Transcript Rotational Motion

Chapters 7 & 8
Rotational Motion
and
The Law of Gravity
Homework
Problems:
CH 7:
1,4,5,7,10,12,14,18,19,21,24,25,29,31,34,
36,40
 CH8:
 17,21, 23

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.
Every point on the object undergoes
circular motion about the point O
 Angles generally need to be measured in

radians

s

r
 s
v


  
t rt
r
s is the length of arc and r is the radius
vt  r
More About Radians

Comparing degrees and radians
360
1 rad 
 57.3
2
 Converting from degrees to radians

 [rad] 
 [deg rees]
180
Angular Displacement, cont.
The angular displacement is defined as
the angle the object rotates through
during some time interval
 Every point on the disc undergoes the
same angular displacement in any given
time interval

Angular Speed

The average angular
speed, ω, of a
rotating rigid object
is the ratio of the
angular
displacement to the
time interval
 f  i 


t f  ti
t
v v
ac 

t
r
2
Angular Speed, cont.
The instantaneous angular speed is defined
as the limit of the average speed as the time
interval approaches zero
 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)

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 f  ti
t
Tangential Acceleration
vt  r
vt  r
vt

r
t
t
at  r
More About Angular
Acceleration
Units of angular acceleration are rad/s²
 When a rigid object rotates about a
fixed axis, every portion of the object
has the same angular speed and the
same angular acceleration

Problem Solving Hints

Similar to the techniques used in linear
motion problems


With constant angular acceleration, the techniques
are much like those with constant linear
acceleration
There are some differences to keep in mind


For rotational motion, define a rotational axis
The object keeps returning to its original
orientation, so you can find the number of
revolutions made by the body
Analogies Between Linear and
Rotational Motion
Rotational Motion
Linear Motion with
About a Fixed Axis with Constant Acceleration
Constant Acceleration
  i  t
v  v i  at
1 2
  i t  t
2
1 2
x  v i t  at
2
    2
v  v  2ax
2
2
i
2
2
i
Relationship Between Angular
and Linear Quantities

Displacements

Speeds

Accelerations
s  r
v  r
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 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

aC   r
2
2
OR
See page 198 for
derivation
v
ac 
r
Total Acceleration
The tangential component of the acceleration
is due to changing speed
 The centripetal component of the acceleration
is due to changing direction
 Total acceleration can be found from these
components

a  a a
2
t
2
C
Pythagorean theorem
Vector Nature of Angular
Quantities


Assign a positive or
negative direction in the
problem
A more complete way is
by using the right hand
rule



Grasp the axis of rotation
with your right hand
Wrap your fingers in the
direction of rotation
Your thumb points in the
direction of ω
Forces Causing Centripetal
Acceleration

Newton’s Second Law says that the
centripetal acceleration is accompanied by a
force
F stands for any force that keeps an object following
a circular path



Tension in a string
Gravity
Force of friction
2

mv
4 r 
Fnet  Fc  m ac 
 m 2 
r
 T 
2
Problem Solving Strategy
Draw a free body diagram, showing and
labeling all the forces acting on the
object(s)
 Choose a coordinate system that has
one axis perpendicular to the circular
path and the other axis tangent to the
circular path

Problem Solving Strategy,
cont.
Find the net force toward the center of
the circular path (this is the force that
causes the centripetal acceleration)
 Solve as in Newton’s second law
problems

The directions will be radial and tangential
 The acceleration will be the centripetal
acceleration

Applications of Forces Causing
Centripetal Acceleration

Many specific situations will use forces
that cause centripetal acceleration
Level curves
 Banked curves
 Horizontal circles
 Vertical circles

f  ma
Level Curves
v2
f m
r
f   n
Friction is the force
f   mg
that produces the
centripetal
acceleration
 Can find the
frictional force, µ, v

v  rg
v2
 mg  m
r

At what maximum
speed can a car
negotiate a turn
on a wet road
with coefficient of
static friction
0.230 without
sliding out of
control?
Banked Curves

