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Chapter 7
Rotational Motion
and the
Law of Gravity
Rotational Motion and the Law of Gravity
Describing Rotational Motion
A Fraction of One Revolution can be
Measured in grads, degrees, or radians.
A Grad is 1/400 of a Revolution
A Degree is 1/360 of a Revolution
A Radian is the Linear Distance around the
Circle Equal to the Radius
A Full Revolution is Equal to 2pr
(Circumference)
Rotational Motion and the Law of Gravity
Describing Rotational Motion
Rotational Motion and the Law of Gravity
Angular Position (q)
Radian
“Angle where the arc length (s) of a circle is
equal to the radius of the circle.”
s  rq
Remember that q must be in Radians
Rotational Motion and the Law of Gravity
Angular Position (q)
Angle Measured from a Reference Line
Units
Radians
Convention
Positive
CCW
Negative
CW
Rotational Motion and the Law of Gravity
Angular Displacement (s)
For rotation through an angle, θ, a point at a
distance, r, from the center moves a distance
given by s = rθ.
s
q
r
Remember that q must be in Radians
Rotational Motion and the Law of Gravity
Angular Velocity (w)
Velocity is Dd/Dt
Angular Velocity is the Change in
Angular Position (q) During an Elapsed
Time
Dq
w
Dt
Rotational Motion and the Law of Gravity
Angular Velocity (w)
Units
Radians / second
Convention
Positive
CCW
Negative
CW
Dq
w
Dt
Rotational Motion and the Law of Gravity
Angular Acceleration (a)
Acceleration
Dv
a
Dt
Angular Acceleration
Dw
a
Dt
Rotational Motion and the Law of Gravity
Angular Acceleration (a)
Units
radians/sec/sec (rad/s2)
Convention
 wf > wi , a is Positive
 wf < wi , a is Negative
Dw
a
Dt
Rotational Speed
If a and w have the Same Sign, Speed Increases
If a and w have Different Signs, Speed Decreases
Rotational Motion and the Law of Gravity
Rotational Kinematics
Let’s Compare
Linear Kinematics
Angular Kinematics
d
q
v
w
a
a
Rotational Motion and the Law of Gravity
Rotational Kinematics
Linear Kinematics
v  vi  at
1
d  d i  vi  v f t
2
1 2
d  d i  vi t  at
2
v  vi  2ad f  di 
2
2
Angular Kinematics
w  wi  at
1
q  q i  wi  w f t
2
1 2
q  q i  wi t  at
2
w  wi  2a qf  qi 
2
2
Rotational Motion and the Law of Gravity
Problem
• A wheel is rotated so that a point on the edge
moves through 1.50m. The radius of the
wheel is 2.50m. Through what angle (in
radians) is the wheel rotated?
Rotational Motion and the Law of Gravity
Solution
s = 1.50m
r = 2.50m
s  rq
s
q
r
1.50m
q
 0.60rad
2.50m
Rotational Motion and the Law of Gravity
Problem
• A steering wheel is rotated through 128°. Its
radius is 22cm. How far would a point on the
steering wheel’s edge move?
Rotational Motion and the Law of Gravity
Solution
s  rq
 q = 128o
r = 22cm
 2prad 
s  (0.22m)(128 )
 0.49m
o 
 360 
o
Rotational Motion and the Law of Gravity
Problem
• A propeller slows from 475rev/min
to 187rev/min in 4.00s. What is its
angular acceleration?
Rotational Motion and the Law of Gravity
Solution
 wi = 475rev/min
 wf = 187rev/min
t = 4.00s
a
Dw
a
Dt
w f  wi
Dt
(187 rev / min  475rev / min)  2prad  1 min 
2
a


  7.54rad / s
4.00s
 rev  60s 
Rotational Motion and the Law of Gravity
• Homework
– Page 269
• Problems
– 6 (4.1m)
– 8 (-1.2rad/s2)
– 12 (41 rad/s2)
Rotational Motion and the Law of Gravity
Tangential Speed (vT)
Linear Velocity of an Object in a Circular
Path
Units
m/s
vT  rw
Rotational Motion and the Law of Gravity
Tangential Acceleration (aT)
Linear Acceleration of an Object in a
Circular Path
Units
m/s2
aT  ra
Rotational Motion and the Law of
Gravity
Centripetal Acceleration
Recall that for acceleration…
Dv v f  vi
a

