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11/18/10
• Finish up Projectile Motion
with test corrections and
extra credit.
• The physics of stunts how
things relate in projectile
motion
• Circular Motion
– Uniform Circular Motion
• Bulls eye lab
Stunt Car
11/19/10

• Extra Credit
• “Extra” extra credit
• Circular Motion
–Uniform Circular
Motion
• Bulls eye lab
What is the difference between rotation and
revolution?

What is the difference between tangential speed and
rotational speed?
Uniform Circular Motion
Tangential speed is how fast a point on a
circular object is moving at a certain distance
from the center.

Rotational speed is how many degrees (or
radians) a point on the circle goes through in a
period of time.
Which part of the
merry-go-round has
the greatest
rotational speed?

The greatest
tangential speed?
Every point on a circle has the same
rotational speed. The further out you go from
the center, the higher the tangential speed is.
Uniform circular motion can be described as the
motion of an object in a circle with a fixed radius,
at a constant speed.

As an object moves in a circle, it is constantly
changing its direction.

It is therefore constantly
accelerating!
At all instances, the object is moving tangent
to the circle. Since the direction of the
velocity vector is the same as the direction of
the object's motion, the velocity vector is
directed tangent to the circle as well. An
object undergoing uniform circular motion is
accelerating due to its change in direction.
The direction of the acceleration is inwards.
An object moving
uniformly in a
circular path always
has centripetal
acceleration which is
directed toward the
center of the circle.

a c=     v2/r
An object moving
uniformly in a
circular path always
has centripetal
acceleration which is
directed toward the
center of the circle.

a c=     v2/r
The force needed to
keep an object moving
in a circular path is
called the centripetal
force. It is the force
that produces the
acceleration and is
always directed toward
the center.

Fc=   mv2/r
ac =  v 2/r

Fc = mac
Fc = mv 2/r
11/22/10
• Turn in “Extra” extra credit
• Homework
• Circular Motion
–Uniform Circular
Motion
–Circular motion problems
• Bulls eye lab
ac =   v 2/r
What is the
minimum tangential
velocity needed to
have the cannon
ball orbit the earth?
Rally Drift
A) Find tangential velocity

B) Find centripetal acceleration (ac) and force (Fc)
Consider the boys:
A) Find tangential
velocity
B) Find centripetal
acceleration
A) Find centripetal
force holding
them in
11/23/10
• Extra Credit Problem
• Homework collection
• Lab is due
• Quiz Problems
• Have a Great Thanksgiving!
Practice #1
A 1200-kg car makes a 90-degree turn
with a speed of 10.0 m/s. The radius of
the circle through which the car is
turning is 25.0 m. Determine the force of
friction and the coefficient of friction
acting upon the car.
Practice #2
A car moving around a curve is acted
upon by a centripetal force, F. If the
speed of the car were twice as great,
what centripetal force would be
necessary to keep it moving in the same
path?
Practice #3
A child having a mass of 30 kg sits 4.0 m
from the center of a merry-go-round that
is rotating with a period of 10 s. What is
the centripetal force acting on the child?
11/29/10
• Labs past due
• Quiz Problems Revisited
• Centripetal Force Lab
Practice #1
A 1200-kg car makes a 90-degree turn
with a speed of 10.0 m/s. The radius of
the circle through which the car is
turning is 25.0 m. Determine the force of
friction and the coefficient of friction
acting upon the car.
Free-Body Diagram

Fc = mv2/r

= Fc = mv2/r
= Fc = mv2/r
Ffrict = 1200 kg(10 m/s)2/25 m
Ffrict = 4800 N

f = Ffrict / FNorm
f = 4800 N / 12000 N
f = 0.4
Practice #2
A car moving around a curve is acted
upon by a centripetal force, F. If the
speed of the car were twice as great,
what centripetal force would be
necessary to keep it moving in the same
path?
Practice #2

F=   mv2/r

If v is doubled to 2v,
Fc = m(2v)2/r = 4mv2/r

Fc = 4F
Practice #3
A child having a mass of 30 kg sits 4.0 m
from the center of a merry-go-round that
is rotating with a period of 10 s. What is
the centripetal force acting on the child?
S = vT
2r = vT
v = 2r/T
v = 2•4 m/10 s = 2.5
m/s    Fc = mv2/r

