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




• The direction of motion changes with time.

r




• The direction of motion changes with time.vv

vv

r


• Period T is the time needed to complete one cycle.

• Frequency f is the number of cycles completed in one second.

r

Tf

1


• Example 1


• 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.


• After one complete cycle,

angular displacement

θ= 2π


• 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


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


r

AB

Δθ

O X

vA

vB


)(. AB vvta

Bv

Av

ta .

|vB| = |vA| = v


r

AB

Δθ

O X

vA

vB


Bv

Av

ta .

Δθ

Note that the triangle is an isosceles triangle.

|vB| = |vA| = v


r

AB

Δθ

O X

vA

vB


Bv

Av

ta .

Δθ

|vB| = |vA| = v

v. Δθ= a. Δt

a = t

v

.


r

AB

Δθ

O X

vA

vB

Bv

Av

ta .

Δθ

|vB| = |vA| = v

r

vaor

ra

vt

va

2

2.

..


r

AB

Δθ

O X

vA

vB

Bv

Av

ta .

Δθ

|vB| = |vA| = v

The acceleration is pointing to the centre of the circle.


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


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


• 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


Top View

There is force Fc acting on the ball alongthe string.There is force F acting at the centre alongthe string.

FcF v


Top View

The two forces Fc and F are action and reaction pair.

Fc

F

v

FcF

v


Top View

What happens if Fc suddenly disappears?(e.g. the string breaks.)

Fc

F

v

FcF

v


What happens if Fc suddenly disappears?(e.g. the string breaks.)

v


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






R

T

mg

The object isunder two forces, the tension Tand the weight mg.


R

T

mg

The net force on the object isthe centripetal force because the object ismoving in a circle.

Fc

cFgmT

.


R

T

mgFc

cFgmT

.

T.cos = mgand

R

mvT

2

sin.


R

T

mgFc

cFgmT

.

Rg

v2

tan


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


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


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


The hanging weight is in equilibrium.T = Mg ----------- (1)

bob

hangingweightMg

T

T

mg

θ


The bob is in circular motion with angular velocity ω.Fc = m.r. ω2 ----------- (2)

bob

hangingweightMg

T

T

mg

r

Lθ


gmTFc

bob

hangingweightMg

T

T

mg

r

L

Fc

θ

The net force on the bob is equal to the centripetal force.

(3)


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)


bob

hangingweightMg

T

T

mg

r

L

Fc

θ

Also cosθ=

L

r(6)


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.


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


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


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


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

http://oldsce.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.



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.



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.


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


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


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


R = mg ------- (1) ----------- (2)

vertical

horizontalW

R

fs

G

r

mvf s

2

R balances W.fs is the centripetalforce.

h


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


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




rf1

f2

R1 R2W acts atthe centre of massG of the car.

W

G




rf1

f2

R1 R2We are goingto compareR1 and R2.

W

G


Let 2L be the separation betweenthe left and right tyres.r

f1f2

R1 R2

W

G

L L


Let h be theheight of the centre of mass G from the ground.

rf1

f2

R1 R2

W

G

L L

h


Without toppling,the moment aboutG must be zero.About G,clockwise moments=anticlockwise moments

rf1

f2

R1 R2

W

G

L L

h


rf1

f2

R1 R2

W

G

L L

h

).( 2112 ffL

hRR

So R2 > R1


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.


• 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.


r

mvFc

2

The centripetal force is

r

Fc

v

O

How comes the centripetal force?


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


T1

mg

rFc

O

v

r

mvmgT

2

1

mgr

mvT

2

1


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


The net forceis the centripetal force Fc.

Ov

Fc

r

θ

r

mvFc

2


Ov

T3

r

θ

mg.cosθ

θ

Along the radial direction,

Fc = T3 – mg.cosθ


Ov

T3

r

θ

mg.cosθ

θ

cos.

cos.

2

3

2

3

mgr

mvT

r

mvmgT

So


• 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


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


• The marble cannot move up the loop and oscillates like a pendulum.

grvo 2

vo rg

vh o

2

2r


• The marble cannot move up the loop and oscillates like a pendulum.

grvo 2

vo rg

vh o

2

2r


• The marble rises up to height more than r and is projected away.

grvgr o 52

vo

rhr 2r


O

r

vo

grvo 5

Whirling freely with a rod

• The ball moves and passes its loop with its own initial energy.

vo


• A light rod would not be loosen.

• The light rod can provide tension or compression depending on the case.

vo


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


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.

Documents

Physics Beyond 2000 Chapter 3 Circular Motion