Physics Beyond 2000 Chapter 3 Circular Motion .
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Transcript of Physics Beyond 2000 Chapter 3 Circular Motion .
![Page 1: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/1.jpg)
Physics Beyond 2000
Chapter 3
Circular Motion
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/ucm.html
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Uniform Circular Motion
• The path is the circumference of a circle with constant radius r.
• The speed is a constant.
• The direction of motion changes with time.
rv
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Uniform Circular Motion
• The path is the circumference of a circle with constant radius r.
• The speed is a constant.
• The direction of motion changes with time.
r
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Uniform Circular Motion
• The path is the circumference of a circle with constant radius r.
• The speed is a constant.
• The direction of motion changes with time.vv
vv
r
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Uniform Circular Motion
• Period T is the time needed to complete one cycle.
• Frequency f is the number of cycles completed in one second.
r
Tf
1
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Uniform Circular Motion
• Example 1
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Uniform Circular Motion
• Angular displacement – It is the angle, in radian, that the
object turns.
r
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Angular displacement
• Angular displacement – It is the angle, in radian, that the
object turns.
r
s
Length of arcs = r.
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Angular displacement
• After one complete cycle,
angular displacement
θ= 2π
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Angular displacement
• Example 2
![Page 11: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/11.jpg)
Angular speed
• Definition of average angular speed, ωav
– Δθis the angular displacement– Δt is the time taken
tav
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Angular speed
tav
If we consider one complete cycle,
Δθ= 2π and Δt = T
then ωav = 2πf
![Page 13: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/13.jpg)
Angular speed
• Example 3
![Page 14: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/14.jpg)
Instantaneous angular speed
dt
d
tt
0
lim
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Instantaneous angular speed
Example 4
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Angular speed and linear speed
• When the object moves from A to B at linear speed v,• Δs = r. Δθ v = r. ω
r
Δs
AB
Δθ
O X
v
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Angular speed and linear speed
• Example 5
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Centripetal acceleration
• In a uniform circular motion, the velocity v changes in direction but not in magnitude.
• It requires an acceleration a to change the direction of the velocity but not the magnitude.
• The acceleration must be always perpendicular to the velocity.
v
a
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Centripetal acceleration
r
AB
Δθ
O X
vA
vB
t
vv
t
va AB
In time Δt, the object moves from A to B.
)(. AB vvta
|vB| = |vA| = v
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Centripetal acceleration
r
AB
Δθ
O X
vA
vB
In time Δt, the object moves from A to B.
)(. AB vvta
Bv
Av
ta .
|vB| = |vA| = v
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Centripetal acceleration
r
AB
Δθ
O X
vA
vB
In time Δt, the object moves from A to B.
Bv
Av
ta .
Δθ
Note that the triangle is an isosceles triangle.
|vB| = |vA| = v
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Centripetal acceleration
r
AB
Δθ
O X
vA
vB
In time Δt, the object moves from A to B.
Bv
Av
ta .
Δθ
|vB| = |vA| = v
v. Δθ= a. Δt
a = t
v
.
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Centripetal acceleration
r
AB
Δθ
O X
vA
vB
Bv
Av
ta .
Δθ
|vB| = |vA| = v
r
vaor
ra
vt
va
2
2.
..
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Centripetal acceleration
r
AB
Δθ
O X
vA
vB
Bv
Av
ta .
Δθ
|vB| = |vA| = v
The acceleration is pointing to the centre of the circle.
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Centripetal acceleration
The acceleration is pointing to the centre of the circle.The magnitude of the acceleration is or r. ω2
rO X
v
a r
v2
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Centripetal acceleration
rO X
v
a
In this motion, though the magnitudeof the acceleration does not change,its direction changes with time.So the motion is of variable acceleration.
rOX
va
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Centripetal acceleration
• Example 6
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Centripetal force
• Force produces acceleration.
amF
.•Centripetal force produces centripetal acceleration.
r
vmFc
2
.