Fc = 30 kg (2.5m/s) 2/4   m
Fc = 47 N
Fc
Fc
Draw free-body
A   diagrams at each
location on the roller
coaster.
C

B
What is the net force on
the rider at point A?
Fnet = Fg + (-FN)
What is Fnet at point A
also called? F      =F                      A
net      centripetal
What is FN at point A
acting on the
rider(apparent weight)?
FN = Fg - Fc
What velocity is needed
at point A to produce an
mg = mv2/r
FN on the rider of 0?
0 = Fg - Fc         Fg = Fc           v = (gr)½
What is the net force on
the rider at point B?
Fnet = FN + (-Fg)
What is Fnet at point B
also called?
Fnet = Fcentripetal
What is FN at point B
acting on the
rider(apparent weight)?
FN = Fc + Fg
What is the “g-force” at          B
point B?
G-force = FN / Fg
What is the net force on
the rider at point C?           C

Fnet = FN + Fg
What is Fnet at point C
also called?
Fnet = Fcentripetal
What is FN at point C
acting on the rider?
FN = Fc - Fg
What does a negative
FN mean?
- FN = “I’m falling”
•
11/30/10
Questions on homework?
• Circular Motion Demos
• A few more centripetal problems
• Things moving in circles about a fixed radius, why does that sound
familiar…?
• Universal gravitation
• Centripetal Force Lab
Practice #4
What centripetal force is needed to keep
a 4-kg mass moving at a constant speed
of 4 m/s in a circle having a radius of 8
m?
Fc = mv /r         2
Fc = (4 kg)(4    m/s)2/8

m F = 8 kg-m/s2
c
Fc = 8 N
Practice #5
A car moves around a circular section
constant speed of 20 m/s. What is the
centripetal acceleration of the car?
ac =   v2/r

ac = (20 m/s) 2/50 m

ac = 8 m/s   2
Practice #6
A stone attached to a string 2.0 m long is
whirled in a horizontal circle.At what
speed must the stone move for its
centripetal acceleration to be equal to
the acceleration of gravity?
Practice #6

ac =    v2/r

v2   = ac•r
v 2   = 9.8   m/s2•   2m
v = 4.4 m/s
Practice #7
On a cold day in January Susan was
walking through the Syracuse University
parking lot when she encountered an
#%@#!

iced-covered surface.
This would
never
happen to
Susan now.
However, in
60 years……
Practice #7
…the purse would go
flying.
If the purse is 0.8 m from
Susan’s shoulder, how
fast does she have to
swing it around so
nothing falls out of it?
Practice #7

ac =    v2/r

v2   = ac•r
v 2   = 9.8   m/s2•   0.8 m
v = 2.8 m/s
Practice #8
A frictionless rollercoaster does a vertical loop with a radius of 6.0m.
What is the minimum speed that the roller coaster must have at the
top of the loop so that it stays in touch with the rail?

mv2/r = mg
g = v2/r
v2 = gr
Fnet = FN + Fg
v2 = 9.8 m/s2 x 6 m
Fc = 0 + Fg
v = 7.7 m/s
Length(m) Period (T) s   Circumference (m) Velocity (m/s)
0.1       0.635             0.628           0.990
0.2       0.898             1.257           1.400
0.3       1.099             1.885           1.715
0.4       1.269             2.513           1.980
0.5       1.419             3.142           2.214
0.6       1.555             3.770           2.425
0.7       1.679             4.398           2.619
Velocity vs. Radius   y = 3.1305x0.5
2
R =1
3.000
2.500
Velocity (m/s)

2.000
1.500
1.000
0.500
0.000
0     0.2       0.4       0.6          0.8
2    2     2
Length(m)                        Velocity (m /s )
0.1                                0.980
0.2                                1.960
0.3                                2.940
0.4                                3.920
0.5                                4.900
0.6                                5.880
0.7                                6.860

Velocity Squared vs Radius            y = 9.8x
R2 = 1
8.000
7.000
Velocity2 (m2/s2)

6.000
5.000
4.000
3.000
2.000
1.000
0.000
0                0.2          0.4       0.6              0.8
M1                   M2

