Download Chapter 6

Chapter 6 Lecture
Chapter 6:
Circular Motion
and Gravitation
© 2016 Pearson Education, Inc.
Goals for Chapter 6
• To understand the dynamics of circular motion.
• To study the unique application of circular
motion as it applies to Newton's law of
gravitation.
• To examine the idea of weight and relate it to
mass and Newton's law of gravitation.
• To study the motion of objects in orbit as a
special application of Newton's law of
gravitation.
© 2016 Pearson Education, Inc.
In Section 3.4
• We studied the kinematics of circular motion.
• Centripetal acceleration
• Changing velocity vector
• Uniform circular motion
• We acquire new terminology.
• Radian
• Period (T)
• Frequency (f)
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Velocity Changing from the Influence of
arad – Figure 6.1
• A review of the relationship between
and arad.
• The velocity changes direction,
not magnitude.
• The magnitude of the centripetal
acceleration is:
2
u
arad =
R
• In terms of the speed and period (time to make
one complete revolution)
© 2016 Pearson Education, Inc.
Circular motion and Gravitation
v2
 FR  maR  m R
Details of Uniform Circular Motion
• An object moves in a circle because of a
centripetal bet force.
• Notice how becomes linear when Frad vanishes.
Fnet = Frad = m
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u2
R
Clicker question
You whirl a ball of mass m in a fast vertical circle
on a string of length R. At the top of the circle, the
tension in the string is five times the ball's weight.
The ball's speed at this point is given by
a) gR
b) 4gR
c)
6gR
d) 6 gR
© 2016 Pearson Education, Inc.
Clicker question
You whirl a ball of mass m in a fast vertical circle
on a string of length R. At the bottom of the circle,
the tension in the string is five times the ball's
weight. The ball's speed at this point is given by
a) gR
b) 4gR
c)
6gR
d) 6 gR
© 2016 Pearson Education, Inc.
The “Giant Swing”:
a) Make a free body diagram
of the seat including the
person on it.
b) Find the time for one
revolution for the indicated
angle of 30o
c) Does the angle depend on
the weight of the
passenger?
a) The person moves on a radius of R=3+5sin30=5.5m
𝑎𝑟 =
b)
𝑣2
𝑅
and 𝑇 =
2𝜋𝑅
𝑣
𝑚𝑔
𝐹𝑦 = 𝑚𝑎𝑦  𝐹 cos 30 = 𝑚𝑔 → 𝐹 = cos 30
𝐹𝑥 = 𝑚𝑎𝑥  𝐹 sin 30 =
𝑣=
𝑚𝑣 2
𝑅
=
𝑚𝑔 sin 30
cos 30
𝑅𝑔 tan 30 = 5.5 ∗ 9.8 ∗ 𝑡𝑎𝑛30 = 5.58
2𝜋5.5
𝑚
𝑠
𝑇 = 5.58 = 6.19 𝑠
c) The net force is proportional to mass that divides out in
𝐹 = 𝑚𝑎 .The angle is independent of mass.
Work out the radial acceleration of the moon around the earth.
𝑇=
𝑣2
𝑎𝑟 = 𝑅
24 ℎ
27.3 𝑑 ∗ 1 𝑑
and 𝑣 =
∗
3600 𝑠
1ℎ
2𝜋𝑅
𝑇
= 2.36𝑥106 𝑠 and 𝑅 = 3.84𝑥108 𝑚
𝑚
So; 𝑎𝑟 = 2.78𝑥10−4 ≈ 3 ∗ 10−3 𝑠^2 g
Ferris wheel
Ferris wheel
Top:
𝑎𝑟 =
𝑣2
𝑅
and 𝑇 =
2𝜋𝑅
𝑣
𝑚𝑣 2
𝐹𝑦 = 𝑚𝑎𝑦 = 𝐹𝑁 − 𝑚𝑔 = − 𝑅
𝑣2
→ 𝐹𝑁 = 𝑚(𝑔 − )
𝑅
The force which the seat applies to the
passenger is smaller than his weight.
For,
off.
