Module_SP015

• In real situation where friction (non conservative force) acts on
the system in a direction opposite to the direction of motion, the
initial mechanical energy of the system equals to the (reduced)
final mechanical energy of the system plus the heat energy due
to work done by friction.

Initial kinetic work done by Final
energy constant friction force kinetic
energy

Initial final
potential potential

energy energy

Non Face-to-face SLT

Problem solving : Conservation of Energy

1) Draw a picture of the physical situation.

2) Determine the system for which energy will be conserved: the object or objects and the forces

acting.

3) Ask yourself what quantity you are looking for and decide what are the initial and final position.

4) If the object under investigation changes its height during the problem, then choose a reference

frame with a convenient y=0 level for gravitational potential energy; the lowest point in the problem

is often a good choice.

If springs are involved, choose the unstretched spring position to be x (or y) = 0

5) Apply conservation of energy,
If no friction acts, then use Ki  Ui  K f  U f
If friction is present, then use Ki Ui W  Kfriction U f
f
What energy you should include at a position given ? Think :

Any height? 

Any motion? 

Any extension or compression of spring 
6) Use the equation(s) you develop to solve for the unknown quantity.

Example : How to apply Conservation of Energy in a problem given

A roller-coaster car shown in the figure is pulled up to point 1 where it is released Non Face-to-face SLT

from rest. Assuming no friction, calculate the speed at points 2 and 3.

Point 1 there is 32 m = Point 2 is the lowest point, no gravitational PE but At point 3, there is
gravitational PE due due to the speed of the car, it has kinetic energy. . gravitational PE
to height ℎ . Since due to height ℎ ,
the car released 3 and also kinetic
from rest, there is no energy due to the
kinetic energy speed at point 3.
possesses by the car.

=14 m

    Treat point 1 as     Treat point 1 as
initial, point 2 as initial, point 3 as
final final
       




Cancelling the
common m

m s−1 m s−1

5.3 Power

Definition of Power o is a scalar quantity.

is defined as the rate o SI unit of the power is
at which work is
done. Watt (W) or J s−1
can also be described
as rate at which o For practical purposes a
energy is transformed larger unit is often used ,
from one type to
another. horsepower (hp)

1 hp = 746 W

o Power tells you how fast
work is being done or
energy is transferred.

o The rate of which work is
done might not be
constant.

Average power, Instantaneous power,

Is defined to be The rate at which work is
done might not be constant.
Work done during We defined Instantaneous
the time interval power as

Duration of time
taken to do the
work

Since work done in a process
involves the transformation of
energy, power can also be defined
as the rate at which energy is
being transformed.

In mechanics, instantaneous power Since
can also express in terms of force
and velocity and
Consider a constant force of
magnitude acting while an object cos
moves through a parallel
displacement of magnitude, in time where power of a force can
interval of : be written in terms of
the force and the
 velocity:

Therefore the instantaneous power: : magnitude of force
: magnitude of velocity
: the angle between and

Topic 6 Circular Motion

6.1 Uniform Circular Motion
6.2 Centripetal Force,

MODE Face to face Non Face to face
Lecture SLT SLT
Tutorial 1
1
2
2

1

Learning Outcomes

At the end of this topic, students should be able to:

6.1 Uniform circular motion

a) Describe uniform circular motion

(Lecture : C2,PLO1, MQF LOD1)

b) Convert unit between degrees, radian and revolution or rotation.

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

6.2 Centripetal force

a) Define centripetal acceleration

(Lecture : C2,PLO1, MQF LOD1)

b) Solve problems related to centripetal force for uniform circular motion cases :
horizontal circular motion, vertical motion and conical pendulum (exclude banked
curve)

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

Circular motion is a motion in which an object move in a

circular path or rotate around something.

