Transcript Circular_Motion

Circular Motion
The Radian
Objects moving in circular (or nearly circular)
paths are often measured in radians rather
than degrees.
 In the diagram, the angle θ, in
radians, is defined as follows


So, if s = r the angle is 1 rad and if s is equal
to the full circumference of the circle, the
angle is 2π rad. (In other words, 360° = 2π
rad.)
Radians and Degrees
Angular Displacement (θ)
If we consider a small body to be moving
round the circle from A to B we say that it
has experienced an angular displacement of
θ radians. The relation between the (linear)
distance moved, d, of the body and the
angular displacement θ, is given by
d = rθ
 Also, if the angle is small, d is very nearly
equal to the magnitude of the linear
displacement of the body.

Angular Velocity (ω)
Suppose that the body moved from P1 to P2
in a time t. The linear speed, v, of the body is
given by v = d/t.
 If we divide d=rθ by t, we have
 The angular velocity ω is defined as
 Units of ω are rad/s
 Therefore,

v = rω
Merry-Go Round
http://animations.50webs.com/free_cartoons_mobile_animation.htm
What happens to your speed
as you go to the middle of
a merry-go round?
a. The speed remains
constant.
b. The speed increases
c. The speed decreases
Merry-Go Round
http://animations.50webs.com/free_cartoons_mobile_animation.htm
What happens to your
angular speed as you go
to the middle of a merrygo round?
a. The angular speed
remains constant.
b. The angular speed
increases
c. The angular speed
decreases
http://mocoloco.com/art/upload/2009/12/biondo_merry_go_round.jpg
Angular Acceleration(α)






Previously we assumed that the body moved from P1 to P2
with constant speed. If the linear speed of the body changes
then, obviously, the angular speed (velocity) also changes.
The angular acceleration, α, is the rate of change of
angular velocity.
So, if the angular velocity changes uniformly from ω1 to ω2
in time t, then we can write:
Now, linear acceleration, a, is given by
Substituting v=rω 
We find,
a  r
a
v f  vo
t
a
v f  vo
t
Circular Motion Definitions
Time Period, T
 The time period of a circular motion is the
time taken for one revolution.
Rotational Frequency, f
 The rotational frequency of a circular motion
is the number of revolutions per unit time.
 Time period is the inverse of frequency
 Also,
and
Rotational Motion
What is the
relationship
between Liner
and Rotational
motion
quantities?
Acceleration Equations Revisited
What is the Acceleration?
a.
b.
c.
d.
No acceleration
Acceleration
outward
Acceleration toward
the center of the
circle
Acceleration points
tangential to the
circle
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
A plane attached to a
string flies in a circle
at constant speed





Even though the speed is
constant, velocity is not constant
since the direction is changing: 
acceleration!
If released the plane would travel
to A at constant speed.
Therefore, the change in velocity
would be a vector from A to P
If θ is very small, this acceleration
(Δv/t) points to the center of the
circle.
This is called centripetal
acceleration
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
What is the Acceleration?
Centripetal Acceleration
Even though the speed is constant, velocity is
not constant since the direction is changing:
must be some acceleration!
 Consider average acceleration in time Δt

a avg

v

t
As we shrink Δ t, Δv / Δt
v
v2
R
dv / dt = a
v1
v2
v1
R
v
v
t
seems like v
v (hence v
v/t )
points at the origin!
Centripetal
Acceleration
v
v2
R
We see that a points
in the - R direction.
But R = vt for small t
v1
v v t

So:
v
R

a = dv
v / dt
v R

Similar triangles:
v
R
v2
R
R
v v 2

t
R
v2
Magnitude: a 
R
Direction:- r (toward center of circle)
2r
Since
and v = ωr
v2
a
=
ω
a

R
Centripetal Acceleration
http://www.lyon.edu/webdata/users/shutton/phy240-fall2003/circle1.gif
Centripetal Acceleration (ac)
‘Centripetal’ means center seeking
 Vector direction always points toward the
center of the circle
 Magnitude:
2

v
ac 
r
v = speed
r = radius of the circle
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/ucm.gif
Centripetal Acceleration (ac)
http://physics.csustan.edu/Astro/Help/NEWTON/cpetal2.gif
Example: Centripetal Acceleration
A fighter pilot flying in a circular turn will pass out if
the centripetal acceleration he experiences is more
than about 9 times the acceleration of gravity g. If
his F-22 is moving with a speed of 300 m/s, what is
the approximate diameter of the tightest turn this
pilot can make and survive to tell about it?
(a) 500 m
(b) 1000 m
(c) 2000 m
Example: Centripetal Acceleration
Using the definition of ac
d = 2R ~ 2000m
2
v
ac   9 g
R
D  2 R  2000 m
Substituting in the values
m2
90000 2
2
v
s
R

9 g 9  9.81 m
s2
Simplifying,
10000
R
m  1000 m
9.81
2km
What is the Net Force?
a.
b.
c.
d.
No net force
Net Force points
outward
Net Force points
toward the center of
the circle
Net Force points
tangential to the
circle
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
A plane attached to a
string flies in a circle
at constant speed
a.
b.
c.
d.
No net force
Net Force points
outward
Net Force points
toward the center of
the circle
Net Force points
tangential to the
circle
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
What is the Net Force?
A plane attached to a
string flies in a circle
at constant speed
Centripetal Force, Fc, is
this net force.
 Recall, Newton's
2nd Law Fnet=ma
 Therefore, Fc = mac
2
v
Fc  m
r
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
What is the Net Force?
A plane attached to a
string flies in a circle
at constant speed


acts toward the center of the circle
depends on mass, speed, and size of
the circle
2
v
Fc  m
r