𝑣2
𝑅
= 𝑔 passenger is starting to fly
Bottom:
𝐹𝑁 − 𝑚𝑔 =
𝑚𝑣 2
𝑅
→ 𝐹𝑁 = 𝑚(𝑔 +
𝑣2
)
𝑅
Passenger in Ferris wheel is pressed into the
seat.
snowboarding
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Clicker question
You're snowboarding down a slope. The free-body
diagram in the figure represents the forces on you
as you
a) go over the top of a mogul.
b) go through the bottom of a hollow
between moguls.
c) go along a horizontal stretch.
d) go along a horizontal stretch or over
the top of a mogul.
© 2016 Pearson Education, Inc.
Conical Pendulum Tether Ball
Problem – Example 6.2
Center-seeking Force: Tension
Circular Motion
Conical Pendulum
What is the period of this conical pendulum ?
𝑎𝑟 =
𝑣2
𝑅
, 𝑅 = 𝐿 sin 𝜃 and 𝑇 =
2𝜋𝑅 1
( 𝑇 )2 𝑅
4𝜋2 𝑅
𝑇2
2𝜋𝑅
𝑣
So; 𝑎𝑟 =
=
𝐹𝑦 = 𝑚𝑎𝑦 = 𝐹 cos 𝜃 = 𝑚𝑔 = 0
𝑚𝑔
 𝐹 cos 𝜃 = 𝑚𝑔 → 𝐹 = cos 𝜃
𝐹𝑥 = 𝑚𝑎𝑥 = 𝑚𝑎𝑟  𝐹 sin 𝜃 =
∴ 𝑎𝑟 = 𝑔 tan 𝜃
So; 𝑎𝑟 =
𝑔 sin 𝜃
cos 𝜃
=
4𝜋2 𝑅
𝑇2
∴ 𝑇 = 2𝜋
 tan 𝜃 =
𝐿 cos 𝜃
𝑔
4𝜋2 𝑅
𝑇2𝑔
=
𝑚𝑣 2
𝑅
=
𝑚𝑔 sin 𝜃
cos 𝜃
4𝜋2 𝐿 sin 𝜃
𝑇2𝑔
(the period is independent of mass)
No Skidding on a Curve (I)
Center-seeking Force : Static Friction
msmin = ?
No Skidding on a Curve (II)
r = 50.0 m
C
m = 1000 kg
v = 14.0 m/s
Un-Banked Curve
Given:
𝑟 = 50 𝑚, m= 1000 𝑘𝑔, 𝑣 = 14
𝑚
𝑠
and 𝜇𝑠 = 0.6
Friction force is larger than radial force
𝐹𝑓𝑟 > 𝐹𝑟 (no skidding)
Friction force is smaller than radial force
𝐹𝑓𝑟 < 𝐹𝑟 (skidding)
𝐹𝑁 = | − 𝐹𝐺 | = 𝑚𝑔 = 1000 ∗ 9.8 = 9800 𝑁
𝐹𝑓𝑟 = 𝜇𝑠 𝐹𝑁 = 0.6 ∗ 9800 = 5900 𝑁
𝑣2
142
𝐹𝑟 = 𝑚𝑎𝑟 = m
= 1000 ∗
= 3900 𝑁
𝑟
50
So, no skidding at 𝜇𝑠 = 0.6
By changing friction to 𝜇𝑠 = 0.25
𝐹𝑓𝑟 = 𝜇𝑠 𝐹𝑁 = 0.25 ∗ 9800 = 2500 𝑁 < 𝐹𝑟 (Skidding)
The maximum safe velocity
𝐹𝑦 = 𝑚𝑎𝑦 = 𝐹𝑁 = 𝑚𝑔 = 0 𝐹𝑁 = 𝑚𝑔
𝐹𝑥 = 𝑚𝑎𝑥 = 𝑚𝑎𝑟 =
𝑣2
m𝑟
𝐹𝑓𝑟,𝑚𝑎𝑥 = 𝜇𝑠 𝐹𝑁 = 𝜇𝑠 𝑚𝑔 =
𝑠𝑎𝑓𝑒
∴ 𝑣𝑚𝑎𝑥 = 𝜇𝑠 𝑔𝑟
2
𝑣𝑚𝑎𝑥
𝑚 𝑟
No Skidding on Banked Curve
Frad
fr
 Fg
f r sin 
f r cos
No Skidding on Banked Curve
The key to this problem is to realize that the net force 𝐹𝑛𝑒𝑡 causes the car to move along the curve.
