GRAVITATIONAL MOTION
Newton’s Law of Universal
Gravitation
states that every particle in the universe
attracts every other particle with a force that is
directly proportional to the product of their
masses and inversely proportional to the
square of the distance between them.
Gravity is a consequence of mass.
The force of gravity between two objects is directly
proportional to the product of the objects’ masses.
The force of gravity between two objects is inversely
proportional to the square of the distance between the
two objects’ centers of mass.
𝑚1 𝑚2
𝐹=𝐺
𝑟2
Newton’s Law of Universal
Gravitation
Universal Constants
Universal Gravitational Constant:
2
𝑁𝑚
−11
𝐺 = 6.67𝑥10
𝑘𝑔2
Mass of Earth:
24
𝑀𝐸 = 6𝑥10 𝑘𝑔
Radius of Earth:
𝑅𝐸 = 6.4𝑥106 𝑚
ACCELERATION DUE TO GRAVITY
GMm
F 
2
R
GMm
mg 
2
R
GM
g 2
R
F  Force of Gravitation
G  Universal Gravitational Constant
M  Mass of the Planet
m  mass of object
R  Radius of Planet
So . . . g = ?
Now work it out.
Now work it out.
Work that problem out.
Get that problem right.
Universal Gravitation #2
The mass of planet Jupiter is 1.9 x 1027 kg and that
of the sun is 1.99 x 1030 kg. The mean distance of
Jupiter from the sun is 7.8 x 1011m. Calculate the
gravitational force which the sun exerts on Jupiter.
m1  1.9 10 kg
27
m2  1.99 10 kg
30
G  6.67 10 11 Nm 2 / kg 2
r  7.8 1011 m
m1 m 2
F G
r2
(1.9  10 kg)(1.99  10 kg)
 (6.67  10 Nm / kg )
(7.8  1011 m ) 2
27
 11
2
 .4145  10 11 27 30 22
 .4145  1024 N
 4.145  1023 N
2
30
Where will the ball go?
What will happen to the Moon?
TORQUE
Torque
Torque
Is the measure of how effectively a
force causes rotation.
τ  Fr sinθ
τ  torque
F  force
r  distance
θ  angle
Torques in equilibrium
When the torques associated with two
masses balance each other.
τ1 - τ2  0
where :
τ  Fgr
Torques in equilibrium
Two people with the same mass. Which
picture shows torques in equilibrium?
A
B
Torques in equilibrium
A – the distances are not the same in B so
the torques will not balance.
A
B
PROBLEM 1
Alfred weighs 400 N. He sits on one end of a
seesaw 1.5 m from the fulcrum. Ann weighs
200 N. How far from the fulcrum must she sit
to balance the seesaw?
?
PROBLEM 1
F1 = 400 N
r1 = 1.5 m
F2 = 200 N
r2 = ?
1 = 2
F1r1 = F2r2
400(1.5) = 200(r2)
r2 = 3 m
3m
1.5 m
CIRCULAR MOTION
Linear Motion
d – distance (in meters)
v – velocity (in meters/second)
a – acceleration (in meters/second2)
Distance = 2r
Linear/Tangential Velocity
Objects moving in a circle still have a
linear velocity = distance/time.
This is often called tangential velocity,
since the direction of the linear velocity
is tangent to the circle.
v
Angular Motion
 – angular displacement (in radians)
 – angular velocity (in radians/second)
 – angular acceleration (in radians/second2)
r – radius of circle (in meters)
Circular Motion Terms
The point or line that is the
center of the circle is the axis
of rotation.
If the axis of rotation is inside
the object, the object is
rotating (spinning).
If the axis of rotation is
outside the object, the object
is revolving.
Angular Velocity
Objects moving in a circle also have a
rotational or angular velocity, which is the
rate angular position changes.
Rotational velocity is measured in
degrees/second, rotations/minute (rpm), etc.
Common symbol,  (Greek letter omega)
∆𝜃
𝜔=
∆𝑡
D
Rotational & Linear Velocity
If an object is rotating:
All points on the object have the same rotational (angular) velocity.
All points on the object do not have the same linear (tangential)
velocity.
Linear velocity of a point depends on:
The rotational velocity of the point.
More rotational velocity means more linear velocity.
The distance from the point to the axis of rotation.
More distance from the axis means more linear velocity
In symbols:
v=r
v
r

