Ch 7 - Circular Motion Circular motion: Objects moving in a circular path.
Physics Beyond 2000 Chapter 3 Circular Motion .
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Transcript of Physics Beyond 2000 Chapter 3 Circular Motion .
Physics Beyond 2000
Chapter 3
Circular Motion
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/ucm.html
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
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
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
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
Uniform Circular Motion
• Example 1
Uniform Circular Motion
• Angular displacement – It is the angle, in radian, that the
object turns.
r
Angular displacement
• Angular displacement – It is the angle, in radian, that the
object turns.
r
s
Length of arcs = r.
Angular displacement
• After one complete cycle,
angular displacement
θ= 2π
Angular displacement
• Example 2
Angular speed
• Definition of average angular speed, ωav
– Δθis the angular displacement– Δt is the time taken
tav
Angular speed
tav
If we consider one complete cycle,
Δθ= 2π and Δt = T
then ωav = 2πf
Angular speed
• Example 3
Instantaneous angular speed
dt
d
tt
0
lim
Instantaneous angular speed
Example 4
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
Angular speed and linear speed
• Example 5
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
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
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
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
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
.
Centripetal acceleration
r
AB
Δθ
O X
vA
vB
Bv
Av
ta .
Δθ
|vB| = |vA| = v
r
vaor
ra
vt
va
2
2.
..
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.
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
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
Centripetal acceleration
• Example 6
Centripetal force
• Force produces acceleration.
amF
.•Centripetal force produces centripetal acceleration.
r
vmFc
2
.
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
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.
Centripetal force
Example 7
Centripetal Force: Example 7
The man is in circular motion.
The net force on the man
= Fc.
Fc
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
Centripetal force• If the provided force =
then the object is kept in a uniform circular motion.
r
vmFc
2
.
r
Centripetal force• If the provided force >
then the object is circulating towards the centre.
r
vmFc
2
.
r
Centripetal force• If the provided force <
then the object is circulating away from the centre.
r
vmFc
2
.
r
Whirling a ball with a string in a horizontal circle
Top View
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
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
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
Whirling a ball with a string in a horizontal circle
What happens if Fc suddenly disappears?(e.g. the string breaks.)
v
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.
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
Centripetal force
r
vmFc
2
. or 2mrFc
Centripetal force
• Fc m
• Fc v2
Fc r
1
Uniform motion in a horizontal circle
R
v
Uniform motion in a horizontal circle
Uniform motion in a horizontal circle
Uniform motion in a horizontal circle
Uniform motion in a horizontal circle
Uniform motion in a horizontal circle
R
T
mg
The object isunder two forces, the tension Tand the weight mg.
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
.
Uniform motion in a horizontal circle
R
T
mgFc
cFgmT
.
T.cos = mgand
R
mvT
2
sin.
Uniform motion in a horizontal circle
R
T
mgFc
cFgmT
.
Rg
v2
tan
Uniform motion in a horizontal circle
R
T
mgFc
Rg
v2
tan
For a faster speed v, the angle θtends to increase.
Experiment
• To verify the equation for centripetal force
r
vmFc
2
. or 2mrFc
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
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
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
To verify the equation for centripetal force
The hanging weight is in equilibrium.T = Mg ----------- (1)
bob
hangingweightMg
T
T
mg
θ
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
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)
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)
To verify the equation for centripetal force
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
Also cosθ=
L
r(6)
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.
To verify the equation for centripetal force
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
Measure M and m before the experiment.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
Measure ω during the experiment.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
ω = Number of revolution × 2π÷ time
Verify the following equation.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
Problem: How to measure L?Refer to the textbook for the
skill.
bob
hangingweightMg
T
T
mg
r
L
Fc
θ
mL
Mg2
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.
Leaning on a vertical cylinder
Place an object on the innerwall of the cylinder. The cylinder starts to rotateabout its axis.
r
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
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
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
Leaning on a vertical cylinder
Solve the two equations.We have
r
g
s
N
f
W
ω
r
r
g
s min
and
Rounding a Bend
http://oldsci.eiu.edu/physics/DDavis/1150/05UCMGrav/Curve.html
Rounding a Bend
r
A car turns rounda corner.It is a circular motionwith a path of radiusof curvature r.
Rounding a Bend
r
A car turns rounda corner.It is a circular motionwith a path of radiusof curvature r.
v
Rounding a Bend
http://plabpc.csustan.edu/general/tutorials/CircularMotion/CentripetalAcceleration.htm
Rounding a Bend
r
It requires a centripetal forcefor the circularmotion.
v
Fc
Rounding a Bend
How comes the centripetal force?
r
v
Fc It may comefrom the frictionor the normalreaction.
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.
Level Road without Banking
r
v
Fc
vmax = grs
Note:vmax is independentof the mass of the car.
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.
Example 9
• To find the coefficient of static friction.
r
v
Fc
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
Banked Road
• Design a banked road which is inclined to the centre.