Circular motion Our focus for this ● ●●
topic
Motion of object moving ● ●
Uniform
in a circular path at ●●●
circular motion
constant speed.
Non uniform
Motion of object moving ● ●
circular motion
●

in a circular path with ● ●
varying speed.
●●●

6.1 Uniform Circular Motion (UCM)

Meaning of UCM:

A motion in with a constant but the direction
speed of velocity is
circular path
● continually
●● changing with
●●
●● time.

●

Terms Angular 4) the linear distance
used displacement,
(arc) is related to the
to Linear as follows:
describe distance,
uniform Angular fixed reference line
circular velocity,
1) Angle undergone by an object from a fixed
motion Linear reference point is known as angular
velocity, displacement,

Frequency, rev rad

Period, 2) SI unit for is radian (rad)
3) = 1 revolution (rev) = rad

5) angular velocity, is 9) Period, is defined as

defined as the rate of time taken to complete one
change of angular
displacement revolution.





10) SI unit for is s

6) SI unit for is rad s‒1 11) Frequency, is defined as

7) Other units used : rps (rev number of revolutions per
per second), rpm (rev
per minute). second.

8) Vector quantity 12) SI unit for is Hz

13) An object moving in uniform circular motion cover the same linear distance, in each second
of time, thus:

Linear velocity,

14) is directed tangentially to the circular path and always perpendicular to the radius of

circular path as shown in figure.

10 m s

10 m s

10 m s

6.2 Centripetal force,

6.2.1 Centripetal acceleration,

Definition Equation Direction

is defined as Since : o Points toward the
centre of the circle.
acceleration of an
object moving in circular (Centripetal means
path whose direction is “centre-seeking”)
towards the centre of
o Perpendicular to
the circular path and
linear velocity.
whose magnitude is
equal to the square of
the speed divided by
the radius.



In UCM, an object move at a constant speed, has an acceleration.

Why object has centripetal acceleration, ?

As object moves in Change in
circle with constant velocity is
towards the
speed, its direction center.
is continually Therefore the
changing with acceleration
time. is towards
the centre.
velocity being a

vector quantity, has a

constant magnitude but
a changing direction
thus there is a

change in velocity
and acceleration
occurs.

6.2.2 Centripetal force,

a) According to Newton’s second law, an d) This net force is in the same direction
object that is accelerating must have
net force acting on   it. as centripetal acceleration thus is known

as Centripetal Force,

 

b) Therefore, an object moving in a circle

must have a force applied to it to keep it

moving in that circle. That is, a net

force is necessary to give it

centripetal acceleration.

c) Applying Newton’s second law to circular

motion:  

 

(toward centre of circle)

e) Centripetal force, is defined as f) The centripetal force is not a new kind
net force acting on a body of mass , of force. The term (centripetal) merely
causing it to move in circular
path of radius, with constant speed, , describes the direction of the net force is
in which the direction of the centripetal directed towards the centre of circular path.
force is towards the centre of circular
g) Any force or components that act
path. along the radius, towards or way
from the centre of circle can be
Since : centripetal force:

i. Friction (example car moving in
circular path)

ii. Tension (example swinging a ball at
the end of a string)

iii. Gravitational force (planets and
moons)

iv. Normal Force

Video: Centripetal Force
Introduction and Demonstration

If there is no more centripetal force If the string is being cut or snaps,
acting on the object, what will
happen to the object? centripetal force suddenly
disappears, then the circular motion stops
Is there any work done by and the object continues its motion in a
centripetal force ? straight path in the direction of
which is tangential to the
circle.(Figure (b)).

is perpendicular
(90°) to the
direction of the
object’s velocity.

No work is done
by the centripetal
force, because

no displacement in
the direction of the
force.

Problem Solving Strategy
for applications of Uniform Circular Motion (UCM)

1. Draw a free body diagram (FBD), showing ALL forces acting on each object under

consideration.

2. Choose a set of coordinate axes. (set radial of the circle as one of the axis)

3. Resolve the forces (if any)

4. Identify which of the force(s) or which of their component(s) point toward or

away from the center of the circle.