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Centripetal force
r
vmFc
2
. or 2mrFc •The force Fc is pointing to the centre of the circle.•The force Fc is perpendicular to the direction of the velocity.
rO X
v
Fc
![Page 30: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/30.jpg)
Centripetal force
r
vmFc
2
. or 2mrFc
rO X
v
Fc
To keep the object moving ina circle of radius r and speedv, it is necessary to have a net force, the centripetal force, acting on the object.
![Page 31: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/31.jpg)
Centripetal force
Example 7
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Centripetal Force: Example 7
The man is in circular motion.
The net force on the man
= Fc.
Fc
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Centripetal Force: Example 7
W – N = Fc
mg – N = Fc
N = mg - FcW N
There are two forceson the man.N = normal contact forceW = weightFc
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Centripetal force• If the provided force =
then the object is kept in a uniform circular motion.
r
vmFc
2
.
r
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Centripetal force• If the provided force >
then the object is circulating towards the centre.
r
vmFc
2
.
r
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Centripetal force• If the provided force <
then the object is circulating away from the centre.
r
vmFc
2
.
r
![Page 37: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/37.jpg)
Whirling a ball with a string in a horizontal circle
Top View
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Whirling a ball with a string in a horizontal circle
Top View
There is force Fc acting on the ball alongthe string.There is force F acting at the centre alongthe string.
FcF v
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Whirling a ball with a string in a horizontal circle
Top View
The two forces Fc and F are action and reaction pair.
Fc
F
v
FcF
v
![Page 40: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/40.jpg)
Whirling a ball with a string in a horizontal circle
Top View
What happens if Fc suddenly disappears?(e.g. the string breaks.)
Fc
F
v
FcF
v
![Page 41: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/41.jpg)
Whirling a ball with a string in a horizontal circle
What happens if Fc suddenly disappears?(e.g. the string breaks.)
v
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Whirling a ball with a string in a horizontal circle
What happens if Fc suddenly disappears?(e.g. the string breaks.)It is moving away tangent to the circle.
![Page 43: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/43.jpg)
Example 8
m=0.4kg
M=0.5kg
r = 0.5 mFc
What is the source of the centripetal force?
In equilibrium.
http://www.dipmat.unict.it/vpl/ntnujava/circularMotion/circular3D_e.html
In circular motion
![Page 44: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/44.jpg)
Centripetal force
r
vmFc
2
. or 2mrFc
![Page 45: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/45.jpg)
Centripetal force
• Fc m
• Fc v2
Fc r
1
![Page 46: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/46.jpg)
Uniform motion in a horizontal circle
R
v
![Page 47: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/47.jpg)
Uniform motion in a horizontal circle
![Page 48: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/48.jpg)
Uniform motion in a horizontal circle
![Page 49: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/49.jpg)
Uniform motion in a horizontal circle
![Page 50: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/50.jpg)
Uniform motion in a horizontal circle
![Page 51: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/51.jpg)
Uniform motion in a horizontal circle
R
T
mg
The object isunder two forces, the tension Tand the weight mg.
![Page 52: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/52.jpg)
Uniform motion in a horizontal circle
R
T
mg
The net force on the object isthe centripetal force because the object ismoving in a circle.
Fc
cFgmT
.
![Page 53: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/53.jpg)
Uniform motion in a horizontal circle
R
T
mgFc
cFgmT
.
T.cos = mgand
R
mvT
2
sin.
![Page 54: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/54.jpg)
Uniform motion in a horizontal circle
R
T
mgFc
cFgmT
.
Rg
v2
tan
![Page 55: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/55.jpg)
Uniform motion in a horizontal circle
R
T
mgFc
Rg
v2
tan
For a faster speed v, the angle θtends to increase.
