# Ch 8 Dynamics II Motion in a Planex

Mechanics

Nov 14, 2013 (4 years and 7 months ago)

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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

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
/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.

(
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.

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
/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

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
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
s
m
.
ω.
 
2 83
1 41
2
What if given v or
ω

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

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
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

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

2
/r

S
F
y

=
F
c

cos
(

-

F
f

sin(

-

0

F
f

=
m
F
c

.

F
c

= mg / [
cos
(

-

m

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

2
/r

S
F
y

=
F
c

cos
(

-

F
f

sin(

-

0

F
f

=
m
F
c

.

F
c

= mg / [
cos
(

-

m

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

2
/r

S
F
y

=
F
c

cos
(

-

F
f

sin(

-

0

F
f

=
m
F
c

.

F
c

= mg / [
cos
(

-

m

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

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
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