𝐹𝑁 sin 𝜃 + 𝑓𝑟 cos 𝜃 = 𝐹𝑛𝑒𝑡
𝐹𝑁 cos 𝜃 − 𝐹𝑔 − 𝑓𝑟 sin 𝜃 = 0 ∴ 𝐹𝑔 = 𝑚𝑔 and 𝑓𝑟 = 𝜇𝑠 𝐹𝑁
𝐹𝑛𝑒𝑡 = 𝐹𝑟𝑎𝑑 = 𝑚𝑎𝑟𝑎𝑑 =
𝑣2
m
𝑟
Use;
𝐹𝑦 = 𝑚𝑎𝑦 = 0
𝐹𝑥 = 𝑚𝑎𝑥 = 𝑚𝑎𝑟𝑎𝑑 = m
𝑣2
m
𝑟
𝑣2
𝑟
𝐹𝑁 sin 𝜃 + 𝜇𝑠 𝐹𝑁 cos 𝜃 = 𝐹𝑛𝑒𝑡 =
= 𝐹𝑁 (sin 𝜃 + 𝜇𝑠 cos 𝜃)....(1)
𝐹𝑁 cos 𝜃 − 𝜇𝑠 𝐹𝑁 sin 𝜃 = 𝐹𝑔 = 𝑚𝑔 = 𝐹𝑁 (cos 𝜃 − 𝜇𝑠 sin 𝜃)..........(2)
Divide equation (1) by equation (2);
(sin 𝜃 + 𝜇𝑠 cos 𝜃) 𝑣 2
=
(cos 𝜃 − 𝜇𝑠 sin 𝜃) 𝑔𝑟
𝑣=
𝑔𝑟
(sin 𝜃 + 𝜇𝑠 cos 𝜃)
(cos 𝜃 − 𝜇𝑠 sin 𝜃)
Note:
For special case 𝜇𝑠 =0;
𝑣𝑑𝑒𝑠𝑖𝑔𝑛 =
𝑔𝑟 tan 𝜃 and tan 𝜃 =
𝑣2
𝑔𝑟
For special case 𝜃=0o;
∴ 𝑣𝑑𝑒𝑠𝑖𝑔𝑛 = 𝜇𝑠 𝑔𝑟 (unbanked curve)
6-54: When the system rotates about the rod the strings are extended as
shown. (The tension in the upper string 𝑇𝐻 is 80 N)
a) The block moves in a circle of radius
𝑟 = (1.259𝑚)2 −(1.00𝑚)2 = 0.75𝑚
Each string makes an angle 𝜃 with the vertical pole
1.00
cos 𝜃 = 1.25 → 𝜃 = 36.9°
This block has an acceleration of 𝑎𝑟 =
𝑣2
𝑟
b) What is the tension in the lower rod?
𝐹𝑦 = 𝑚𝑎𝑦 = 0 𝑇𝐻 cos 𝜃 − 𝑇𝐿 cos 𝜃 − 𝑚𝑔 = 0
𝑚
4.00𝑘𝑔
∗
9.8
𝑚𝑔
2
𝑠
∴ 𝑇𝐿 = 𝑇𝐻 −
= 80𝑁 −
= 31𝑁
cos 𝜃
𝑐𝑜𝑠36.9
c) What is the speed of the block?
𝐹𝑥 = 𝑚𝑎𝑥 (𝑇𝐻 + 𝑇𝐿 ) sin 𝜃 =
∴𝑣=
𝑟(𝑇𝐻 +𝑇𝐿 ) sin 𝜃
𝑚
=
𝑣2
m𝑟
0.75 80+31 𝑠𝑖𝑛36.9
4
= 3.53
𝑚
𝑠
6-55 . As the bus rounds a flat curve at constant speed, a package
suspended from the luggage rack on a string makes an angle with the
vertical as shown.
𝐹𝑦 = 𝑚𝑎𝑦 = 0 𝑇 cos 30 − 𝑚𝑔 = 0
𝑚𝑔
∴𝑇=
cos 30
𝐹𝑥 =
∴𝑣=
𝑣=
𝑣2
𝑚𝑎𝑥 𝑇 sin 30 = m
𝑟
𝑟𝑇 sin 30
𝑚𝑔
𝑟 sin 30
=
∗
=
𝑚
cos 30
𝑚
𝑚
9.8 ∗ 50 ∗ 𝑡𝑎𝑛30 = 16.8
𝑠
𝑔𝑟 tan 30
Gravitation
Newton’s Law of Gravitation
Gm1m2
Fg 
r2
G=gravitational constant = 6.673(10) 10-11 Nm2 / kg 2
Note: The weight  of a body of mass m on the earth's surface with
radius R E is
GmE  m
  mg 
RE2
GmE
or g  2
RE
Gravitational attraction
Note: Two particles of different mass exert equally
strong gravitational force on each other
Gravitational Forces (I)
MM ME
MM
FG = G
r2
‘‘Attractive Force”
ME
Why is the Aggie not falling off the earth?