![Page 56: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/56.jpg)
Experiment
• To verify the equation for centripetal force
r
vmFc
2
. or 2mrFc
![Page 57: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/57.jpg)
To verify the equation for centripetal force
• Whirl the bob in a horizontal circle with string.• The other end of the string is tied to some hanging weight.• Maintain the hanging weight in equilibrium.
bob
hangingweight
hollow plastictube
![Page 58: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/58.jpg)
To verify the equation for centripetal force
L = length of the string in motionm = mass of the bobM = mass of the hanging weightω= angular velocity of the bobθ= angle that the string makes with horizontal
bob
hangingweight
Lθ
m
M
![Page 59: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/59.jpg)
To verify the equation for centripetal force
T = tension on the string (tensions on both ends are equal if thereis not any friction between the string and the tube.)Mg = weight of the hanging weightmg = weight of the bob
bob
hangingweightMg
T
T
mg
θ
![Page 60: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/60.jpg)
To verify the equation for centripetal force
The hanging weight is in equilibrium.T = Mg ----------- (1)
bob
hangingweightMg
T
T
mg
θ
![Page 61: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/61.jpg)
To verify the equation for centripetal force
The bob is in circular motion with angular velocity ω.Fc = m.r. ω2 ----------- (2)
bob
hangingweightMg
T
T
mg
r
Lθ
![Page 62: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/62.jpg)
To verify the equation for centripetal force
gmTFc
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
The net force on the bob is equal to the centripetal force.
(3)
![Page 63: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/63.jpg)
To verify the equation for centripetal force
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
Resolve the forces on the bob into vertical and horizontal components.
Fc = T.cosθ ------------- (4)and mg = T.sinθ------------ (5)
![Page 64: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/64.jpg)
To verify the equation for centripetal force
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
Also cosθ=
L
r(6)
![Page 65: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/65.jpg)
To verify the equation for centripetal force
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
From equations (1), (2), (4), (5) and (6),find ωin terms of L, m, M and g.
![Page 66: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/66.jpg)
To verify the equation for centripetal force
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
![Page 67: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/67.jpg)
Measure M and m before the experiment.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
![Page 68: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/68.jpg)
Measure ω during the experiment.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
ω = Number of revolution × 2π÷ time
![Page 69: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/69.jpg)
Verify the following equation.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
![Page 70: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/70.jpg)
Problem: How to measure L?Refer to the textbook for the
skill.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
![Page 71: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/71.jpg)
bobT
mg
r
L
Fc
θ
Note that the bob must have a net force, the centripetal force, on it in order to keep it in acircular motion.As a matter of fact, the bob is not in equilibrium.It is in a motion with variable acceleration.
![Page 72: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/72.jpg)
Leaning on a vertical cylinder
Place an object on the innerwall of the cylinder. The cylinder starts to rotateabout its axis.
r
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Leaning on a vertical cylinder
As the cylinder rotates, the object performs a circularmotion. At a certain angularvelocity ω, the static friction may be sufficient to supportthe object on the wall withouttouching the ground.
ω
r
![Page 74: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/74.jpg)
Leaning on a vertical cylinderThere are 3 forces on theobject.
N = normal reactionW = weight of the object = mgf = static friction = μs.Nwhere μs is the coefficientof static friction.
N
f
W
ω
r
![Page 75: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/75.jpg)
Leaning on a vertical cylinder
As the object is in a circularmotion, the net force mustbe the centripetal force.
N = mrω2
andμsN mg≧
Note that the static friction (f) cancels the weight (W).But the left hand side on the second equation is thelimiting static friction which is the maximum friction.
N
f
W
ω
r
![Page 76: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/76.jpg)
Leaning on a vertical cylinder
Solve the two equations.We have
r
g
s
N
f
W
ω
r
r
g
s min
and
![Page 77: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/77.jpg)
Rounding a Bend
http://oldsci.eiu.edu/physics/DDavis/1150/05UCMGrav/Curve.html
![Page 78: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/78.jpg)
Rounding a Bend
r
A car turns rounda corner.It is a circular motionwith a path of radiusof curvature r.