Dt t f  ti
Rotational Motion and the Law of Gravity
Centripetal Acceleration
Velocity is a Vector
It has Magnitude and Direction
Either can Change the Velocity
So Using Rotational Kinematics
vT
ac 
r
2
Rotational Motion and the Law of Gravity
Centripetal Acceleration (ac)
2
Of an Object on a Circular Path
Units
vt  rw
m/s2
vt
ac 
r
vt
( rw )
rw
2
ac 


 rw
r
r
r
2
2
2
ac  rw
2
2
Rotational Motion and the Law of Gravity
Centripetal Acceleration (ac)
Direction
Always Toward the Center of Rotation
ac  rw
2
Rotational Motion and the Law of Gravity
Forces of Rotational Motion
Centripetal Force
F  ma
Fc  mac
Rotational Motion and the Law of Gravity
Forces of Rotational Motion
Centripetal Force
2
vt
mvt
Fc  mac  m

r
r
2
mvt
2
Fc 
 mrw
r
2
Rotational Motion and the Law of Gravity
Forces of Rotational Motion
Centripetal Force
mvt
Fc 
r
2
Fc  mrw
2
Rotational Motion and the Law of Gravity
Forces of Rotational Motion
Centripetal Force
Direction
Always toward the Center of Rotation
mvt
Fc 
r
2
Fc  mrw
2
Rotational Motion and the Law of Gravity
Problem
• A pebble breaks loose from the
treads of a tire with a radius of
23cm. If the pebble’s angular speed
is 213 rad/s, what is the tire’s
tangential speed?
Rotational Motion and the Law of Gravity
Solution
r = 23cm
w = 213rad/s
vt
w
r
vt  wr
vt  wr  (213rad / s)(0.23m)  49m / s
Rotational Motion and the Law of Gravity
Problem
• A washing machine’s two spin
cycles are 328rev/min and
542rev/min. The diameter of the
drum is 0.43m. What is the ratio of
the centripetal accelerations for the
fast and slow spin cycles?
Rotational Motion and the Law of Gravity
2
v
ac 
r
v  rw
Solution
 wslow = 328rev/min
 wfast = 542rev/min
d= 0.43m (r = 0.215m)
ac  fast
ac  slow
ac  fast
ac  slow
2

rw fast
rw
2
2
slow
(542rev / min)

 2.73
2
(328rev / min)
Rotational Motion and the Law of Gravity
Problem
• A washing machine’s two spin
cycles are 328rev/min and
542rev/min. The diameter of the
drum is 0.43m. What is the ratio of
the linear velocity of an object at
the surface of the drum for the fast
and slow spin cycles?
Rotational Motion and the Law of Gravity
2
v
ac 
r
v  rw
Solution
 wslow = 328rev/min
 wfast = 542rev/min
d= 0.43m (r = 0.215m)
v fast
vslow
v fast
vslow

rw fast
rwslow
w fast

wslow
(542rev / min)

 1.65
(328rev / min)
Rotational Motion and the Law of Gravity
Problem
• A laboratory ultracentrifuge is
designed to produce a centripetal
acceleration of 3.5x105g at a
distance of 2.50cm from the axis.
What angular velocity in rev/min is
required?
Rotational Motion and the Law of Gravity
2
Solution
ac = 3.5x105g
r = 2.50cm
v  rw
v
ac 
r
ac  w r
2
ac
w
r
(3.5 x105 )(9.8m / s 2 )  1rev  60s 
5
w


  1.1x10 rev / min
0.025m
 2prad  1 min 
Rotational Motion and the Law of Gravity
Problem
• A bicycle wheel has an angular
acceleration of 1.5rad/s2. If a point
on it’s rim has a tangential
acceleration of 48cm/s2, what is the
radius of the wheel?
Rotational Motion and the Law of Gravity
aT  ra
Solution
 a = 1.5rad/s2
at = 48cm/s2
r
at
2
at
a
0.48m / s
r 

0
.
32
m
2
a 1.5rad / s
Rotational Motion and the Law of Gravity
Problem
• Mr. Funchess twirls a set of keys in
a circle at the end of a cord. If the
keys have a centripetal acceleration
of 145m/s2 and the cord has a length
of 0.34m, what is the tangential
speed of the keys?
Rotational Motion and the Law of Gravity
2
Solution
vt
ac 
r
r = 0.34m
ac = 145m/s2
vt  rac
vt rac  (0.34m)(145m / s )  7.0m / s
2
Rotational Motion
• Homework
– Page 270 - 271
• Problems
–22 (1.8x10-3rad/s )
–24 (0.023m)
–26 (2.7m/s)
–37 (a, 515kg b, 12.1m/s)
Rotational Motion and the Law of Gravity
Newton’s Law of Universal Gravitation
Newton found that the magnitude of the
force, F, on a planet due to the Sun varies
inversely with the square of the distance, r,
between the centers of the planet and the Sun.
That is, F is proportional to 1/r2. The force, F,
acts in the direction of the line connecting the
centers of the two objects.
Rotational Motion and the Law of Gravity
Newton’s Law of Universal Gravitation
The sight of a falling apple made Newton
wonder if the force that caused the apple to
fall might extend to the Moon, or even
beyond.
He found that both the apple’s and the
Moon’s accelerations agreed with the 1/r2
relationship.
Rotational Motion and the Law of Gravity
Newton’s Law of Universal Gravitation
According to his own third law, the force
Earth exerts on the apple is exactly the same
as the force the apple exerts on Earth.
The force of attraction between two objects
must be proportional to the objects’ masses,
and is known as the gravitational force.
Rotational Motion and the Law of Gravity
Newton’s Law of Universal Gravitation
The law of universal gravitation states that
objects attract other objects with a force that
is proportional to the product of their masses
and inversely proportional to the square of
the distance between them.
m1m2
F G 2
r
Rotational Motion and the Law of Gravity
Newton’s Law of Universal Gravitation
The gravitational force is equal to the
universal gravitational constant, times the
mass of object 1, times the mass of object 2,
divided by the square of the distance between
the centers of the objects.
m1m2
F G 2
r
Rotational Motion and the Law of Gravity
Inverse Square Law
According to Newton’s equation, F is
inversely related to the square of the distance.
F r
2
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
Newton stated his law of universal gravitation
in terms that applied to the motion of planets
about the Sun. This agreed with Kepler’s
third law and confirmed that Newton’s law fit
the best observations of the day.
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
Consider a planet orbiting the Sun. Newton's
second law of motion, Fnet = ma, can be
written as Fnet = mpac.
Fnet is the gravitational force
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
For an elliptical orbit, ac = 4π2r/T2. This
means that Fnet = mpac may now be written as
Fnet = mp4π2r/T2.
T is the time required for the planet to make
one complete revolution about the Sun.
Period
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
F G
ms m p
r
2