Linear to Angular
d = r
vT = r
aT = r
r
Angular & Linear Velocity
If an object is rotating:
All points on the object have the same
Angular velocity.
All points on the object do not have the
same linear (tangential) velocity.
Angular & Linear Velocity
Linear velocity of a point depends on:
The Angular velocity of the point.
More rotational velocity means more
linear velocity.
The distance from the point to the axis of
rotation.
More distance from the axis means more
linear velocity.
Acceleration
As an object moves around a circle, its
direction of motion is constantly changing.
Therefore its velocity is changing.
Therefore an object moving in a circle is
constantly accelerating.
Centripetal Acceleration
The acceleration of an object moving in a
circle points toward the center of the circle.
This is called a centripetal (center pointing)
acceleration.
ac
Centripetal Acceleration
The centripetal acceleration depends on:
The velocity of the object.
The radius of the circle.
2
T
v
ac 
r
2
a c  r
vT  Tangential Speed
a c  centripetal acceleration
r  radius of circular path
  angular velocity
Circular Motion #8
A spinning ride at a carnival has an angular
acceleration of 0.50 rad/s2. How far from the
center is a rider who has a tangential
acceleration of 3.3 m/s2?
aT  rα
3.3  r (.50)
r  6.6m
Circular Motion #9
What is the tire’s angular acceleration if the
tangential acceleration at a radius of 0.15m is
9.4 x 10-2 m/s2?
aT  r
.094  .15
  .63rad / s
2
Circular Motion #10
A test car moves at a constant speed of 10 m/s
around a circular road of radius 50 m. Find the
car’s A) centripetal acceleration and B) angular
speed.
2
A)
B)
2
v
(10)
2
ac 

 2m / s
r
50
v 10
 
 0.2rad / s
r 50
Circular Motion #11
A test car moves at a constant speed around a circular
track. If the car is 48.2 m from the track’s center and
has a centripetal acceleration of 8.05 m/s2, what is its
tangential speed?
v  rac  8.05  48.2
Givens :
r  48.2m
ac  8.05m / s
vT  ?
2
T
2
vT  8.05  48.2  19.7 m / s
Circular Motion #12
A race car moves along a circular track at an angular
speed of 0.512 rad/s. If the car’s centripetal acceleration
is 15.4 m/s2, what is the distance between the car and
the center of the track?
Givens :
  .512rad / s
ac  15.4m / s
r ?
2
v
( r )
2
ac 

 r
r
r
ac
15.4
r 2 
 58.7 m
2

(.512)
2
2
Circular Motion #13
A piece of clay sits 0.20 m from the center of a potter’s
wheel. If the potter spins the wheel at an angular speed
of 20.5 rad/s, what is the magnitude of the centripetal
acceleration of the piece of clay on the wheel?
Givens :
  20.5rad / s
r  .20m
ac  ?
a c  r
2
 (.20)(20.5)
 84.05m / s
2
2
Tangential vs. Centripetal
tangential and centripetal acceleration are
perpendicular to one another.
atotal  a  a
2
T
 ac
  tan 
 aT
1



2
c
ac
aT
θ
atotal
Equations for Circular Motion
 f  i   t
1 2
  i t   t
2
2
2
 f  i  2
i  initial angular velocity(rad/s)
 f  final angular velocity(rad/s)
  angular displacement(rad)
  angular acceleration(rad/s )
2
t  time(s)
Angular Acceleration
=
𝜟
𝜟𝒕
 – angular velocity (radians/second)
 – angular acceleration (radians/second2)
t – time (seconds)
Circular Motion #14
A ventilator fan is turning at 600 rev/min when
the power is cut off, and it turns 1000 rev while
coasting to a stop. Calculate the angular
acceleration and the time required to stop.
i  600rev / min
 f  0rev / min
  1000rev
t ?  ?
    2
 f  i   t
0  (600)  2 (1000)
 2000  360000
0  600  (180)t
t  3.33 min
2
f
2
i
2
  180rev / min
2
Circular Motion #15
A bicycle wheel rotates with a constant angular
acceleration of 3.5 rad/s2. If the initial speed of the
wheel is 2 rad/s at t = 0 s. a) Through what angle does
the wheel rotate in 2 s? b) what is the angular speed at
t = 2 s?
1 2
  i t   t
2
1
2
 2(2)  (3.5)(2)
2
 4  7  11rad
 f  i   t
 2  (3.5)2
 2  7  9rad / s
Circular Motion #16
A potter’s wheel moves from rest to an angular
speed of 0.20 rev/s in 30s. Find the angular
acceleration in rad/s2.
rev 2rad
 f  0.20