5. Equating summation of t  hese force(s) w  ith centripetal force.

Tips : Horizontal circle : ;
;   
Vertical circle :   



  

Uniform Circular Motion: Horizontal circle

Case : object revolving in a horizontal circle with constant speed.


FBD





1. 2 forces acting on the object : 3. Consider component:

a) Weight,
 

b) Tension,

2. Tension in string, is the only  
component in the radial direction
that provides the centripetal force.

Uniform Circular Motion: Horizontal circle

Case : Motion of car round a flat curve Along the horizontal
(radial) axis ( ),
FBD

Along friction force,
vertical between the tires
axis ( ), and the road, points
toward the center of
normal
the circle provides
force
the centripetal
balances
the car’s ( ) = force.

weight. The maximum speed at
Center of the curve = which the car can

( ) =

=  
    negotiate the curve is :

 
   
Note, this does not
depends on the mass of
the car.

Uniform Circular Motion: Horizontal circle

Case : Conical Pendulum

o The object is in equilibrium in vertical direction and undergoes uniform
circular motion in horizontal direction.

FBD  
 

 
 

θ o Speed of the object is:
=
 

 

Horizontal component of tension =
( ) provides the centripetal
force. =

: length of string Note, is independent of mass,
: ∡ string makes with vertical
: radius of circle

Uniform Circular Motion: Vertical circle

Case : object revolving in vertical circle

When a body moves in a vertical At the top (point 1):
circle at the end of a string, the
Both and point towards
tension in the string varies
with the body’s position. center. FBD

 

①
 
At the bottom (point 2):
point towards center and
point away from center.

  FBD


 

② Note : Tension is maximum at
bottom and minimum at the
top of the circle.



Uniform Circular Motion: Vertical circle

Case : Roller coaster / go around Loop to loop

When a body moves loop and loop ① FBD ① At the top ( ):
in vertical circle, the normal reaction ① Both and
point towards
force of the body varies with the center.
body’s position.
 
②
At the bottom ( ): ②
point towards center and A jet executing a vertical loop  
point away from center.

  = −

Note: Normal reaction
  force also known as
② apparent weight of the

A roller coaster traveling around a circular track body
= +

Why don’t stone fall down as it goes Why don’t you fall out a roller coaster when it goes
upside down at top? upside down ?

AAAAAhhhhhh
hhhhhhhhhhh
hhhh!!!!!!!!!!!!!

Concept: If the Concept : If fall down,
string is sagging, lost contact with the
track, =
=

There is a minimum speed at top of its arc so that There is a minimum speed at top so that object

object continue moving in a circle. maintain contact with the track at the top.


= − = −

= −   = −


Uniform Circular Motion: Vertical circle

Case : Ferris wheel

When a person rides on ferris wheel, the normal reaction force varies with the body’s

position. FBD At the top ( ):

At the bottom ( ): Top point towards center
① ①
point towards center and point away from
center.
and point away from
center.  

 
 

 

②
= + bottom = −


②

rider feels heavier Note: Normal reaction force also known as apparent rider feels lighter
weight of the body

LEARN MORE THROUGH VIDEOS

(Scan QR codes given, WATCH the videos
before attending tutorial class)

There is a maximum speed at top so that Video 1: Car over Video 2: Car turn Video 3: object on a
rider will not leave the top of the track: a hill a flat corner turntable

  Video 4: conical Video 5: Rotor Video 6:
pendulum ride Ferris wheel

Video 7: Spinning Non Face-to-face SLT
Refer equation at top (position ①): bucket
Video 8: min speed for Water in a Bucket
Revolving in a Vertical Circle
= −

If the speed is greater that this ,
would have to be negative.
This is impossible, so the rider would leave
the seat.

