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Transcript
Circular motion
A particle P travels in circular path.
It performs circular motion about O with radius r.
s
w
P
A
If P travels in constant speed, the circular motion
is known as uniform.
1
Mathematics – angles
1 cm
1 cm
q
1 cm

Arc length = circumference x q /360o
1 = 2 p x q / 360o
q =180o / p ≈ 57o
2
Angle in radian (another unit for angle)
s
r
q
r




If s = r, then q is 1 radian.
1 radian = 180o / p
180o = p radian
360o = 2 p radian
3
q = s/r
s

Relation between arc length s, angle at
centre q and radius r is
s = rq
4
Describing circular motion

Consider a body moving uniformly from A to B in time t
so that it rotates through an angle q .
s
w
B
A
Angular displacement (q / rad)
* Anticlockwise as positive direction
Angular velocity (w / rad s-1)
= Angular displacement / time
q
* w= t
distance s
=
Speed of the body v =
time
t
By s = rq
rq
q
v=
= r = rw
t
t
5
Speed and angular speed

2m
A
1m

B

Consider carts A and B in a
mechanical game.
Do they have the same angular
speed?
Which of them moves faster?
w = 3 rad s-1
6
Describing circular motion
s
w
B
A
Period of the motion
= distance / speed
2pr 2pr 2p
T=
=
=
v
rw
w
Period of the motion
= angular displacement /
angular speed
T=
2p
w
7
Three useful expressions
(1) s = rq
(2) v = rw
(3) 180o = p radians
8
Example 1
Find the angular velocity in radian s-1if a motor makes 3000
revolutions per minute.

Solution:
1 revolution = 360o = 2p radians.
Angular velocity = 3000 x 2p / 60 = 314 rad s-1
9
Rotational motion and Translational
motion
a
w
q
q: angular displacement
 w: angular velocity
 a: angular acceleration
Equations of motion for uniform
acceleration:
w1 = w0 + at
v = u + at
q = w0t + ½ at2
s = ut + ½ at2
v2 –u2 = 2as
w12 – w02 = 2aq

10
Example




A turntable is rotating at 1 rev / s initially. A motor is
turned on such that the turntable rotates at an angular
acceleration of 0.3 radian s-2. Find the angular
displacement covered by the turntable for 5 s just after
the motor is turned on.
Solution:
w0 = 1 rev / s = 2p radian s-1 (∵360o = 2p radians)
a = 0.3 radian s-2
t=5s
By q = w0t + ½ at2
q = (2p)(5) + ½ (0.3)(5)2
= 35.2 radians
11
Centripetal acceleration


A body which travels equal distances in equal times
along a circular path has constant speed but not
constant velocity.
Since the direction of the velocity changes from time
to time, the body has acceleration.
12
Centripetal acceleration (Deriving a = v2/r)

Consider a body moving with constant speed v in a circle of
radius r. It travels from A to B in a short interval of time dt.
vB
vA
dv
w
B
dq
O
r
vBdq vA
A
Change in velocity = vB – vA = (– vA) + vB = dv
Since d t is small, dq will also be small, so dv = vdq .
But v = rw,
Therefore,
v2
a=
r
2
d v vdq
or
a
=
r
w
=
= vw
dt
dt
Since dq is small, dv will be perpendicular to vA and points towards the centre.
Hence, the direction of acceleration is towards the centre. i.e. centripetal.
13
Acceleration =
Centripetal force




Since a body moving in a circle (or a circular arc) is
accelerating, it follows from Newton’s first law of motion
that there must be a force acting on it to cause the
acceleration.
The direction of this force is also towards the centre,
therefore this force is called centripetal force.
By Newton’s second law, the magnitude of centripetal
force is
v2
F = ma = m = or = mrw 2
r
Since the centripetal force  displacement, NO work is
done by the centripetal force
⇒ K.E. of the body in uniform circular motion remains
unchanged.
14
A stone of mass 2 kg is tied by a string and moving in a
horizontal circular path of radius 0.5 m. Find the tension in
the string if the speed of the stone is 4 ms-1.

Solution:
Tension
=Centripetal force
= mv2/r
= (2)(4)2/0.5
= 64 N
Smooth table
15
Examples of circular motion

q

T
w
mg
A bob is tied to a string and whirled
wound in the horizontal circle.
In vertical direction:
T cos q = mg  (1)

In horizontal direction:
v2
T sin q = m  (2)
r
2
2
v
r
w
( 2)
or tan q =
 tan q =
rg
g
(1)
16
Rounding a bend

Case 1:
Without banking
r



f
When a car of mass m moving with speed v goes around a
circular bend with radius of curvature r, the centripetal force
v2
F =m
r
The frictional force between the tires and the road provides the
centripetal force.
v2
If limiting friction < m
, the car slips on the road and it
r
may cause an accident.
17
The coefficients of friction between the tires and the road in
rainy days and in sunny days are 0.4 and 0.6 respectively.
If a car goes around a circular bend without banking of
radius of curvature 50 m, find the maximum speeds of the
car without slipping in rainy days and sunny days.