Remember there is equally strong attraction between the earth and the
Aggie and vice versa
Acceleration of Aggies compare to the acceleration of the Earth
𝑚𝐴 𝑚𝐸
𝐹 = 𝐺 2 = 𝑚𝐴 𝑔
𝑔 =
𝑟𝐸
𝑚𝐸
𝐺 2
𝑟𝐸
Forces are equal between Aggies and Earth
𝑚𝐴 𝑚𝐸
𝐹 = 𝐺 2 = 𝑚𝐴 𝑎𝐴 = 𝑚𝐸 𝑎𝐸
𝑎𝐴

𝑎𝐸
=
𝑟𝐸
𝑚𝐸
≈
𝑚𝐴
1023 and 𝑚𝐸 = 6𝑥1024 𝑘𝑔
So; 𝑚𝐴 = 60 𝑘𝑔 (Aggie’s mass)
Cavendish balance
Cavendish(1798) announced that he has weighted the earth
Cavendish Tension balance (1798)
Air current in the room is negligible to the gravitational attraction force
𝑀𝑚
𝐺 2
𝑟
𝐹=
= 1.33𝑥10−10 𝑁 (Torsion force)
and 𝑀 = 0.5 𝑘𝑔; 𝑚 = 0.01 𝑘𝑔 𝑎𝑛𝑑 𝑟 = 0.05𝑚
When torsion and gravitational forces are in equilibrium;
0.5∗0.01
−10
1.33𝑥10
=𝐺
2
𝐺 =
0.05
𝑚2 𝑁
−11
6.6𝑥10
𝑘𝑔2
Molecular motors (kinetics); F = 1.33𝑥10−12 𝑁
Gravitational Force Falls off Quickly –
Figure 6.15
• The gravitational force is proportional to 1/r2, and
thus the weight of an object decreases inversely
with the square of the distance from the earth's
center (not distance from the surface of the earth).
© 2016 Pearson Education, Inc.
What is the magnitude of the
gravitational force inside, on
the surface, and outside the
earth??
Earth mass 𝑀𝐸 = 6𝑥1024 𝑘𝑔 and radius 𝑅𝐸 = 6.37𝑥106 𝑚
𝑀𝐸 𝑚
𝐹=𝐺 2 =
𝑅𝐸
𝑔𝑅𝐸 2
𝑀𝐸 =
𝐺
𝑚𝑔
When radius is variable like 𝑟 with variable mass 𝑚𝑖𝑛𝑠𝑖𝑑𝑒 of Earth.
Then;
𝐹=
𝑚𝑖𝑛𝑠𝑖𝑑𝑒 𝑚
𝐺
𝑟2
∴𝐹=
and 𝑚𝑖𝑛𝑠𝑖𝑑𝑒 =
𝑀𝐸 𝑚
𝐺 3 𝑟
𝑅𝐸
4
3
𝑀𝐸 𝜋 𝑟 3
4
3
𝜋
𝑅
𝐸
3
=
𝑀𝐸 𝑟 3
3
𝑅𝐸
Average Density of the Earth
g = 9.80 m/s2
RE = 6.37 x 106 m
ME = 5.96 x 1024 kg
 rE = 5.50 x 103 kg/m3
= 5.50 g/cm3 ~ 2 x rRock
Projectile
Circular orbit
G  mE  m mv

2
r
r
v  GmE / r
2
2
2r
r
2r 3
T
 2r

v
Gme
Gme
The larger r then slower the speed
and the larger the period
Weather Satellite
Example 6.10:
Earth mass 𝑀𝐸 = 5.98𝑥1024 𝑘𝑔 and radius 𝑅𝐸 = 6380𝑘𝑚
𝑟 = 6380𝑘𝑚 + 300𝑘𝑚 = 6.68𝑥106 𝑘𝑚
a) What is the speed?
