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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
Also, if the angle is small, d is very nearly equal to the magnitude of the linear displacement of the body.
d = rθ
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
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
http://animations.50webs.com/free_cartoons_mobile_animation.htm
Merry-Go Round
What happens to your angular speed as you go to the middle of a merry-go round?
a. The angular speed remains constant.
b. The angular speed increases
c. The angular speed decreases
http://animations.50webs.com/free_cartoons_mobile_animation.htm
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,
t
vva of
t
vva of
ra
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. No accelerationb. Acceleration
outwardc. Acceleration toward
the center of the circle
d. Acceleration points tangential to the circle
A plane attached to a string flies in a circle at constant speed
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
What is the Acceleration?
Even though the speedspeed is constant, velocityvelocity is notnot 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
Cut
nell
& J
ohns
on,
Wil
ey P
ub
lish
ing
, P
hysi
cs 5
th E
d.
Centripetal Acceleration
Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing: must be some acceleration!
Consider average acceleration in time Δt
As we shrink Δ t, Δv / Δt dv / dt = aat
vaavg
vv2
t
vv1
vv1vv2
vv
vv1vv2
vv
RRRR vvvv
seems like vv (hence vv/t )
points at the origin!
Centripetal Acceleration
Magnitude:
Direction:- r r (toward center of circle)Since and v = ωr
points toward the center of the circle a = ω2r Since and v = ωr
aa = dvv / dt
We see that a a points
in the - RR direction.
RR
vv2
vv1
vv2
vv
RRRR
vv
RR
Similar triangles: vv
RR
Similar triangles:
But R = vt for small t
So:vt
vR
2 v
vv tR
R
va
2
R
va
2
a = ω2r
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:
v = speed
r = radius of the circle
r
vac
2
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
Substituting in the values
Simplifying,
d = 2R ~ 2000m
gR
vac 9
2
2
2
2
2
s
m8199
s
m90000
g9
vR
.
mmR 100081.9
10000
D R m 2 2000
2km
D R m 2 2000
2km
What is the Net Force?
a. No net forceb. Net Force points
outwardc. Net Force points
toward the center of the circle
d. Net Force points tangential to the circle
A plane attached to a string flies in a circle at constant speed
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
What is the Net Force?
a. No net forceb. Net Force points
outwardc. Net Force points
toward the center of the circle
d. Net Force points tangential to the circle
A plane attached to a string flies in a circle at constant speed
Cut
nell
& J
ohns
on,
Wil
ey P
ub
lish
ing
, P
hysi
cs 5
th E
d.
What is the Net Force?
Centripetal Force, Fc, is this net force.
Recall, Newton's 2nd Law Fnet=ma
Therefore, Fc = mac
A plane attached to a string flies in a circle at constant speed
Cut
nell
& J
ohns
on,
Wil
ey P
ub
lish
ing
, P
hysi
cs 5
th E
d.
r
vmFc
2
Centripetal Force (Fc)
acts toward the center of the circle depends on mass, speed, and size of
the circle
Net force, provided by another force or interactions of forces
r
vmFc
2
http
://m
otiv
ate.
mat
hs.o
rg/c
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renc
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otio
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if
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/hand/lawsCentripetalForce_files/image004.jpg
http://www.vast.org/vip/book/LOOPS/HOME7.GIF
Centripetal Force Source
What provides the centripetal force (Fc) in the following scenarios?
The planets orbiting around the sun.
The sun’s force of gravity
on the planets.
http
://q
uest
.nas
a.go
v/ae
ro/p
lane
tary
/orb
it/I
mag
e1.jp
g
Centripetal Force Source
What provides the centripetal force (Fc) in the following scenarios?
A motorcycle stuntman going around a loop.
Both the normal force of the track and gravity.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
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:
m = 0.25 kg
v = 2 m/s
r = 0.5 m
r
vmFc
2
cnet FFTF 5.0
225.0 2
cFT T = 2 N
http://ww
w.frontiernet.net/~jlkeefer/centacc.gif
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.
Moving in a Straight line on a Horizontal Surface
Moving in a Straight line on a Horizontal Surface
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.
If the angle, , 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.Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Angle of Banking
The magnitude of the horizontal component of the normal force is
This force causes the centripetal acceleration, so, the magnitude of NX is also given by
So,
The vertical forces acting on the
vehicle are in equilibrium. Therefore, summing the vertical forces
r
mvF xN
2
sinNxFNF
mgFN cos
r
mvFN
2
sin
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Angle of Banking
Solving for FN and substituting
into gives:
Simplifying,
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.
rg
v 2
tan
mgFN cos
r
mvFN
2
sin r
mvmg 2
cos
sin
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
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 Therefore, recall So,
at the top of the loop
r
mvFFF cnetg
2min mgFg
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
grv min
Kepler’s Law of Periods Kepler’s Law of Periods ProofProof
Assumptions:• Must conform to equations for circular
motion• Newton’s Universal Law of Gravity
•Planet rotates in a circular (elliptical) path
Newton’s 3rd Law symmetry
Recall, so
Therefore, rearranging
r
mvmaFF ccnet
2
cnetgravity FFF r
mv
r
GMmFg
2
2
T
rrv
2
T
2
2
2
2
4
T
rm
r
GMm
GMR
T 2
3
2 4
Law of Periods
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 acceleration due to gravity should be
The centripetal acceleration of the moon is given by where the radius of the moon’s orbit is r = 3.84 × 108 m and the time period of the moon’s orbital motion is T = 27.3 days.
23
21072.2
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8.9
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m
r
vac
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T
rv
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2
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T
rac
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)1084.3(4
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hdays
ac
231072.2
s
mac