TOPIC 7
GRAVITATION

7.1 Gravitational Force and Field Strength
7.2 Gravitational Potential Energy
7.3 Satellite Motion in a Circular Orbit

MODE Face to face Non Face to face
Lecture SLT SLT
Tutorial 1
1
4
4
Prepared by ChongYL/KMJ/session 20192020

Learning outcomes

At the end of this topic, students should be able to:

7.1 Gravitational Force and Field Strength b) Use the gravitational potential energy,

a) State and use the Newton’s law of U   GMm
r
gravitation, F  G Mm
r2 (Tutorial : C4, PLO4, CTPS3, MQF LOD6)

7.3 Satellite motion in a circular orbit

b) Define and use the gravitational field a) Derive and use escape velocity,

strength,  G M vesc  2GM  2gR (Lecture : C2,PLO1, MQF LOD1)
r2 R
ag

(Lecture : C2,PLO1, MQF LOD1) (Tutorial : C4, PLO4, CTPS3, MQF LOD6)

(Tutorial : C4, PLO4, CTPS3, MQF LOD6) b) Derive and use equation for satellite motion.

7.2 Gravitational potential energy (i) velocity, (ii) Period,

a) Define gravitational potential energy. v  GM T  2 r3
r GM
(Lecture : C2,PLO1, MQF LOD1)

(Lecture : C2,PLO1, MQF LOD1)
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

7.1 Gravitational Force ( ) and Field Strength ( )

7.1.1 Gravitational Force ( ) Newton’s Law of Gravitation

o Any 2 bodies with masses can
attract each other. This universal effect is
known as Gravitation. is measured from center to center




Gravitational force Any two particles attract each other through gravitational force.
exerted on satellite Even if the particles have very different masses, the gravitational
by Earth forces they exert on each other are equal in magnitude (strength)

Gravitational force states that :
exerted on Earth
by satellite Every particle in the universe attracts
every other particle with a force,
o The force with which one body attracts the
which is directly proportional to the
other due to their masses is known as product of their masses ( × ) and

Gravitational force, inversely proportional to the square
distance ( ) between their centers.

OR mathematically 1
2
∝ and ∝

∝ 1 21 12 2
2

= = = =
2

12 ∶ gravitational force by particle 1 on particle 2

where 21 ∶ gravitational force by particle 2 on particle 1

F : Gravitational force o m1 attracts m2 with a gravitational force
which is equal in magnitude but
G : Universal Gravitational constant
( 6.67×10‒11 N m2 kg‒2 ) opposite in direction to the force of

m, M : masses of the two particles attraction of m2 on m1.

r : distance between 2 particles o The gravitational forces between 2
particles form an action – reaction
o This gravitational force, F acts along the pair ( Newton’s 3rd law of motion)
line joining the center of mass of

the bodies.

How Gravitational force, varies with ? Example (TLO 7.1 (b))
(N)
What must the separation be between a 5.2 kg
Force is inversely proportional to . particle and a 2.4 kg particle for their
Doubling the distance between the gravitational attraction to have a magnitude of
2.3×10–12N?
masses causes the force to decrease

4 by a factor of 4
2
9 =

(m) 2.3 × 10−12 = 6.67 × 10−11(5.2)(2.4)
2
0 2 3
2 = 361.92
(N)
= . m

Gradient = 1 (m)
2
0

7.1.2 Gravitational Field Strength ( ) Gravitational Field Strength at a
point ( ) is defined as gravitational
o A gravitational field represents a region
where a body (test mass) experiences force per unit mass of a body (test
gravitational force.
mass) place at a point.

Test mass  FOR : gravitational field strength
a  mg : gravitational force
: mass (test mass)

o SI Unit of : N kg–1 or m s–2
o vector quantity.

o It is also known as gravitational
acceleration (free-fall acceleration).

o The body (test mass) can be an atom, Gravitational field strength on earth’s surface = 9.81 N kg‒1
an object or even a planet. (produce acceleration of 9.81 m s‒2)

o Its direction is always in the same
direction as gravitational force.

