Chapter 6,9,10
Circular Motion, Gravitation,
Rotation, Bodies in
Equilibrium
Circular Motion



Ball at the end of a string revolving
Planets around Sun
Moon around Earth
The Radian


The radian is a unit
of angular measure
The radian can be
defined as the arc
length s along a
circle divided by
the radius r

s
 
r
57.3°
More About Radians

Comparing degrees and radians

360
1 rad 
 57.3
2
Converting from degrees to radians

 [rad] 
 [deg rees]
180
Angular Displacement



Axis of rotation is
the center of the
disk
Need a fixed
reference line
During time t, the
reference line
moves through
angle θ
Angular Displacement, cont.




The angular displacement is defined
as the angle the object rotates
through during some time interval
   f i
The unit of angular displacement is
the radian
Each point on the object undergoes
the same angular displacement
Average Angular Speed

The average
angular speed, ω,
of a rotating rigid
object is the ratio
of the angular
displacement to
the time interval


t

 f i
t
Angular Speed, cont.


The instantaneous angular speed
Units of angular speed are
radians/sec
• rad/s


Speed will be positive if θ is
increasing (counterclockwise)
Speed will be negative if θ is
decreasing (clockwise)
Average Angular Acceleration

The average angular acceleration
of an object is defined as the ratio of
the change in the angular speed to
the time it takes for the object to
undergo the change:
 
 f  i
t
Angular Acceleration, cont



Units of angular acceleration are rad/s²
Positive angular accelerations are in the
counterclockwise direction and negative
accelerations are in the clockwise direction
When a rigid object rotates about a fixed
axis, every portion of the object has the
same angular speed and the same angular
acceleration
Angular Acceleration, final


The sign of the acceleration does not
have to be the same as the sign of
the angular speed
The instantaneous angular
acceleration
Analogies Between Linear and
Rotational Motion
Linear Motion with constant
acc.
(x,v,a)
v  v0  at
vaverage 
1
v0  v 
2
1 2
x  x0  v0t  at
2
v 2  v02
a
2( x  x0 )
Rotational Motion with fixed axis
and constant 
,,
  0  t
1
 average   0   
2
1
   0   0t  t 2
2
 2   02

2(   0 )
Examples




78 rev/min=?
A fan turns at a rate of 900 rpm
Tangential speed of tips of 20cm long
blades?
Now the fan is uniformly accelerated
to 1200 rpm in 20 s
Relationship Between Angular and
Linear Quantities



Displacements
Speeds

x  R
v  R
Accelerations
a//  R

Every point on the
rotating object has
the same angular
motion
Every point on the
rotating object
does not have the
same linear motion
Centripetal Acceleration


An object traveling in a circle, even
though it moves with a constant
speed, will have an acceleration
The centripetal acceleration is due to
the change in the direction of the
velocity
Centripetal Acceleration, cont.



Centripetal refers
to “center-seeking”
The direction of the
velocity changes
The acceleration is
directed toward the
center of the circle
of motion
Centripetal Acceleration, final

The magnitude of the centripetal
acceleration is given by
v2
a
R
• This direction is toward the center of the
circle
Centripetal Acceleration and
Angular Velocity


The angular velocity and the linear
velocity are related (v = ωR)
The centripetal acceleration can also
be related to the angular velocity
a  R
2
Forces Causing Centripetal
Acceleration

Newton’s Second Law says that the
centripetal acceleration is accompanied by
a force
2
• F = ma

v
F m
R
• F stands for any force that keeps an object
following a circular path



Tension in a string
Gravity
Force of friction
Examples


Ball at the
end of
revolving
string
Fast car
rounding a
curve
More on circular Motion


Length of circumference = 2R
Period T (time for one complete
circle)
2R

v
2
2
v
(2R)
a

R
R 2
a
4 2 R

2
Example

200 grams mass revolving in uniform
circular motion on an horizontal
frictionless surface at 2
revolutions/s. What is the force on
the mass by the string (R=20cm)?
Newton’s Law of Universal
Gravitation

Every particle in the Universe
attracts every other particle with a
force that is directly proportional to
the product of the masses and
inversely proportional to the square
of the distance between them.
m1m2
F G 2
R
Universal Gravitation, 2



G is the constant of universal
gravitational
G = 6.673 x 10-11 N m² /kg²
This is an example of an inverse
square law
Universal Gravitation, 3


The force that
mass 1 exerts on
mass 2 is equal
and opposite to the
force mass 2
exerts on mass 1
The forces form a
Newton’s third law
action-reaction
Universal Gravitation, 4

