Circular motion

1
Circular motion
A particle P travels in circular path. It
performs circular motion about O with radius r.
If P travels in constant speed, the circular
motion is known as uniform.
2
Mathematics angles

• Arc length circumference x q /360o
• 1 2 p x q / 360o
• q 180o / p 57o

3
Angle in radian (another unit for angle)

• If s r, then q is 1 radian.
• 1 radian 180o / p
• 180o p radian
• 360o 2 p radian

4

• Relation between arc length s, angle at centre q
  and radius r is

s rq
5
Describing circular motion

• Consider a body moving uniformly from A to B in
  time t so that it rotates through an angle q .

• Angular displacement (q / rad)
• Anticlockwise as positive direction
• Angular velocity (w / rad s-1)
• Angular displacement / time

Speed of the body

By s rq
6
Speed and angular speed

• Consider carts A and B in a mechanical game.
• Do they have the same angular speed?
• Which of them moves faster?

2 m
A
1 m
B
w 3 rad s-1
7
Describing circular motion
Period of the motion

• distance / speed

Period of the motion
angular displacement / angular speed

8
Three useful expressions

• (1) s rq
• (2) v rw
• (3) 180o p radians

9
Example 1Find the angular velocity in radian
s-1if a motor makes 3000 revolutions per minute.

• Solution1 revolution 360o 2p
  radians.Angular velocity 3000 x 2p / 60 314
  rad s-1

10
Rotational motion and Translational motion
a

• q angular displacement
• w angular velocity
• a angular acceleration
• Equations of motion for uniform acceleration
• w1 w0 at
• q w0t ½ at2
• w12 w02 2aq

w
q
v u at s ut ½ at2 v2 u2 2as
11
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 5 s
• By q w0t ½ at2
• q (2p)(5) ½ (0.3)(5)2
• 35.2 radians

12
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.

13
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.

B
w
dq
A
O
r
But v rw, Therefore,
Change in velocity vB vA ( vA) vB dv
Since d t is small, dq will also be small, so
dv vdq . Acceleration
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.
14
Centripetal force

• Since a body moving in a circle (or a circular
  arc) is accelerating, it follows from Newtons
  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 Newtons second law, the magnitude of
  centripetal force is

• Since the centripetal force ? displacement, NO
  work is done by the centripetal force
• ? K.E. of the body in uniform circular motion
  remains unchanged.

15
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

16
Examples of circular motion

• A bob is tied to a string and whirled wound in
  the horizontal circle.
• In vertical direction

• In horizontal direction

17
Rounding a bend

• Case 1 Without banking
• When a car of mass m moving with speed v goes
  around a circular bend with radius of curvature
  r, the centripetal force

• The frictional force between the tires and the
  road provides the centripetal force.

18
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.

In rainy days, Maximum speed without slipping
ms-1 (50.9 km h-1)
In sunny days, Maximum speed without slipping
ms-1 (62.4 km h-1)
19
Rounding a bend

• In order to travel round a bend with a higher
  safety speed, the road is designed banked.
• The centripetal force does not rely on friction,
  but provided by the horizontal component of the
  normal reaction.

• Case 2 with banking

q
In vertical direction
In horizontal direction
The safety speed of rounding the bank is
which can be increased by increasing the degree
of banking of the road.
20
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

In vertical direction
In horizontal direction
.
21
Aircraft turning in flight
straight level flight
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.
22
Aircraft turning in flight
L cos q
L
q
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.
mg
L sin q
turning in flight
The weight of the aircraft is supported by the
vertical component of L. i.e.
23
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.
Solution Speed of the aircraft 360 /3.6 ms-1
100 ms-1 Resolve horizontally, L sin q mv2 /
r --- (1) Resolve vertically, L cos q mg ---
(2)

• (1) / (2)
• tan q v2 / rg
• tan q 1002 / (2000 x 10)q 26.6o

24
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

However, f also has a moment about C.G which
tends to turn the rider outwards.
25
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..

• In vertical direction

• Sub (2) and (3) into (1)

• In horizontal direction

26
The rotor

• A mechanical game in amusement parks.

• It consists of an upright drum, inside which
  passengers stand with their backs against the
  wall.
• 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 by the normal
  reaction from the wall. i.e.

• Therefore, the normal reaction increases as the
  spinning speed increases.
• As normal reaction increases, the weight of the
  passenger can be balanced by the frictional
  force. i.e.

When the floor is pulled downwards, the passenger
will not fall but remains stuck against the wall
of the rotor.
27
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

In horizontal direction
In vertical direction
By the law of friction
Sub. (1) and (2) into (3)
The minimum speed required is
7.07 ms-1
28
Looping the loop

• At C, the centripetal force is provided by the
  normal reaction and the weight of the object.

• If the object does not leave the track, R ? 0.
• The minimum speed at C to just complete the loop
  is given by

A

• By conservation of energy, the minimum speed at A
  for the object to complete the loop is given by

Sub. (1) into (2)
29
Notice that
and the remaining centripetal force is provided
by the normal reaction.
30
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.

31
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.

v
mg
r
Note g lt 10 ms-2 because the astronaut is now
far away from the surface of the earth.
32
Summary
Weightlessness Experience the sensation of weightlessness
W 0 W gt 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
33
Centrifuge (???)

• 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.

34
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.

35
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

36

• 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
37

• 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 mg mv2/r
T mg mv2/r
T mv2/r
B
38
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
C

• When the ball is at A,

• Centripetal force
• RA - mg mu2/r
• RA mg m(6gr)/r 7mg

B
D
r
u

• When the ball is at B,

A

• 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

• 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