When a test mass is placed at ag  GM
point X in space around the mass r2

it experiences the existance of the

field in the form of gravitational

force.

M ag GM X where
r2
 G : Universal Gravitational constant

Planet of mass (6.67×10‒11 N m2 kg‒2)
creates a
is measured from center of planet M : mass of the planet
gravitational
r : distance between center of planet and the
field in space
point
around it.

o Refer fig. above, another formula for the o Refer to eq. above,
 is directly proportional to the mass
gravitational field strength at a of planet, . Planet of greater mass have
point (e.g point X) is given by greater value of .

ag  F and F  GMm  is inversely proportional to distance
m r2 . As increases, decreases.

ag  1  GMm 
m  r2 

Graph against

varies with altitude (height) I’m feeling earth
gravity, it is
and depth from the surface of a getting strong as I
fall towards earth
planet. Let’s take earth as
example:

Explanation

① < : ∝

(N kg−1) ② = : = = . −

③ > :
GM
ag  R2  9.81 = , ∝


As ↑, ↓

2 3 (m)



0

7.2 Gravitational Potential Energy (U)

Gravitational potential energy, U o Formula

o Gravitational potential energy, stored in = −
a test mass placed at a distance from

the center of planet is defined as the where

work done by gravitational force G : Universal Gravitational constant

in bringing a test mass from infinity to (6.67×10‒11 N m2 kg‒2)

that point. M : mass of the planet
: mass of the test mass placed at a point
Final Initial
position position r : distance between center of planet and the

point

o Scalar quantity

Planet Test o SI unit : Joule (J)
(point mass
mass) o chosen to be zero when the body of mass
is infinitely far from the earth ( = ∞) , thus

potential energy is never positive.

Graph against Example (TLO 7.2 (b))

Graph below shows variation of gravitational 1000 kg is measured
ℎ = 9.4× 105 m
potential energy, when the test mass, from center of

move away from the planet’s surface. planet
= + ℎ

= 6.38106 m


U(J) 1 = 5.981024 kg
0
∝ at infinity is zero. satellite’s gravitational potential

(m) due to the Earth is

= − = −
( + )

is always negative, = − , × − ( . × )( )
( . × + . × )
but it becomes less
= − = − . × J
negative with increasing
distance .

7.3 Satellite Motion in a Circular Orbit

7.3.1 Escape velocity ( ) Derivation of escape velocity

Definition o The gravitational potential energy of an object
of mass m on the surface of the planet is :
o Escape velocity is defined as a
UR   GMm
minimum velocity required by a R
mass to escape completely
o For the object to escape completely from the
from gravitational field to infinity in gravitational pull of the planet, its final distance
outer space. would be infinity ( ∞ ) and the final gravitational
potential energy is :

U  0
Formula
o If sufficient energy in form of kinetic
= = energy is supplied to the object during
launching, then the object can be escaped
Radius of planet, from the planet.

Mass of planet, KE  U Usupplied  R

1 mv2  0  ( GMm) o The escape velocity from surface of a planet
2R
depends on
2GM where : mass of planet  Mass of the planet
R : radius of planet  Radius of the planet or
 gravitational acceleration of the planet

vesc 

since ag  g  GM ; thus we get : GM  gR2
R2

 vesc 
2GM  2 gR2
R R

vesc  2gR

where : gravitational acceleration or
gravitational field strength at the
surface of the planet

: radius of planet

7.3.2 Tangential (linear/orbital) velocity of a satellite ( )

o Consider a satellite of mass o Gravitational force acting on satellite :
traveling round a planet of mass ,
and radius in a circular orbit of =
radius . 2

o This gravitational force provided

the centripetal force that required

by the satellite to travel in a circular

path. =

= 2
2

o Let the tangential linear speed of the =
satellite be .

where 7.3.3 Period of revolution of
satellite ( )
G : Universal Gravitational constant
o Period, of the satellite orbits around the
(6.67×10‒11 N m2 kg‒2)
planet is given by:
M : mass of the planet
r : distance between center of planet to the satellite = 2 and =

o satellite velocity depends on the radius of 2 2
= =
its orbit.
4 2 2
o satellite
2 =
velocity NOT
depends on
the mass of
satellite. = 2 3 ∝ 3
Larger orbits
1 > 2 → 1 < 2 correspond to longer
Satellite in smaller periods
orbital radius moves
faster.