![Page 79: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/79.jpg)
Rounding a Bend
r
A car turns rounda corner.It is a circular motionwith a path of radiusof curvature r.
v
![Page 80: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/80.jpg)
Rounding a Bend
http://plabpc.csustan.edu/general/tutorials/CircularMotion/CentripetalAcceleration.htm
![Page 81: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/81.jpg)
Rounding a Bend
r
It requires a centripetal forcefor the circularmotion.
v
Fc
![Page 82: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/82.jpg)
Rounding a Bend
How comes the centripetal force?
r
v
Fc It may comefrom the frictionor the normalreaction.
![Page 83: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/83.jpg)
Level Road without Banking
r
v
Fc
Fc comes from the static friction fs between tyres and the road.
The speed v of thecar must not exceed .grs
where μs is the coefficient of static friction.
![Page 84: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/84.jpg)
Level Road without Banking
r
v
Fc
vmax = grs
Note:vmax is independentof the mass of the car.
![Page 85: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/85.jpg)
Force on the passenger
• The passenger needs a centripetal force for
turning round the corner with the car.
• The normal contact force from the car is
the centripetal force.
![Page 86: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/86.jpg)
Example 9
• To find the coefficient of static friction.
r
v
Fc
![Page 87: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/87.jpg)
Don’t rely on friction!
• When the road condition changes (e.g.
on a rainy day), μs becomes very small.
Even a slow speed may exceed the safety
limit. vmax = grs
![Page 88: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/88.jpg)
Banked Road
• Design a banked road which is inclined to the centre.
![Page 89: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/89.jpg)
Banked Road
• Design a banked road which is inclined to the centre.
R
Wθ
The car is movingforward (into theplane) withvelocity v and is turning left.The radius ofcurvature is r.
r
![Page 90: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/90.jpg)
Banked Road• The centripetal force comes from
the normal contact force R.
R
W
Fc
θ
r
![Page 91: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/91.jpg)
Banked Road• The centripetal force comes fromthe normal contact force R.
Note that Fc is the horizontal component of R.
R
W
Fc
θ
r
![Page 92: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/92.jpg)
Banked Road• In ideal case, friction is not necessary.
The ideal banking angle is
R
W
Fc
θ
rg
v2
tan
r
![Page 93: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/93.jpg)
Example 10
• Find the ideal banking angle of a road.
• The ideal speed is
rg
v2
tan
tan.rg
![Page 94: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/94.jpg)
Banked Road• In non-ideal case, friction f is needed.
• Speed is too slow, less than the ideal speed.
R
W
Fc
θ
fr
![Page 95: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/95.jpg)
Banked Road• In non-ideal case, friction f is needed.
• Speed is too fast, more than the ideal speed.
R
W
Fc
θ
fr
![Page 96: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/96.jpg)
Railway
• When there is a bend, the railway is banked.
• This avoids having lateral force on the rail.
![Page 97: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/97.jpg)
Aircraft
rFL
W
Back view ofthe car.The aircarft is movingforward (into theplane) withvelocity v.
• When an airplane moves in a horizontal circular path in air, it must tilt about its own axis an angle θ.
• The horizontal component of the lift force
FL is the centripetal force Fc.
θ
![Page 98: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/98.jpg)
Aircraft• When an airplane moves in a horizontal
circular path in air, it must tilt about its own axis an angle θ.
![Page 99: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/99.jpg)
Aircraft• When an airplane moves in a horizontal circ
ular path in air, it must tilt about its own axis an angle θ.
• The horizontal component of the lift force
FL is the centripetal force Fc.
rFL
W
Fc Note thatFc is horizontal.
θ
![Page 100: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/100.jpg)
Aircraft• When an airplane moves in a horizontal circ
ular path in air, it must tilt about its own axis an angle θ.