Solution:
Limiting friction is attained to achieve maximum speed.
v2
By F = m
r 2
v
mg = m  v 2 = rg
r
Maximum speed without
slipping = rg
In rainy days,
Maximum speed without slipping = 0.4  50  10 = 14.1 ms-1
-1)
(50.9
km
h
In sunny days,
Maximum speed without slipping = 0.6  50  10 = 17.3 ms-1
(62.4 km h-1)
18
Rounding a bend
In order to travel round a
 Case 2 with banking
bend with a higher safety
R
speed, the road is designed
banked.
r
 The centripetal force does not
rely on friction, but provided
mg
by the horizontal component
q
of the normal reaction.
In vertical direction: R cos q = mg  (1)
In horizontal direction:
(2)
v2
: tan q =
(1)
gr

v2
R sin q = m (2)
r
The safety speed of rounding the bank is gr tan q
which can be increased by increasing the degree of banking of the road.
19
The figure above shows a car moving round a corner with a
radius of 8 m on a banked road of inclination 20o. At what
speed would there be no friction acting on the car along OA?

Solution:
R cos 20 = mg  (1)
2
v
In horizontal direction: R sin 20 = m (2)
8
2
(2)
v
: tan 20 =
 v = 5.4 ms-1
(1)
8g
In vertical direction:
Note: If the car travels at a higher speed say 6 ms-1 round the
above corner, some frictional force would act on the car to
prevent the car from slipping upward along the road provided
that the limiting friction is not exceeded i.e. f  R
20
Aircraft turning in flight
L
L q L cos q
L sin q
mg
straight level flight
mg
turning in flight
In straight level flight, the wings provide a lifting force L that
balances its weight mg. i.e. L = mg.
To make a turn at v with radius r, the flight banks and the
horizontal component of the lifting force (L sin q) provides the
centripetal force for circular motion.
21
Aircraft turning in flight
L q L cos q
L sin q
mg
turning in flight
To make a turn at v with radius r,
the flight banks and the horizontal
component of the lifting force
(L sin q) provides the centripetal force
for circular motion.
v2
L sin q = m (1)
r
The weight of the aircraft is supported by the vertical
component of L.
i.e. L cos q = mg (2)
22
Find the angle of inclination of the wings of an aircraft
which is traveling in a circular path of radius 2000 m at a
speed of 360 km h-1.
L Lcos q
Solution:
q
Speed of the aircraft
= 360 /3.6 ms-1 = 100 ms-1
Lsin q
Resolve horizontally,
L sin q = mv2 / r --- (1)
mg
Resolve vertically,
L cos q = mg --- (2)
(1) / (2):
tan q = v2 / rg
tan q = 1002 / (2000 x 10)
q = 26.6o
23
Cyclist rounding a corner

The frictional force
provides the centripetal
force.

By the law of friction, the
maximum speed of the
cyclist without slipping is
given by
C.G.
r
mg
f
R
v2
f = mg = m
r
 v = gr
However, f also has a moment about C.G which tends to turn
the rider outwards.
24
The rider must lean inwards so that the moment of f is
counterbalanced by the moment of R about C.G..


For no overturning of the
cyclist, C.G.
Take moment about C.G..:
r
fh = Ra  (1)

In horizontal direction:
v2
f = m  (3)
r
q
R
h
f
a
In vertical direction:
R = mg  (2)

C.G.
mg

Sub (2) and (3) into (1):
v2
m h = mga
r
a v2
=
∴
h gr
v2
tan q =
gr
2
v
1
For given v and r, the cyclist must bend at the correct angle tan ( )
gr
in order not to overturn.
25
The rotor
A mechanical game in amusement parks.
It consists of an upright drum, inside which
passengers stand with their backs against the wall.
f
The drum spins at increasing speed about its
central vertical axis. Hence, the centripetal force
required for circular motion also increases.
The centripetal force is provided by2the normal
reaction from the wall. i.e. R = m v
r
Therefore, the normal reaction increases as the
spinning speed increases.
R
mg
As normal reaction increases, the weight of the
passenger can be balanced by the frictional force. i.e. f = mg
When the floor is pulled downwards, the passenger will not fall but remains
stuck against the wall of the rotor.
26
It is given that the radius of the drum is 2 m and the
coefficient of static friction between clothing and the wall is
0.4 m. Find the minimum speed v of the passenger before
the floor is pulled downwards?