R
Fg = mg

GM1 M2
Fg =
R2
m1                           m2

R

Gm1 m2             Gm1 m2
Fg =                m2g =
R2
R2
Fg = m2g            g= Gm1
R2
Me               m2

R
G Me
g=
R 2
Me                                    m2

R

6.7 x 10-11N•m2/kg2(6.0 x 1024 kg)
g=
(6.4 x 106 m)2
Me                    m2

R

g = 9.8   m/s2
50 kg                   6 kg

2 .0 m

GM1 M2
Fg =
R2
50 kg                              6 kg

2 .0 m

6.7 x 10-11N•m2/kg2(50 kg)(6 kg)
Fg =
(2 m)2
50 kg                        6 kg

2 .0 m

Fg =   5.0 x 10-8N
m1                           m2

R

Gm1 m2             Gm1 m2
Fg =                m2g =
R2
R2
Fg = m2g            g= Gm1
R2
M
100 1                       M
50 kg2
kg

1.0 m
R

GM1 M2
Fg =
R2
M
100 1                              M
50 kg2
kg

1.0 m
R

6.7 x 10-11N•m2/kg2(50 kg)(100
Fg =
kg)
(1.0 m)2
Fg =   3.35 x 10-7 N
M
100 1                              M
50 kg2
kg

1.0 m
R

Fg = 3.35 x 10-7 N = 50 kg x g
g = 6.7 x 10-9 m/s2
M
100 1                              M
50 kg2
kg

1.0 m
R

Fg = 3.35 x 10-7 N = 100 kg x g
g = 3.35 x 10-9 m/s2
Concept Problem #5
A radioactive cesium nucleus emits a beta particle of mass 9.1 x 10-31 kg and
transmutes (changes) into a barium nucleus that has a mass of 2.2 x 10-25 kg.
What is the gravitational force of attraction between the barium nucleus and
the beta particle when they are 2.0 x 10-8 m apart? Based on your answer, is
the force of gravity important in holding subatomic particles together?
Explain.
Concept Problem #5

3.3 x 10-50 N
GM
g=   2
R
Orbiting the Earth
• To maintain a constant
distance around the Earth, a
satellite must maintain a
certain speed.
• If it did not it would fall into
the atmosphere.
• We can determine the speed
with which something orbits
the Earth by the radius of its
orbit.
Orbiting the Earth
• We can also use the
orbit to determine the
period of its orbit.
• A satellite is in orbit around a small planet.
The orbital radius is 6.7 X 104 km and its speed
is 2.0 X 105 m/s. What is the mass around
which the satellite orbits?
• Building Newton’s Cannon on the Moon
The Moon’s mass is 7.3 X 1022 kg and has a radius of
1785 km. If we took Newton’s cannon to the
moon and fired it from a high mountain, how fast
would the cannonball have to be fired to keep it
in orbit? How long would it take the cannonball to
Jupiter and Earth
Jupiter is 5.2 times farther from the Sun than
Earth. Find Jupiter’s orbital period in Earth yrs.
Gravity is all around
A moon in orbit around a planet, like ours, experiences a
gravitational force not only from the planet, but also from
the Sun. The illustration below shows a moon during a
solar eclipse, when the planet, the moon and the Sun
are aligned. The moon has a mass of 3.9x1021 kg, the
planet is 2.4x1026 kg and the Sun is 2.0x1030 kg. The
distance from the moon to the center of the planet is
6.0x108 m. The moon to the Sun is 1.5x1011 m. What is
the ratio of the gravitational force on the moon due to the
planet compared to the gravitational force on the moon
due to the Sun?
Gravity is all around
What is the deal with
weightlessness?
When do you feel weightless?

How do you sense weight?

Can you think of what the main difference is
between gravitational force and every other force
we have talked about so far this year?
It does not depend on two objects touching to
impart a force, it acts in a field:
If you had to draw vector arrows to
describe it, what do you think the Earths
gravitational field would look like?
Two Types of Mass
Einstein’s Theory of Gravity
Rotational Motion
• HOW do we describe rotational motion?
– By breaking down its revolution into fractions.
– You know most of the ways, but here they are:
Angular displacement
Angular Velocity
Angular Acceleration

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