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
Banked Road• The centripetal force comes from
the normal contact force R.
R
W
Fc
θ
r
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
Banked Road• In ideal case, friction is not necessary.
The ideal banking angle is
R
W
Fc
θ
rg
v2
tan
r
Example 10
• Find the ideal banking angle of a road.
• The ideal speed is
rg
v2
tan
tan.rg
Banked Road• In non-ideal case, friction f is needed.
• Speed is too slow, less than the ideal speed.
R
W
Fc
θ
fr
Banked Road• In non-ideal case, friction f is needed.
• Speed is too fast, more than the ideal speed.
R
W
Fc
θ
fr
Railway
• When there is a bend, the railway is banked.
• This avoids having lateral force on the rail.
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
Aircraft• When an airplane moves in a horizontal
circular path in air, it must tilt about its own axis an angle θ.
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
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
Example 11
• Find the speed of the aircraft.
gr
v2
tan
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.
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
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
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
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
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
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
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
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
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
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
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
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
Tilt of a Car in Circular Motion
rf1
f2
R1 R2
W
G
L L
h
).( 2112 ffL
hRR
So R2 > R1
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
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.
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.
Uniform Motion in a Vertical Circle
r
mvFc
2
The centripetal force is
r
Fc
v
O
How comes the centripetal force?
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.
At the highest position
T1
mg
T1 + mg = Fc
andrFc
O
v
r
mvFc
2
r
mvmgT
2
1
At the highest position
T1
mg
rFc
O
v
r
mvmgT
2
1
mgr
mvT
2
1
At the highest position
T1
mg
rFc
O
v
mgr
mvT
2
1
What would happen
if v2 = ?r
m
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
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
At any other positions
The net forceis the centripetal force Fc.
Ov
Fc
r
θ
r
mvFc
2
At any other positions
Ov
T3
r
θ
mg.cosθ
θ
Along the radial direction,
Fc = T3 – mg.cosθ
At any other positions
Ov
T3
r
θ
mg.cosθ
θ
cos.
cos.
2
3
2
3
mgr
mvT
r
mvmgT
So
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
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.
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
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
Looping the loop
• The speed v of the marble changes on the
loop.
• The centripetal force changes on the loop.
v
O
r
vv
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
At the lowest position
O
r
vo
N1
mg
r
mvmgN o
2
1 mgr
mvN o
2
1
Below the centre
O
r
v2
N2
mg
r
mvmgN
22
2 cos. cos.22
2 mgr
mvN
θ
Below the centre
O
r
v2
N2
mg
θ
From conservation of energy, )cos1(222
2 grvv o
vo
h = r(1-cosθ)
Above the centre
O
r
v3
N3
mg
r
mvmgN
23
3 cos. cos.23
3 mgr
mvN
ψ
Above the centreFrom conservation of energy, )cos1(222
3 grvv o
O
r
v3
N3
mgψ
h=r(1+cosψ)vo
At the highest position
v4
O
rN4mg
r
mvmgN
24
4 mgr
mvN
24
4
At the highest positionFrom conservation of energy, grvv o 422
4
O
rN4mg
h = 2rvo
v4
Completing the Circle
• For the marble to reach the highest position,
024
4 mgr
mvN grv 4
and
grgrvv o 4224 grvo 5
Completing the Circle
• The marble cannot move up the loop and oscillates like a pendulum.
grvo 2
vo rg
vh o
2
2r
Completing the Circle
• The marble cannot move up the loop and oscillates like a pendulum.
grvo 2
vo rg
vh o
2
2r
Completing the Circle
• The marble rises up to height more than r and is projected away.
grvgr o 52
vo
rhr 2r
Completing the Circle
O
r
vo
grvo 5
Whirling freely with a rod
• The ball moves and passes its loop with its own initial energy.
vo
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
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
Whirling freely with a rod
min vo
2r
v
θ
h=r(1+cosθ)F mg
r
mvmgF
2
cos.
θ
)cos1(22 grv
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).
Changing from tension to compression
min vo
2r
vθo
h=r(1+cosθo)mg
θo
Prove that θo = 48.2o when F = 0.
Example 12
• Whirling a bucket of water in a vertical
circle.
Example 12
• Water does not flow out when the bucket is at the top position.
v
rmg
Centrifuge
• It is a device to
separate solid or
liquid particles of
different densities
by rotating
them in a tube in a
horizontal circle.
Centrifuge
r
axis ofrotation
ω
• ω is the angular velocity.• r is the distance of the small portion of liquid from the axisof rotation.
Centrifuge
r
axis ofrotation
ω
The small portion is in uniformcircular motion.The centripetal force comesfrom the pressure difference ΔP.
ΔP
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
Centrifuge
r
axis ofrotation
ω
As a result, this portion of smallerdensity moves towards the centralaxis.
ΔP
Centrifuge
r
axis ofrotation
ω
Portion of larger density moves away from the central axis.
ΔP
Centrifuge
• Study p.51 and 52 for the mathematical
deduction.