 1.256rad / s
s
1rev
i  0rad / s
 f  i   t
t  30 s
  ? rad / s
2
1.256  0   (30)
  .041rad / s
2
Circular Motion #16
A dentist’s drill starts from rest. After 3.20 seconds of
constant angular acceleration it turns at a rate of 2.51 x
104 rev/min. a) find the drill’s angular acceleration. b)
Determine the angle (radians) through which the drill
rotates during this period.
rev 2rad 1 min
 f  25100


 2627.13rad / s
min 1rev
60 s
i  0rad / s
t  3.2 s
 ?  ?
 f  i   t
2627.13  0   (3.2)
  821rad / s
2
1 2
  i t   t
2
1
2
 0  (821)(3.2)
2
 4203.52rad
Circular Motion #18
A floppy disk in a computer rotates from rest up to an
angular speed of 31.4 rad/s in a time of 0.892 s.
A) What is the angular acceleration of the disk, assuming
angular acceleration is uniform? B) How many
revolutions does the disk make while coming up to
speed? C) Find v if r = 4.45cm D) Find at if r = 4.45cm.
i  0rad / s
 f  31.4rad / s
t  .892 s
 ?  ?
A) f  i  t
31.4  0   (.892)
  35.2rad / s
2
1 2
1
2
B)  i t  t  0  (35.2)(.892)  14rad
2
2
1rev
14rad 
 2.23rev
2rad
C )v  r  (.0445m)(31.4rad / s)  1.4m / s
D)at  r  (.0445)(35.2rad / s 2 )  1.57m / s 2
Centripetal Force
Newton’s Second Law says that if an object is
accelerating, there must be a net force on it.
For an object moving in a circle, this is called
the centripetal force.
The centripetal force points toward the
center of the circle.
Centripetal Force
In order to make an object revolve about an
axis, the net force on the object must pull it
toward the center of the circle.
This force is called a centripetal (center
Fc
seeking) force.
Centripetal Force
Centripetal force on an object depends on:
The object’s mass - more mass means
more force.
The object’s velocity - more speed means
more force.
And…
Centripetal Force
The centripetal force on an object also
depends on:
The object’s distance from the axis
(radius).
If linear velocity is held constant, more
distance requires less force.
If rotational velocity is held constant,
more distance requires more force.
Centripetal Force
2
t
v
2
Fc  mac  m
 mrω
r
Fc  centripeta l force (N)
m  mass (kg)
v t  tangential speed(m/s)
r  radius of circular path(m)
ω  angular speed(rad/ s)
Fc
vt
Circular Motion #19
A test car moves at a constant speed of 10 m/s around a
circular road of radius 50 m. Find the car’s A) centripetal
acceleration and B) angular speed.
vt  10m / s
r  50m
2
t
v
2
ac 
 2m / s
r
v 10
 
 .2rad / s
r 50
Circular Motion #20
A 0.015 kg rubber stopper is attached to 0.93 m length
string. The stopper is swung in a horizontal circle, making
one revolution in 1.18 seconds. A) find the speed of the
stopper. B) find the force the string exerts on it.
m  0.015kg
r  .93m
1rev 2rad


 5.32rad / s
1.18s 1rev
A)v  r  .93m  5.32rad / s  4.95m / s
B) Fc  mr 2  .015  .93  (5.32) 2  .4 N
Circular Motion #21
A 50 kg runner moving at a speed of 8.8 m/s rounds
a bend with a radius of 25 m. A) Find the centripetal
acceleration of the runner B) what supplied the force
needed to give this acceleration to the runner?
m  50kg
r  25m
v  8.8m / s
2
t
2
v
(8.8)
2
A)ac 

