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
  r s 

 t
 r

 s t
  v r s is the length of arc and r is the radius v t
 r


More About Radians

Comparing degrees and radians

1 rad

360
2



57 .
3

Converting from degrees to radians

[ rad ]


180


[deg rees ]

Angular Displacement, cont.

The angular displacement during some time interval is defined as the angle the object rotates through

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

 t f f

 t
 i i


 t
 a c

 v
 t
 v
2 r

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

, 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 t f
  i
 t i

 
 t

Tangential Acceleration v t
 r

 v t
 r
 
 v t
 t a t

 r
 
 t 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
About a Fixed Axis with
Constant Acceleration
Linear Motion with
Constant Acceleration
   i
  t v
 v i
 at
    i t

1
2
 t 2  x
 v i t

1 at 2
2
 2   2 i

2
   v 2  v i
2 
2 a
 x

Relationship Between Angular and Linear Quantities



Displacements s
  r
Speeds 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 velocity direction of the

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 a
C
  2 r
OR a c
 v
2 r
See page 198 for derivation

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 t
2  a 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
F net

F c
 ma c
 mv r
2
 m


4

T
2
2 r



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


Friction is the force that produces the centripetal acceleration
Can find the frictional force, µ, v f f f



 v
2 m r
 n
  mg v
  rg
  mg
 m v
2 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 tan
  v
2 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
 a
The horizontal component of the tension causes the centripetal acceleration
 g tan

C
See next page for derivation

Derivation
T cos
  mg

0
T
F
F a c
 mg cos


T sin
  mg sin cos


 ma c

F m
 m g tan
 m

 mg tan
 g tan


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 10 2 m/s executes a vertical loop with a radius
5.00 x 10 2 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.
F

G m
1 m
2 r
2 or
F

G
G m
1 m
2 r
2

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 ~ R
E
M
E
 gR
E
2
G

Applications of Universal
Gravitation


Acceleration due to gravity g will vary with altitude g

G
M
E r
2

Gravitational
Field Lines

Gravitational Field
Strength is considered force per unit mass
F

Gm earth m testmass r
2
F m testmass

GM earth r
2 g

F m

GM earth r
2

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
M
E m

PE
 
G r
Zero reference level is infinitely far from the earth
F
F
PE

Gm
 mg r
2
1 m
2
 mgh mg mgr


Gmm
2 r
2
PE mgr

Gmm
2 r
PE

Gmm
2 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, K
S
= 2.97x10
-19 s 2 /m 3
K is independent of the mass of the planet
T
2 
Kr
3



T a
T b


 2



 r a r b


 3

Derivation
So…
F
G
T
2

M s r
2
G

M s
M p
2 r
2
 
T
2
 r ma c




4

2
GM s


 r
3 

K s r
3
M p v
2 r
K s
4

2

GM s

2 .
97 x 10

19 s
2 m
3

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