Check-point 1 (TLO 7.1) Check-point 2 (TLO 7.1) Check-point 4 (TLO 7.2)

The gravitational force Above the surface of the The gravitational potential
acting on the Earth by a earth if we go to a energy can have positive
stone is much smaller distance equal to double value.
than that acting on the the radius of the earth,
stone by the Earth. the value of will (True / False)
become?
(True / False) A. one-ninth Check-point 5 (TLO 7.3)
B. one-third
C. one-fourth In defining orbital velocity of
D. one-half satellite, thing which is
unimportant is
Check-point 3 (TLO 7.1)
A. mass of satellite
The gravitational field B. mass of Earth
vector always point C. gravitational force
towards the center of the D. radius of Earth

planet. (True / False)

Types of satellite Extra note for better understanding

Natural satellite Artificial satellite

Any celestial body in space that orbits An object that people have made and
around a larger body. launched into orbit using rockets.

e.g : Earth is satellite of the Sun Different satellites serve different
purposes.
Moon is a satellite of the Earth

Synchronous (Geostationary) satellite 3. It moves directly above the Extra note for better understanding
equator.
o Figure below shows a synchronous
(geostationary) satellite which stays 4. The centre of a synchronous
above the same point on the equator satellite orbit is the centre of the
of the Earth. Earth.

o It purposes are for communication,
coordination, TV broadcasting and
earth observation

o The satellite have the following An observer at any
characteristics: place where the
satellite is visible will
1. It revolves in the same direction as always see it in exactly
the Earth. the same spot in the
sky.
2. It rotates with the same period of
rotation as that of the Earth (24
hours).

TOPIC 8
Rotational of rigid body

8.1 Rotational kinematics
8.2 Equilibrium of a uniform rigid body
8.3 Rotational dynamics
8.4 Conservation of angular momentum

MODE Face to face Non Face to face
Lecture SLT SLT
Tutorial 1
1
7
7

Prepared by ChongYL/KMJ/session 20192020

LEARNING OUTCOMES

At the end of this topic, students should be able to:

8.1 Rotational kinematics (c) Solve problem related to rotational motion
with constant angular acceleration
(a) Define and use :
ω = ω0 + t
angular displacement () θ = ½ ( ω0 + ω ) t
average angular velocity (av) θ = ω0 t + ½ t 2
instantaneous angular velocity () ω 2 = ω0 2 + 2 θ
average angular acceleration (av)
instantaneous angular acceleration () (Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) 8.2 Equilibrium of a uniform rigid body
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6) (a) Define torque

(b) Relate parameters in rotational motion (Lecture : C1&C2,CLO1, PLO1, MQF LOD1)

with their corresponding quantities in (b) Solve problems related to equilibrium of a
uniform rigid body.
linear motion. Write:
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)
s  r ;v  r, at  r; ac  r 2  v2
r
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

LEARNING OUTCOMES

At the end of this topic, students should be able to:

8.3 Rotational dynamics 8.4 Conservation of angular momentum

(a) Define and use moment of inertia of a (a) Define and use angular momentum,
uniform rigid body (sphere, cylinder, ring,
disc and rod) (Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6) (b) State and use principle of conservation of
angular momentum.
(b) State and use torque,
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) (Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

Rotational of rigid body

Rotational motion Rigid body

o motion about a o Object with a definite shape that
fixed axis passing doesn’t change, so that the
through the body. particles composing it, stay in
fixed positions relative to one
o As the body rotate, another.
all points on the
body move in
circular path.