• The horizontal component of the lift force
FL is the centripetal force Fc.
rFL
W
Fc θgr
v2
tan
![Page 101: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/101.jpg)
Example 11
• Find the speed of the aircraft.
gr
v2
tan
![Page 102: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/102.jpg)
Bicycle on a Level Road
• When turning round a corner, it needs centripetal force.
• Like a car bending round a corner, the centripetal force comes from the static friction between the tyres and the road.
![Page 103: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/103.jpg)
Bicycle on a Level Road
Unlike a car, the bike inclines towards the centre to avoid toppling.
The bike is movinginto the plane atspeed v and is turningleft.The radius of curvatureis r.
vertical
horizontal
r
![Page 104: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/104.jpg)
Bicycle on a Level Road
What is the angle of tilt ?
The bike is movinginto the plane atspeed v and is turningleft.The radius of curvatureis r.
vertical
horizontal
r
![Page 105: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/105.jpg)
Bicycle on a Level Road
Forces on the bike: weight W, normal contact force R and static friction fs.
vertical
horizontalW
R
fs
Note that Wacts at the centreof mass G of the bike.h is the height ofG from the ground.
G
h
![Page 106: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/106.jpg)
Bicycle on a Level Road
R = mg ------- (1) ----------- (2)
vertical
horizontalW
R
fs
G
r
mvf s
2
R balances W.fs is the centripetalforce.
h
![Page 107: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/107.jpg)
Bicycle on a Level Road
R = mg ------- (1) ----------- (2)
vertical
horizontalW
R
fs
G
r
mvf s
2
In order not totopple, the momentabout G must be zero.
About G, clockwise moment= anticlockwise moment
h
![Page 108: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/108.jpg)
Bicycle on a Level Road
R = mg ------- (1) ----------- (2)
vertical
horizontalW
R
fs
G
r
mvf s
2
About G, clockwise moment= anticlockwise moment fs.h = R.ah
rg
v2
tan
![Page 109: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/109.jpg)
Tilt of a Car in Circular Motion
Forces on the car: frictions f1 and f2, normal
contact forces R1 and R2, weight W.
The car is movinginto the plane atspeed v and is turningleft.The radius of curvatureis r.
r
![Page 110: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/110.jpg)
Tilt of a Car in Circular Motion
Forces on the car: frictions f1 and f2, normal
contact forces R1 and R2, weight W.
rf1
f2
R1 R2W acts atthe centre of massG of the car.
W
G
![Page 111: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/111.jpg)
Tilt of a Car in Circular Motion
Forces on the car: frictions f1 and f2, normal
contact forces R1 and R2, weight W.
rf1
f2
R1 R2We are goingto compareR1 and R2.
W
G
![Page 112: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/112.jpg)
Tilt of a Car in Circular Motion
Let 2L be the separation betweenthe left and right tyres.r
f1f2
R1 R2
W
G
L L
![Page 113: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/113.jpg)
Tilt of a Car in Circular Motion
Let h be theheight of the centre of mass G from the ground.
rf1
f2
R1 R2
W
G
L L
h
![Page 114: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/114.jpg)
Tilt of a Car in Circular Motion
Without toppling,the moment aboutG must be zero.About G,clockwise moments=anticlockwise moments
rf1
f2
R1 R2
W
G
L L
h
![Page 115: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/115.jpg)
Tilt of a Car in Circular Motion
rf1
f2
R1 R2
W
G
L L
h
).( 2112 ffL
hRR
So R2 > R1
![Page 116: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/116.jpg)
Tilt of a Car in Circular Motion
rf1
f2
R1 R2
W
G
As R2 > R1, the springs on theright are compressed more.
The car tilts rightwhen it turnsleft.
http://www.sciencejoywagon.com/physicszone/lesson/03circ/centrif/centrif.htm
![Page 117: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/117.jpg)
Uniform Motion in a Vertical Circle
• The path is the circumference of a vertical circle with constant radius r.
• The speed is v, a constant.