A component of the
normal force adds to
the frictional force to
allow higher speeds
2
v
tan  
rg
remember
tan  
from
See p. 204
sin 

cos 

A race track is to have a banked curve
with a radius of 25m. What should be the
angle of the bank if the normal force alone
is to allow safe travel around the curve at
58.0 m/s?
Horizontal Circle

The horizontal
component of the
tension causes the
centripetal
acceleration
aC  g tan 
See next page for derivation
Derivation
T cos  m g  0
mg
T
cos
m g sin 
F  T sin  
 m g tan
cos
F  m ac
F mg tan
ac  
 g tan
m
m
Vertical Circle
Look at the forces at
the top of the circle
 The minimum speed
at the top of the
circle can be found

v top  gR
See ex. 7.9 on page 205

A jet traveling at a speed of 1.20 x 102
m/s executes a vertical loop with a radius
5.00 x 102 m. Find the magnitude of the
force of the seat on a 70.0 kg pilot (a) at
the top (b)the bottom of the loop.
Forces in Accelerating
Reference Frames
Distinguish real forces from fictitious
forces
 Centrifugal force is a fictitious force
 Real forces always represent
interactions between objects

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
or
m1m 2
FG  G 2
r
Law of Gravitation
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
Gravitation Constant
Determined
experimentally
 Henry Cavendish



1798
The light beam and
mirror serve to
amplify the motion
Applications of Universal
Gravitation

Mass of the earth

Use an example of
an object close to
the surface of the
earth

r ~ RE
gR
ME 
G
2
E
Applications of Universal
Gravitation
Acceleration due to
gravity
 g will vary with
altitude

ME
gG 2
r
Gravitational
Field Lines

Gravitational Field
Strength is
considered force per
unit mass
Gmearthmtestmass
F
2
r
GM earth
F

2
mtestmass
r
F GM earth
g 
m
r2
Gravitational Potential
Energy


PE = mgy is valid
only near the
earth’s surface
For objects high
above the earth’s
surface, an
alternate
expression is
needed
MEm
PE  G
r

Zero reference
level is infinitely
far from the earth
F
Gm1 m2
r2
F  mg
PE  m gh
Gm m2
mg
r2
m gr  PE
Gm m2
m gr 
r
Gm m2
PE 
r
Einstein’s view of Gravity
Space-Time
Kepler’s Laws
All planets move in elliptical orbits with
the Sun at one of the focal points.
 A line drawn from the Sun to any planet
sweeps out equal areas in equal time
intervals.
 The square of the orbital period of any
planet is proportional to cube of the
average distance from the Sun to the
planet.

Kepler’s Laws, cont.
Based on observations made by Brahe
 Newton later demonstrated that these
laws were consequences of the
gravitational force between any two
objects together with Newton’s laws of
motion

Kepler’s First Law

All planets move in
elliptical orbits with
the Sun at one
focus.


Any object bound to
another by an
inverse square law
will move in an
elliptical path
Second focus is
empty
Kepler’s Second Law

A line drawn from
the Sun to any
planet will sweep
out equal areas in
equal times

Area from A to B and
C to D are the same
Kepler’s Third Law

The square of the orbital
period of any planet is
proportional to cube of the
average distance from the
Sun to the planet.


For orbit around the Sun, KS
= 2.97x10-19 s2/m3
K is independent of the
mass of the planet
T  Kr
2
2
3
 Ta   ra 
    
 Tb   rb 
3
Derivation
So…
F G
M sM p
2
r
2r

2
v
 m ac  M p
r

2
Ms
T
G 2 
r
r
2
 4  3
2
3


T 
r

K
r
s

 GM s 
4 2
19 s 2
Ks 
 2.97x10
3
m
GM s
Kepler’s Third Law application
Mass of the Sun or
other celestial body
that has something
orbiting it
 Assuming a circular
orbit is a good
approximation
