R
HORIZ
mv
sin  
R
v  Rg tan 
v  (450m)(9.80m / s ) tan 33
2
V = 54 m/s
2
(120 MPH)
o
SATELLITES
The gravitational force acting on
a satellite is the centripetal force
responsible for its “circular” path.
Fgrav
SATELLITES
Fc = Fgrav
mv
r
v
2
2
mM
G
2
r
M
G
r
Where m is the mass of the
satellite and
M is the mass of the “planet”
GM
v
r
Remember this !!!
SATELLITES
M
2G
v
r
rv  GM
2
So, for satellites orbiting the
same body :
rv constant
2
Orbital radius is inversely proportional to the speed squared
SATELLITES
2r
v
T
GM 2r
GM
v
r
r

T
SATELLITES
GM 2r
GM 4 r

 2
r
r
T
T
3
r GM


constant
2
2
T 4
2 2
REMEMBER THIS !!!
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st
1
LAW
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nd
2 LAW
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a
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nd
2 LAW
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b
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rd
3 LAW
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SOUND FAMILIAR ?
END OF CIRCULAR MOTION