Net force, provided by another force
or interactions of forces
http://motivate.maths.org/conferences/conf14/images/circular_motion3.gif
Centripetal Force (Fc)
Centripetal Acceleration Example


When you are driving a car,
and you turn the steering wheel
sharply to the right in order to
turn the car to the right, you
"feel" as if a force is pushing
your body to the left against the
door.
In order for your body to follow
the car in the tight circular path,
something has to push your
body toward the center of the
circle-- in this case it is the
driver's-side door-- and your
tendency otherwise is to travel
in a straight line tangent to the
circular path
Date Physics
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/rht.html
Centripetal Force Source
What provides the centripetal force (Fc) in the
following scenarios?
A car going around a corner at a constant speed.

Friction force between the
tires and the pavement
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Centripetal Force Source
What provides the centripetal force (Fc) in the
following scenarios?
 A ball on a string being
twirled in a circle.

Tension force of the string
on the ball
http://www.mansfieldct.org/schools/mms/staff/ha
nd/lawsCentripetalForce_files/image004.jpg
http://www.vast.org/vip/book/LOOPS/HOME7.GIF
Centripetal Force Source

The sun’s force of gravity
on the planets.
http://quest.nasa.gov/aero/planetary/orbit/Image1.jpg
What provides the centripetal force (Fc) in the
following scenarios?
 The planets orbiting around the sun.
Centripetal Force Source
What provides the centripetal force (Fc) in the
following scenarios?
A motorcycle stuntman going around a loop.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Both the normal force of the track and
gravity.
Centripetal Force Example
A ball attached to a string is twirled in a circle of
radius 0.5 m with a speed of 2 m/s. If the ball has a
mass of 0.25 kg, what is the tension of the string.
 Knowns:
http://www.frontiernet.net/~jlkeefer/centacc.gif
m = 0.25 kg
v = 2 m/s
r = 0.5 m
 F  T  Fnet  Fc
2
v
Fc  m
r
0.252 
T  Fc 
0 .5
2
T=2N
Moving in a
Straight line
on a
Horizontal
Surface
Moving in a
Straight line
on a
Horizontal
Surface
The normal reaction, FN,
has no component
acting towards the
center of the circular
path.
Therefore the required
centripetal acceleration
is provided by the force
of friction, Ff, between
the wheel and the road.
If the force of friction is not strong enough, the vehicle
will skid.
Turning on a Banked Surface
The normal reaction, FN, now has
a component acting towards the
center of the circular path.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
If the angle, q, is just right, the
correct centripetal acceleration can
be provided by the horizontal
component of the normal reaction.
This means that, even if there is
very little force of friction the
vehicle can still go round the curve
with no tendency to skid.
Angle of Banking



The magnitude of the horizontal component of the
normal force is
FN x  FN sin q
This force causes the centripetal acceleration, so,
the magnitude of NX is also given by
mv2
FN x 
r
So, F sin q  mv 2
N


r
The vertical forces acting on the
vehicle are in equilibrium.
Therefore, summing the vertical forces
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
FN cos q  mg
Angle of Banking

Solving
into
FN cos q  mg
mv
FN sin q 
r
Simplifying,
2
for FN and substituting
mg sin q mv 2
gives: cos q  r
v2
tan q 
rg
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

This equation allows us to calculate the angle θ
needed for a vehicle to go round the curve at a
given speed, v, without any tendency to skid.
Indy Physics
Example:
 What is the minimum speed an Indy 500 car
must go not to slide down the banked curves
of the Indianapolis Motor Speedway if the
banked angle is 9.2° and the radius is 280.0
m?

NASCAR Physics

What is the minimum speed a NASCAR car
must go not to slide down the banked curves
of the Bristol Motor Speedway if the banked
angle is 36° and the radius is 73.5m?
Loopy
The minimum speed to complete a loop requires:
 Speed large enough to reach the top of the
loop.
 At the top of the loop Fnet = Fg + FN
 For the minimum speed FN = 0
2
mv
 Therefore, F  F  F  min recall Fg  mg
g
net
c
r
 So,
vmin  gr
at the top of the loop
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Kepler’s Law of Periods
Proof
Assumptions:
• Must conform to equations for circular motion
• Newton’s Universal Law of Gravity
• Planet rotates in a circular (elliptical) path
Fnet
mv 2
 Fc  mac 
r
Newton’s 3rd Law  symmetry
Fgravity  Fnet  Fc
Recall,
GMm mv 2
Fg  2 
r
r
so
2

T
v  r 
Therefore, GMm  m4 r
2
r2
T2
2r
T
Law of Periods
T
4
rearranging 3 
R
GM
2
2
Testing the Inverse Square Law of
Gravitation


The acceleration due to gravity at the surface of
the earth is 9.8m/s2.
If the inverse square relationship for gravity
(Fg~1/r2) is correct , then, at a distance ~60 times
further away from the center of the earth, the
9.8
m
 2.72  10
acceleration due to gravity should be 60
s
The centripetal acceleration of the moon is given
by a  vr where the radius of the moon’s orbit is r
= 3.84 × 108 m and the time period of the moon’s
2r
orbital motion is T = 27.3 days.
v
3
2

2
c
ac 
4 r
T2
2
ac 
4 2 (3.84  10 8 )

24h (3600s) 
(27.3days )(1day ) 1h 


T
2
a c  2.72  10 3
m
s2
2