Chapter Eight
Dynamics II
Motion in a Plane
Uniform Circular Motion
Uniform Circular Motion:
Movement at constant speed
around a circle of radius, r.
Period (T)
–
the time it takes to go
around the circle once.
1circumference
2
v
1period t
Uniform Circular Motion
rad
s
r
CCW > 0
CW < 0
Angular Velocity
avg
omega
t
t 0
d
lim
t dt
f i
t
v
t
vs
ω
v
t
=
ds
/
dt
s = r
θ
ds
/
dt
=
r d
θ
/
dt
v
t
= r d
θ
/
dt
ω
=
d
θ
/
dt
v
t
=
ω
r
(with
ω
in
rad
/s)
Centripetal Acceleration
v
0
=
v
r
r
v
f
=
v
v
f
v
0
a
c
=
v
f

v
0
r
r
v t
By similar triangles as
θ
=> 0
2
c f t
f
c
t
a v v
v
v
a
v r r r
Kinematics of
UCM
a
c
=
a
r
Dynamics
of
UCM
c c
F ma
2
c
v
F m
r
Rank in order, from largest to
smallest, the centripetal accelerations
(
a
c
)
a
to (
a
c
)
e
of particles a to e.
Test Your Understanding
(
a
c
)
b
>
(
a
c
)
e
>
(
a
c
)
a
=
(
a
c
)
c
>
(
a
c
)
d
Rank in order, from largest to smallest, the centripetal
accelerations (
a
c
)
a
to (
a
c
)
e
of particles a to e.
Test Your Understanding
Acceleration for UCM
We have shown that:
a
c
= v
t
2
/r and
v
t
=
ω
r
Substituting gives
a
c
=
ω
2
r
2
/r=
ω
2
r
Nonuniform Circular Motion
2 2
c t
a a a
If the particle velocity changes,
there is
tangential acceleration,a
t
.
If a
t
is constant, 1D kinematic
equations may be applied.
Nonuniform
Circular Motion
For non

uniform circular
motion:
•
the acceleration vector
no longer points
towards the center of
the circle.
•
a
t
can be positive (ccw)
or negative (cw) but a
c
is
always positive.
Angular
vs
Tangential
Acceleration
a
t
=
dv
/
dt
where v =
v
t
v
t
= r
ω
a
t
= r d
ω
/
dt
d
ω
/
dt
=
angular acceleration
:
α
= d
ω
/
dt
, in
rad
/s
2
a
t
= r
α
or
α
= a
t
/r
Angular Acceleration
Angular Acceleration
if
α
is constant,
θ
and
ω
can be found with the angular
equivalents of the kinematic equations.
f i
2
1
2
f i
2 2
f i
t
t t
2
Newton & the Moon
•
What is the acceleration of the Moon due to its
motion around the Earth?
•
What we know (Newton knew this also):
–
T
= 27.3 days = 2.36 x 10
6
s
(period ~ 1 month)
–
R
= 3.84 x 10
8
m
(distance to moon)
–
R
E
= 6.35 x 10
6
m
(radius of earth)
R
R
E
UCM Dynamics
5/1/2009
36
AP Physics C
UCM Dynamics
5/1/2009
37
AP Physics C
net
mv
F ma
r
2
If a 25 kg object on a 2 m rope is moving in a circle at
uniform speed and the tension in the rope is 100 N.
What is the angular velocity?
Known
: Mass, radius,
F
net
Unknown
:
ω
Plan
: Find v, then
ω
Problem 1
5/1/2009
38
AP Physics C
net
mv
F
r
2
v
2
25
100
2
m
v.
s
2 83
Find V:
Find
ω
:
v
ω
r
m
s
rad
s
m
.
ω.
2 83
1 41
2
What if given v or
ω
and asked for F
r
?
5/1/2009
AP Physics C
39
Problem 2
A car is going around a curve which has a 50 m radius.
If the mass of the car is 1500 kg, and the coefficient of
friction is 1, how fast can the car make the curve without
sliding? The road is unbanked.
a=v
2
/r
W=mg
F
c
F
f
5/1/2009
AP Physics C
40
Problem 2
con’t
Car going around a turn
What are the forces that cause this
radial
acceleration?
a=v
2
/r
W=mg
F
c
Car is heading into the screen.
Car going around a turn
The
fastest the car can go
around the
turn without
sliding is
when the
friction is maximum:
F
f
=
m
F
c
.
a=v
2
/r
W=mg
F
c
F
f
m
F
c
Max Velocity on Curve
Apply N2L

S
F
x
= F
f
= ma =
mv
2
/r
For max speed,
we need
max
force of friction:
F
f
=
m
F
c
=
m
浧
Substituting:
m
mg
=
mv
2
/r
Solving for v:
a=v
2
/r
W=mg
F
c
F
f
m
F
c
max
v gr
m
Banked turn
W=mg
F
c
a=v
2
/r
F
f
GSP
Banked turn
S
F
x
=
F
c
sin(
⤠⬠+
f
cos
(
⤠㴠
浶
2
/r
S
F
y
=
F
c
cos
(
⤠