𝑚𝑣 2
𝑟
=
𝑣 =
𝑀𝐸 𝑚
𝐺 2
𝑟
𝑀𝐸
𝐺
𝑟
=
𝑚
7730
𝑠
b) When is the period?
𝑇=
2𝜋𝑟
𝑣
=
2𝜋(6.68𝑥106 )
7730
= 5430 𝑠
c) What is the radial acceleration?
𝑎𝑟𝑎𝑑 =
𝑣2
𝑟
=
(7730)2
6.68𝑥106
= 8.95
𝑚
𝑠2
Geo-synchronous Satellite (at the equator of Earth)
a) Height above the surface of Earth.
ℎ = 𝑟 − 𝑟𝐸
Earth mass 𝑀𝐸 = 6𝑥1024 𝑘𝑔 and radius 𝑟𝐸 = 6380𝑘𝑚
𝑚𝑠 𝑣 2
𝑟
𝑀 𝑚
2𝜋𝑟
= 𝐺 𝐸2 𝑠  𝑣 =
and 𝑇 = 1 𝑑𝑎𝑦 = 86400 𝑠𝑒𝑐
𝑟
𝑇
2
𝑚𝑠 (2𝜋𝑟)2
𝑀𝐸 𝑚𝑠
𝐺𝑀
𝑇
𝐸
3 =
22 𝑚3
=
𝐺
→
𝑟
=
7.54𝑥10
𝑟𝑇 2
𝑟2
4𝜋 2
∴ 𝑟 = 4.23𝑥107 𝑚
 ℎ = 𝑟 − 𝑟𝐸 =36000 km≈ 6𝑟𝐸
b) What is the velocity?
𝑣 =
𝐺
𝑀𝐸
𝑟
= 3070
𝑚
𝑠
If an Object is Massive, Even Photons
Cannot Escape
• A "black hole" is a collapsed sun of immense
density such that a tiny radius contains all the
former mass of a star.
• The radius to prevent light from escaping is
termed the "Schwarzschild Radius."
• The edge of this radius has even entered pop
culture in films. This radius for light is called the
"event horizon."
© 2016 Pearson Education, Inc.
Black hole
Sun mass 𝑀𝑆 = 1.99𝑥1030 𝑘𝑔 and radius 𝑅 = 6.96𝑥108 𝑚
Average density of Sun;
ρ=
𝑀𝑠
𝑉
=4
3
𝑀𝑠
𝜋𝑅3
=4
3
1.99𝑥1030
𝜋(6.96𝑥108 )3
= 1.41
𝑔
𝑐𝑚3
 40% denser than water
Temperature: 5800o K at surface and (1.5x107)o K in the interior of
Sun. (highly ionize plasma gas)
Steven Hawkins is associated with the department of
Physics and Astronomy at TAMU
Escape velocity from the Sun
(mass 𝑀𝑆 = 1.99𝑥1030 𝑘𝑔 and radius 𝑅 = 6.96𝑥108 𝑚)
𝑣=
𝑀𝑠
2𝐺
𝑅
=
𝑚
5
6.18𝑥10
𝑠
 𝑀𝑆 =
4
𝜋𝜌𝑅3
3
and 𝑣 =
4
2𝐺 𝜋𝜌
3
𝑅
How can the sun become a Black Hole?
𝑐
𝑣
3𝑥108
6.18𝑥105
=
= 500
If the radius of the sun become 500 times larger for the same
density, then the light could not escape. This would increase the
sun’s mass. In another word sun became Black Hole.
There is a second way, which is to decrease the radius of sun.
c=
𝑀𝑠
2𝐺
𝑅𝑠
→ 𝑅𝑠 =
𝑀𝑠
2𝐺 2
𝑐
(Schwarzschild radius)
For 𝑅 > 𝑅𝑠  light can be emitted
For 𝑅 < 𝑅𝑠  no light can be emitted (Black hole)
To what fraction of sun’s current radius would the sun have to be
compressed to become a black hole?
𝑀𝑠 2.6𝑥10−11 𝑥2𝑥1030
𝑅𝑠 = 2𝐺 2 =
= 2.95 𝑘𝑚
8
2
𝑐
(3𝑥10 )

𝑅𝑠
𝑅
=
2.95𝑥103
6.95𝑥108
= 4.2𝑥10−6