The gravitational force exerted by a
uniform sphere on a particle outside
the sphere is the same as the force
exerted if the entire mass of the
sphere were concentrated on its
center
Gravitation Constant


Determined
experimentally
Henry Cavendish
• 1798

The light beam and
mirror serve to
amplify the motion
Applications of Universal
Gravitation

Weighing the Earth
mmE
w  Fg  G 2
RE
mmE
mg  G 2
RE
mE
g G 2
RE
gRE2
mE 
G
take g  9.8 m / s
RE  6380 km
2
 mE  6 10 24 kg
Applications of Universal
Gravitation

“g” will vary with
altitude
mE
" g"  G 2
r
Escape Speed

The escape speed is the speed
needed for an object to soar off into
space and not return
vesc


2GmE

RE
For the earth, vesc is about 11.2 km/s
Note, v is independent of the mass of
the object
Various Escape Speeds


The escape speeds
for various
members of the
solar system
Escape speed is
one factor that
determines a
planet’s
atmosphere
Motion of Satellites



Consider only
circular orbit
Radius of orbit r:
Gravitational force
is the centripetal
force.
mmE
F  ma  G
r
r2
r  RE  h
v2
mE
2
m
G
v
r
r
GmE
v
r
Motion of Satellites

Period 
2r

v
2r 3 2

GmE
Kepler’s 3rd Law
  86400s, G  6.67 10 ,
11
mE  6 10  r  4.23 10 m  2.6 10 miles
24
7
4
Communications Satellite

A geosynchronous orbit
• Remains above the same place on the earth
• The period of the satellite will be 24 hr


r = h + RE
Still independent of the mass of the satellite
  86400s, G  6.67 10 ,
11
mE  6 10  r  4.23 10 m  2.6 10 miles
24
7
4
Satellites and Weightlessness





weighting an object in an elevator
Elevator at rest: mg
Elevator accelerates upward:
m(g+a)
Elevator accelerates downward:
m(g+a) with a<0
Satellite: a=-g!!
Force vs. Torque



Forces cause accelerations
Torques cause angular accelerations
Force and torque are related
Torque


The door is free to rotate about an axis through
O
There are three factors that determine the
effectiveness of the force in opening the door:
• The magnitude of the force
• The position of the application of the force
• The angle at which the force is applied
Torque, cont

Torque, , is the tendency of a force
to rotate an object about some axis
  Fl


 is the torque
F is the force
• symbol is the Greek tau



l is the length of lever arm
SI unit is N.m
Work done by torque W=
Direction of Torque


If the turning tendency of the force
is counterclockwise, the torque will
be positive
If the turning tendency is
clockwise, the torque will be
negative
Multiple Torques


When two or more torques are acting
on an object, the torques are added
If the net torque is zero, the object’s
rate of rotation doesn’t change
Torque and Equilibrium

First Condition of Equilibrium

The net external force must be zero
F  0 or
Fx  0 and Fy  0
• This is a necessary, but not sufficient,
condition to ensure that an object is in
complete mechanical equilibrium
• This is a statement of translational equilibrium
Torque and Equilibrium, cont


To ensure mechanical equilibrium,
you need to ensure rotational
equilibrium as well as translational
The Second Condition of Equilibrium
states
• The net external torque must be zero
  0
Equilibrium Example



The woman, mass m,
sits on the left end of
the see-saw
The man, mass M, sits
where the see-saw will
be balanced
Apply the Second
Condition of
Equilibrium and solve
for the unknown
distance, x
Moment of Inertia


The angular acceleration is inversely
proportional to the analogy of the
mass in a rotating system
This mass analog is called the
moment of inertia, I, of the object
• SI units are kg m2
I  mr
2
Newton’s Second Law for a
Rotating Object
  I


The angular acceleration is directly
proportional to the net torque
The angular acceleration is inversely
proportional to the moment of inertia
of the object
More About Moment of Inertia


There is a major difference between
moment of inertia and mass: the
moment of inertia depends on the
quantity of matter and its
distribution in the rigid object.
The moment of inertia also depends
upon the location of the axis of
rotation
Moment of Inertia of a Uniform
Ring


Image the hoop is
divided into a
number of small
segments, m1 …
These segments
are equidistant
from the axis
I  mi ri2  MR2
Other Moments of Inertia
Example
Wheel of radius R=20 cm and
I=30kg·m². Force F=40N acts along
the edge of the wheel.
1. Angular acceleration?
2. Angular speed 4s after starting
from rest?
3. Number of revolutions for the 4s?
4. Work done on the wheel?
Rotational Kinetic Energy