• The mass of the object is m.
![Page 118: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/118.jpg)
Uniform Motion in a Vertical Circle
• The path is the circumference of a vertical circle with constant radius r.
• The speed is v, a constant.
• The mass of the object is m.
![Page 119: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/119.jpg)
Uniform Motion in a Vertical Circle
r
mvFc
2
The centripetal force is
r
Fc
v
O
How comes the centripetal force?
![Page 120: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/120.jpg)
Uniform Motion in a Vertical Circle
r
Fc
v
O
The centripetal force comesfrom the tension T and the weightof the mass W or mg.
![Page 121: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/121.jpg)
At the highest position
T1
mg
T1 + mg = Fc
andrFc
O
v
r
mvFc
2
r
mvmgT
2
1
![Page 122: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/122.jpg)
At the highest position
T1
mg
rFc
O
v
r
mvmgT
2
1
mgr
mvT
2
1
![Page 123: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/123.jpg)
At the highest position
T1
mg
rFc
O
v
mgr
mvT
2
1
What would happen
if v2 = ?r
m
![Page 124: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/124.jpg)
At the lowest position
T2
mg
T2 - mg = Fc
and
r
mvFc
2
O
v
Fc
Note that T2 is always positive.
r
mgr
mvT
2
2
![Page 125: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/125.jpg)
At any other positions
O
vT3
r
θ
mg
F
There are three forces on the mass.T3 is the tension from the rod,F is force from the rodand mg is the weight ofthe mass
![Page 126: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/126.jpg)
At any other positions
The net forceis the centripetal force Fc.
Ov
Fc
r
θ
r
mvFc
2
![Page 127: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/127.jpg)
At any other positions
Ov
T3
r
θ
mg.cosθ
θ
Along the radial direction,
Fc = T3 – mg.cosθ
![Page 128: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/128.jpg)
At any other positions
Ov
T3
r
θ
mg.cosθ
θ
cos.
cos.
2
3
2
3
mgr
mvT
r
mvmgT
So
![Page 129: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/129.jpg)
Uniform Motion in a Vertical Circle
• At the highest position,
• At the lowest position,
• At any other
positions,
mgr
mvT
2
1
mgr
mvT
2
2
cos.2
3 mgr
mvT
![Page 130: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/130.jpg)
Non-uniform Motion in a Vertical Plane
• The motion of an object coasting
along a vertical “ looping-the-loop”.
• Its speed would change at different
positions.
• The principle is also applied to whirling
mass with a string in a vertical plane.
![Page 131: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/131.jpg)
Looping the loop• Mass of the marble is m.• Radius of the loop is r.• The marble starts at the lowest position
with speed vo.
voO
r
![Page 132: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/132.jpg)
Looping the loop• Note that there is change in kinetic energy
and gravitational potential energy.
• Assume that energy is conserved.
voO
r
vo
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/ce.html
![Page 133: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/133.jpg)
Looping the loop
• The speed v of the marble changes on the
loop.
• The centripetal force changes on the loop.
v
O
r
vv
![Page 134: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/134.jpg)
At the lowest position• The speed v1 of the marble is vo.• The centripetal force Fc comes from the norm
al contact force N1 and the weight of the marble mg.