F
f
sin(
⤠

浧m㴠
0
F
f
=
m
F
c
.
F
c
= mg / [
cos
(
⤠

m
獩渨
⥝
F
c
W=mg
a=v
2
/r
F
f
F
c
θ
θ
F
f
mg
F
f
sin
θ
F
c
cos
θ
F
c
sin
θ
F
f
cos
θ
gr{sin(
θ) +μcos(θ)}
v =
cos(
θ)  μsin(θ)
Banked turn
S
F
x
=
F
c
sin(
⤠⬠+
f
cos
(
⤠㴠
浶
2
/r
S
F
y
=
F
c
cos
(
⤠

F
f
sin(
⤠

浧m㴠
0
F
f
=
m
F
c
.
F
c
= mg / [
cos
(
⤠

m
獩渨
⥝
F
c
W=mg
a=v
2
/r
F
f
F
c
θ
θ
F
f
mg
F
f
sin
θ
F
c
cos
θ
F
c
sin
θ
F
f
cos
θ
gr{sin(
θ) +μcos(θ)}
v =
cos(
θ)  μsin(θ)
Banked turn
S
F
x
=
F
c
sin(
⤠⬠+
f
cos
(
⤠㴠
浶
2
/r
S
F
y
=
F
c
cos
(
⤠

F
f
sin(
⤠

浧m㴠
0
F
f
=
m
F
c
.
F
c
= mg / [
cos
(
⤠

m
獩渨
⥝
F
c
W=mg
a=v
2
/r
F
f
F
c
θ
θ
F
f
mg
F
f
sin
θ
F
c
cos
θ
F
c
sin
θ
F
f
cos
θ
gr{sin(
θ) +μcos(θ)}
v =
cos(
θ)  μsin(θ)
Banked turn

Minimum Speed
Find
an equation for
the
minimum speed necessary
?
What changes in what we did
for
max
speed?
F
c
W=mg
a=v
2
/r
F
f
5/1/2009
AP Physics C
49
Minimum Speed
Con’t
F
c
W=mg
a=v
2
/r
F
f
Banked Curve
•
For a car traveling with speed
v
around a curve of radius
r
,
determine a formula for the angle at which a road should be
banked so that no friction is required.
∑
F
y
= ma
y
F
N
cos
θ
–
mg = 0
F
N
= mg/
cos
θ
F
N
sin
θ
–
mv
2
/r
mg(sin
θ
)/
cos
θ
= mv
2
/r
tan
θ
= v
2
/
gr
θ
= tan

1
(v
2
/
gr
)
5/1/2009
50
AP Physics C
Problem #1
•
What is the maximum speed with which a 1200 kg car can
round a turn of
radius 80.0 m is
on a flat road, if the coefficient
of friction between tires and road 0.65?
5/1/2009
51
AP Physics C
g
m
fr
F
N
F
R f
F F
N
mv
μF μmg
r
2
v
μrg (.)(.m)(.m/s ).m/s
2
65 80 0 9 8 22 5
Problem #2
•
Highway curves are marked with a suggested speed. If this
speed is based on what would be safe in wet weather,
μ
= .7,
estimate the radius of curvature for a curve marked 50 km/h
curve. See the free

body diagram, which assumes the center of
the curve is to the right in the diagram.
5/1/2009
52
AP Physics C
g
m
fr
F
N
F
2
R fr s N s
F =F mv r =
μ F =μ mg
2
s
r = v
μ g
2
2
1m s
30km h
3.6km h
= = 28m
0.7 9.80m s
Problem #3
•
What is the minimum speed that a roller coaster must be
traveling when upside down at the top of a circular loop so that
the passengers do not fall out? Assume a radius of curvature
of 7.6 m.
5/1/2009
53
AP Physics C
g
m
N
F
min
2
min
 g= 0 v = rg
v
r
2
= 9.80m s 7.6m = 8.6m s
2
v
r
N
F =F +mg=ma =m
N
2
F =m  g
v
r
At minimum speed F
N
= 0
5/1/2009
AP Physics C
54
Problem #4
You swing a pail of water in a vertical circle of radius
r. The speed of the pail is
v
t
at the top of the circle.
(a) Find the force exerted on the water by the pail at
the top of the circle. (b) Find the minimum value of
v
t
for the water to remain in the pail. (c) Find the
force exerted by the pail on the water at the bottom
of the
circle, where the pail’s speed is v
b
.
(a)
Two forces act on the water: F
P
and mg.
The a
c
is
–
v
2
/r.
(b)
t
Z P
v
F F mg ma m
r
2
t
P
v
F m g
r
2
5/1/2009
AP Physics C
55
Problem #4
Con’t
Similar to top except that force of pail is positive
b
P
v
F m g
r
2
What would be the minimum speed to swing a
bucket containing 3kg of water in a vertical circle
with a 1 m radius without spilling the water?
b
P
v
F m g
r
2
0
m
s
v rg.
9 8
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