An object rotating about some axis
with an angular speed, ω, has
rotational kinetic energy KEr=½Iω2
Energy concepts can be useful for
simplifying the analysis of rotational
motion
Units (rad/s)!!
Total Energy of a System

Conservation of Mechanical Energy
(KEt  KEr  PEg )i  (KEt  KEr  PEg )f
• Remember, this is for conservative
forces, no dissipative forces such as
friction can be present
• Potential energies of any other
conservative forces could be added
Rolling down incline


Energy conservation
Linear velocity and angular speed are
related v=R
1 2 1 2
mgh  mv  I
2
2
1 2 1 I 2 1
I 2
mgh  mv  ( 2 )v  (m  2 )v
2
2 R
2
R

Smaller I, bigger v, faster!!
Work-Energy in a Rotating
System


In the case where there are
dissipative forces such as friction,
use the generalized Work-Energy
Theorem instead of Conservation of
Energy
(KEt+KER+PE)i+W=(KEt+KER+PE)f
Angular Momentum


Similarly to the relationship between
force and momentum in a linear
system, we can show the relationship
between torque and angular
momentum
Angular momentum is defined as
•L = I ω
• and
L
 
t
Angular Momentum, cont


If the net torque is zero, the angular
momentum remains constant
Conservation of Angular Momentum
states: The angular momentum of a
system is conserved when the net
external torque acting on the
systems is zero.
• That is, when
  0, Li  Lf or Iii  If f
Conservation Rules, Summary

In an isolated system, the following
quantities are conserved:
• Mechanical energy
• Linear momentum
• Angular momentum
Conservation of Angular
Momentum, Example

With hands and
feet drawn closer
to the body, the
skater’s angular
speed increases
• L is conserved, I
decreases, 
increases
Example
A 500 grams uniform sphere of 7.0 cm
radius spins at 30 rev/s on an axis
through its center.
 Moment of inertia
 Rotational kinetic energy
 Angular momentum
Example
Find work done to open 30 a 1m wide
door with a steady force of 0.9N at
right angle to the surface of the
door.
Example
A turntable is a uniform disk of metal
of mass 1.5 kg and radius 13 cm.
What torque is required to drive the
turntable so that it accelerates at a
constant rate from 0 to 33.3 rpm in
2 seconds?
Center of Gravity


The force of gravity acting on an
object must be considered
In finding the torque produced by
the force of gravity, all of the weight
of the object can be considered to be
concentrated at a single point
Calculating the Center of
Gravity


The object is
divided up into a
large number of
very small particles
of weight (mg)
Each particle will
have a set of
coordinates
indicating its
location (x,y)
Calculating the Center of
Gravity, cont.


We wish to locate the point of
application of the single force whose
magnitude is equal to the weight of
the object, and whose effect on the
rotation is the same as all the
individual particles.
This point is called the center of
gravity of the object
Coordinates of the Center of
Gravity

The coordinates of the center of
gravity can be found
xcg
mi xi
mi yi

and ycg 
mi
mi
Center of Gravity of a Uniform
Object


The center of gravity of a
homogenous, symmetric body must
lie on the axis of symmetry.
Often, the center of gravity of such
an object is the geometric center of
the object.
Example
Find the center of mass (gravity) of
these masses: 3kg (0,1), 2kg (0,0)
And 1kg (2,0)
Example
Find the center of mass (gravity) of
the dumbbell, 4 kg and 2 kg with a
4m long 3kg rod.
Torque, review
  Fl


 is the torque
F is the force
• symbol is the Greek tau


l is the length of lever arm
SI unit is N.m
Direction of Torque


If the turning tendency of the force
is counterclockwise, the torque will
be positive
If the turning tendency is
clockwise, the torque will be
negative
Multiple Torques


When two or more torques are acting
on an object, the torques are added
If the net torque is zero, the object’s
rate of rotation doesn’t change
Example
A 2 m by 2 m square metal plate
rotates about its center. Calculate
the torque of all five forces each with
magnitude 50N.
Torque and Equilibrium

First Condition of Equilibrium

The net external force must be zero
F  0 or
Fx  0 and Fy  0
The Second Condition of Equilibrium
states

•The net external torque must be zero
  0
Example
The system is in equilibrium. Calculate
W and find the tension in the rope
(T).
Example
A 160 N boy stands on a 600 N
concrete beam in equilibrium with
two end supports. If he stands one
quarter the length from one support,
what are the forces exerted on the
beam by the two supports?