 3.09m / s
r
25
vt2
8.82
B) Fc  m  50 
 154.88 N
r
25
“Centrifugal Force”
“Centrifugal force” is a fictitious force - it is
not an interaction between 2 objects, and
therefore not a real force.
Nothing pulls an object away from the
center of the circle.
What is erroneously attributed to
“centrifugal force” is actually the action of
the object’s inertia - whatever velocity it has
(speed + direction) it wants to keep.
As a car makes a turn, the force of
friction acting upon the turned wheels of
the car provide the centripetal force
required for circular motion.
As a bucket of water is tied to a string and
spun in a circle, the force of tension acting
upon the bucket provides the centripetal
force required for circular motion.
As the moon orbits the Earth, the force
of gravity acting upon the moon
provides the centripetal force required
for circular motion.
Without a centripetal force, an
object in motion continues
along a straight-line path.
With a centripetal force, an object
in motion will be accelerated and
change its direction
The tendency of a moving object to continue in a straight line in the absence
of an unbalanced force and to turn in a circle in the presence of a inwarddirected force (i.e., centripetal force) has been experienced by any
passenger in an automobile. When the car makes a sudden turn, the
passengers tend to continue in their straight line path. This straight line
motion continues until the presence of a side door or another passenger
pushes upon the passenger in order to accelerate him/her towards the
center of the turn. The force experienced by the passenger is an inward
force; without it, the passenger would slide out of the car.
Feelings of weightlessness and heaviness are associated with the
normal force; they have little to do with the force of gravity. A
person who feels weightless has not lost weight; the force of
gravity acting upon the person is the same magnitude as it always
is. Witness in the animation above that the force of gravity is
everywhere the same. The normal force however has a small
magnitude at the top of the loop (where the rider often feels
weightless) and a large magnitude at the bottom of the loop
(where the rider often feels heavy). The normal force is large at
the bottom of the loop because in order for the net force to be
directed inward, the normal force must be greater than the
outward gravity force. At the top of the loop, the gravity force is
directed inward and thus, there is no need for a large normal force
in order to sustain the circular motion. The fact that a rider
experiences a large force exerted by the seat upon her body when
at the bottom of the loop is the explanation of why she feels
heavy. In actuality, she is not heavier; she is only experiencing the
large magnitude of force which is normally exerted by seats upon
heavy people while at rest.
Banking Curve
When an automobile is driven around a
sharp turn on a perfectly level road, friction
between the tires and the road provides the
centripetal force. If this centripetal force is
not adequate, the car may slide off the road.
The maximum speed with which a car can
negotiate a turn of a given radius.
Fs  Force of Static Friction
Fs  Fc
mv
 s mg 
r
2
v
s g 
r
v   s rg
2
Fc  Centripetal Force
s  Coefficient of Static Friction
g  gravity (9.8m/s )
r  Radius
m  mass
2
Banking of a Cyclist
A cyclist provides himself the necessary centripetal
force by leaning inward on a horizontal track, while
going around a curve. The cyclist should bend
through an angle  to have the necessary centripetal
force while going around a curved path.
2


v
1
  tan  
 rg 
v  velocity
r  radius
g  gravity (9.8m / s 2 )
Circular Motion #22
What is the maximum speed at which an automobile
can negotiate a curve of radius 100m with out
sliding, if the coefficient of static friction is 0.7?
Fs  Fc
2
mv
 s mg 
r
2
v
s g 
r
v   s Rg
 (0.7)(100)(9.8)
 26.1m / s
Circular Motion #23
A car travels at a constant speed of 13.4 m/s on a
level circular turn of radius 50m. What is the
minimum coefficient of static friction between the
tires and roadway in order that the car makes the
circular turn without sliding?
Fs  Fc
2
mv
 s mg 
r
2
v
s g 
r
2
(13.4)
 s (9.8) 
50
2
(13.4)
s 
 .37
50  9.8
Circular Motion #24
Find the required banking angle for a curve of radius
300m if the curve is to be negotiated at a speed of 80
km/h without the need of a frictional force.
km 1000m
1h
v  80


 22.2m / s
h
1km 3600 s
2
2



22.2 
1 v
1
  9.50
  tan    tan 
 rg 
 300(9.8) 
Minimum Velocity Required
Motion in a Vertical Circle
The minimum velocity that the object should
possess at the top so that the string doesn’t
slack is
vtop  gr
The minimum velocity that the object should
possess at the bottom so that the string
doesn’t slack is
vbottom  5 gr
Circular Motion #25
A person swings a pail of water in a vertical circle
with a radius of 1 m. What is the minimum speed the
person must maintain at the top and bottom to keep
the water from falling out?
v top  gR  9.8  3.13m / s
v bottom  5gR  5(9.8)  7m / s