O
r
vo
N1
mg
![Page 135: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/135.jpg)
At the lowest position
O
r
vo
N1
mg
r
mvmgN o
2
1 mgr
mvN o
2
1
![Page 136: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/136.jpg)
Below the centre
O
r
v2
N2
mg
r
mvmgN
22
2 cos. cos.22
2 mgr
mvN
θ
![Page 137: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/137.jpg)
Below the centre
O
r
v2
N2
mg
θ
From conservation of energy, )cos1(222
2 grvv o
vo
h = r(1-cosθ)
![Page 138: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/138.jpg)
Above the centre
O
r
v3
N3
mg
r
mvmgN
23
3 cos. cos.23
3 mgr
mvN
ψ
![Page 139: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/139.jpg)
Above the centreFrom conservation of energy, )cos1(222
3 grvv o
O
r
v3
N3
mgψ
h=r(1+cosψ)vo
![Page 140: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/140.jpg)
At the highest position
v4
O
rN4mg
r
mvmgN
24
4 mgr
mvN
24
4
![Page 141: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/141.jpg)
At the highest positionFrom conservation of energy, grvv o 422
4
O
rN4mg
h = 2rvo
v4
![Page 142: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/142.jpg)
Completing the Circle
• For the marble to reach the highest position,
024
4 mgr
mvN grv 4
and
grgrvv o 4224 grvo 5
![Page 143: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/143.jpg)
Completing the Circle
• The marble cannot move up the loop and oscillates like a pendulum.
grvo 2
vo rg
vh o
2
2r
![Page 144: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/144.jpg)
Completing the Circle
• The marble cannot move up the loop and oscillates like a pendulum.
grvo 2
vo rg
vh o
2
2r
![Page 145: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/145.jpg)
Completing the Circle
• The marble rises up to height more than r and is projected away.
grvgr o 52
vo
rhr 2r
![Page 146: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/146.jpg)
Completing the Circle
O
r
vo
grvo 5
![Page 147: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/147.jpg)
Whirling freely with a rod
• The ball moves and passes its loop with its own initial energy.
vo
![Page 148: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/148.jpg)
Whirling freely with a rod
• A light rod would not be loosen.
• The light rod can provide tension or compression depending on the case.
vo
![Page 149: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/149.jpg)
Whirling freely with a rod
The minimum vo is for the marble to just
reach the top.
From conservation of energy, minimum vo= 2
min vo
2r
v = 0
h
gr
![Page 150: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/150.jpg)
Whirling freely with a rod
min vo
2r
v
θ
h=r(1+cosθ)F mg
r
mvmgF
2
cos.
θ
)cos1(22 grv
![Page 151: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/151.jpg)
Changing from tension to compression
min vo
2r
vθo
h=r(1+cosθo)mg
θo
When F = 0, the force changes fromtension (F>0) to compression (F<0).
![Page 152: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/152.jpg)
Changing from tension to compression
min vo
2r
vθo
h=r(1+cosθo)mg
θo
Prove that θo = 48.2o when F = 0.
![Page 153: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/153.jpg)
Example 12
• Whirling a bucket of water in a vertical
circle.
![Page 154: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/154.jpg)
Example 12
• Water does not flow out when the bucket is at the top position.
v
rmg
![Page 155: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/155.jpg)
Centrifuge
• It is a device to
separate solid or
liquid particles of
different densities
by rotating
them in a tube in a
horizontal circle.
![Page 156: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/156.jpg)
Centrifuge
r
axis ofrotation
ω
• ω is the angular velocity.• r is the distance of the small portion of liquid from the axisof rotation.
![Page 157: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/157.jpg)
Centrifuge
r
axis ofrotation
ω
The small portion is in uniformcircular motion.The centripetal force comesfrom the pressure difference ΔP.
ΔP
![Page 158: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/158.jpg)
Centrifuge
r
axis ofrotation
ω
The small portion is replaced by another portion of smaller density.The centripetal force is not enoughto support its uniform circular motion.
ΔP
![Page 159: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/159.jpg)
Centrifuge
r
axis ofrotation
ω
As a result, this portion of smallerdensity moves towards the centralaxis.
ΔP
![Page 160: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/160.jpg)
Centrifuge
r
axis ofrotation
ω
Portion of larger density moves away from the central axis.
ΔP
![Page 161: Physics Beyond 2000 Chapter 3 Circular Motion .](https://reader038.fdocuments.us/reader038/viewer/2022110209/56649dfe5503460f94ae64fb/html5/thumbnails/161.jpg)
Centrifuge
• Study p.51 and 52 for the mathematical
deduction.