Solution:
v2
In horizontal direction: R = m (1)
f
r
In vertical direction: f = mg (2)
R
By the law of friction: f = R  (3)
Sub. (1) and (2) into (3):
mg
v2
gr
mg = m
 v2 =
r

The minimum speed required is
v=
10  2
= 7.07 ms-1
0.4
27
Looping the loop

At C, the centripetal force is provided by the normal reaction and
the weight of the object.
v2
R  mg = m
r


If the object does not leave the
track, R  0.
The minimum speed at C to just
complete the loop is given by
v2
mg = m
r

C
v
R mg
D
 v = gr (1)
r
u
B
A
By conservation of energy, the minimum speed at A for the object to
complete the loop is given by
1
1
mu 2 = mv 2  mgh  u 2 = v 2  4 gr (2)
2
2
Sub. (1) into (2): u 2 = gr  4 gr = 5gr  u = 5gr
28
v
C
R mg
D
r
u
B
Notice that:
If the ball passes point A with speed  5gr , the ball will fall from the track.
If the ball passes point A with speed  5gr , the ball will complete the loop
and the remaining centripetal force is provided by the normal reaction.
29
Weightlessness

Weightlessness means the weight of an object is equal
to zero. This happens at a place where there is no
gravitational field (g = 0).

We are aware of our weight because the ground exerts
an upward push (normal reaction) on us.

If our feet are completely unsupported, for example, in a
free falling lift, we experience the sensation of
‘weightlessness’.

An astronaut orbiting the earth in a space vehicle with its
rocket motors off also experiences the sensation of
‘weightlessness’.
30
An astronaut orbiting the earth


An astronaut orbiting the
earth in a space vehicle with
its rocket motors off also
experiences the sensation of
‘weightlessness’. Why?
The weight of the astronaut
provides the centripetal force
and the walls of the vehicle
exert no force on him.
i.e.
R=0
v
mg
r
v2
v2
mg = m  g =
r
r
Note: g < 10 ms-2 because the astronaut is now far away
from the surface of the earth.
31
Summary


Weightlessness
Experience the sensation of
‘weightlessness’
W=0
W > 0 but R = 0
At a place where there is
no gravitational field
(g = 0).



In a free falling lift.
In a space vehicle with its
rockets off orbiting the earth.
Normal reaction R = 0
32
Centrifuge (離心機)

Centrifuges separate solids suspended in liquids, or liquids
of different densities.

The mixture is in a tube, and when it is rotated at high
speed in a horizontal circle, the less dense matter moves
towards the centre of rotation.

On stopping the rotation, the tube returns to the vertical
position with the less dense matter at the top. Cream is
separated from milk in this way.
33
Working principle of a centrifuge

Consider the part of liquid between A and
B inside the tube.

The pressure at B is greater than that at A.
This provides the necessary centripetal
force acting inwards.

For this part of liquid, the force due to
pressure difference supplies exactly the
centripetal force required.

If this part of liquid is replaced by matter
of small density or mass, the centripetal
force is too large and the matter is pushed
inwards.

On stopping the rotation, the tube returns
to the vertical position with the less dense
matter at the top.
pA
pB
A
B
pB > pA
34
H.W. Chapter 2 (1, 2, 3, 5,
6(a)(b)(c) )
Due date: 27/11 Day 1

1.
A particle moves in a semicircular path AB of radius
5.0 m with constant speed 11 ms-1. Calculate
(a) the time taken to travel from A to B (p = 22/7)
(b) the average velocity,
(c) average acceleration
11 ms-1
A
5m
5m
B
35

2
A turntable of a record player makes 33 revolutions
per minutes. Calculate
(a)
its angular velocity in rad s-1,
(b)
the linear velocity of a point 0.12 m from the
centre.
33 rev / min
v
180o = p radians
v= rw
36

3
What is meant by a centripetal force? Why does
such a force do no work in a circular orbit?
(a) An object of mass 0.5 kg on the end of a string is
whirled round a horizontal circle of radius 2.0 m with
a constant speed of 10 ms-1. Find it angular velocity
and the tension in the string.
(b) If the same object is now whirled in a vertical circle
of the same radius with the same speed, what are
the maximum and minimum tensions in the string?
A
T = mv2/r
T + mg = mv2/r
T – mg = mv2/r
B
37
Referring to the above diagram, find the normal reaction
acting on the ball at A, B and C if the speed of the ball at A
is 6 gr
vc C





When the ball is at A,
Centripetal force:
RA - mg = mu2/r
RA = mg + m(6gr)/r = 7mg
When the ball is at B,
By conservation of energy,
½ mu2 = ½ mvB2 + mgr
vB2 = u2 – 2gr = 6gr – 2gr = 4gr
Centripetal force = RB = mvB2/r
RB = m(4gr)/r = 4mg
R mg
vB
RB
D
RA
B
r mg
u
A mg



When the ball is at C,
By conservation of energy,
½ mu2 = ½ mvc2 + mg(2r)
vc2 = u2 – 4gr = 6gr – 4gr = 2gr
Centripetal force:
RC + mg = mvB2/r
RC = m(2gr)/r – mg = mg
38