m p 4p 2 r
2

 3
4
p
2
r
T  
 Gms 
T2
 4p  3
r
T  
 Gms 
2
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third Law
The period of a planet orbiting the Sun can be
expressed as follows.
 r3 

T  2p 
 Gms 
The period of a planet orbiting the Sun is equal to 2
times the square root of the orbital radius cubed,
divided by the product of the universal gravitational
constant and the mass of the Sun.
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
Cavendish’s Experiment
Often is called “weighing Earth,” because his
experiment helped determine Earth’s mass. Once
the value of G is known, not only the mass of
Earth, but also the mass of the Sun can be
determined.
Rotational Motion and the Law of Gravity
Why is “g” 9.8m/s2?
Newton’s Law of Universal Gravitation
m1m2
F G 2
r
“G” = Universal Gravitational Constant
6.67259x10-11 N*m2/kg2
Rotational Motion and the Law of Gravity
Weight
Equals Gravitational Force on Earth
Force Toward the Center of the Earth
(Center of Mass)
M E m GM E m
W G 2 
2
r
r
Rotational Motion and the Law of Gravity
Mass vs. Weight
GM E m
W
2
r
W mg
Rotational Motion and the Law of Gravity
Universal Gravitation and Kepler’s Third
Law
Cavendish’s Experiment
ME
g G 2
rE
2
grE
ME 
G
2
6
2
2
grE
(9.8m / s )(6.38 x10 m)
24
ME 

 5.98 x10 kg
11
2
2
G
6.67 x10 N * m / kg
Gravitation
Problem
• Two balls have their centers 2.0m apart. One
ball has a mass of 8.0kg. The other has a mass
of 6.0kg. What is the gravitational force
between them?
Gravitation
Solution
m1 = 8.0kg
m2 = 6.0kg
r = 2.0m
F  (6.67 x10
11
m1m2
F G 2
r
(8.0kg)(6.0kg)
10
)
 8.0 x10 N
2
(2.0m)
Gravitation
Problem
• Assume that you have a mass of 50.0 kg.
Earth has a mass of 5.97x1024 kg and a radius
of 6.38x106 m. What is the force of
gravitational attraction between you and
Earth?
Gravitation
Solution
m1 =
m2 = 50.0kg
r = 6.38x106m
5.97x1024kg
m1m2
F G 2
r
24
(
5
.
97
x
10
kg)(50.0kg)
11
F  (6.67 x10 )
 489 N
6
2
(6.38x10 m)
Gravitation
Problem
• Assume that you have a mass of 50.0 kg.
Earth has a mass of 5.97x1024 kg and a radius
of 6.38x106 m. What is your weight?
Gravitation
Solution
m1 = 5.97x1024kg
m2 = 50.0kg
r = 6.38x106m
W  Fg  mg
W  (50.0kg)(9.8m / s )  490 N
2
Gravitation
Problem
• Two spheres are placed so that their centers
are 2.6m apart. The force between the two
spheres is 2.75x10-12 N. What is the mass of
each sphere if one sphere is twice the mass of
the other sphere?
Gravitation
Solution
F = 2.75x10-12 N
r = 2.6m
m1 
m1m2
F G 2
r
(m1 )( 2m1 )
F G
r2
Fr 2
2G
(2.75 x1012 N )( 2.6m) 2
m1 
 0.37kg
11
2(6.67 x10 )
m2  0.75kg
Rotational Motion
• Homework
– Page 271 - 272
• Problems
– 39 (2.50m)
– 40 (0.10nm)
– 50