Module_SP015

8.1 Rotational kinematics

8.1.1 Term(s) used to describe Rotational Motion

Terms Definition Equation unit

Angle undergoes from a fixed SI unit : rad
reference line. Other units :
revolution (rev),
Angular degree (
displacement,
: angular Sign Convention
displacement rotational motion

Reference : arc length Positive Anticlockwise
line : distance from Negative Clockwise

rotation axis Unit Conversion

Degree Radian No of
revolution
360° 2
(rev)

1 revolution

Terms Definition Equation unit

Angular velocity, Rate of change of angular SI unit : rad s−1
displacement.
Other units :
Angular velocity at a rpm (rev per min)
specified position or at a rps (rev per second)
particular of time (time
Average interval, ∆t approaches zero ). Vector quantity
angular
velocity,
: final angular Its direction can be
Instantaneo displacement determine by using
us angular right hand grip rule
velocity, : initial angular
displacement If you curl the
finger of your
: time interval right hand in
the direction of
rotation …

→

… your right
thumb will
points in the
direction of


Terms Definition Equation unit

Angular acceleration, Rate of change of angular Vector quantity
velocity SI unit : rad s−2

Average Note:
angular If is positive  is
acceleration, At At increasing with time
(rotation speeding up)

: final angular If is negative  is
velocity decreasing with time
(rotation slowing
: initial angular down)
velocity
Instantaneous Angular acceleration at a
angular specified position or at a : time interval
particular of time (time
acceleration, interval, ∆t approaches ( = − )

→

zero ).

8.1.2 Rotational motion with uniform (constant) angular acceleration

Consider the rotational motion for a rigid Kinematic equation of
body under uniform angular acceleration rotational motion
motion:


Let )
= initial angular velocity (at time
= final angular velocity (after time )
= angular acceleration
= angular displacement/angle turned
through/revolutions undergo
= time taken
Rotation Linear

Note:
if the rotation slow down −
if the rotation speed up  +

Derivation of kinematic equation of rotational motion

When the angular velocity of rigid body Substitute (1) into (2) : Extra note for better understanding
increases uniformly from 0 to in time
, then uniform angular acceleration, is:   1 ( 0   0   t)t
2
change in angular velocity
  time taken   0t  1  t2
2 … (3)

    0 Squaring equation (1):

t  2  (o   t) 2

 0  t … (1)  2  2ot  2t 2
0

The angular displacement in time t :  2  2 (ot  1 t 2 )
0 2
θ = ( Average angular velocity )(time )

  1 ( 0 )t … (2)  2  0 2  2  … (4)
2

8.1.3 Relationship between Linear and Rotational motion

Quantity Linear Rotational Relationship Refer slide 11 on how
Displacement we obtain these relation

Velocity Note : is the distance from the point of rotation
Acceleration

o Every point on a rotating body has same value of Since

BUT >

o Two points on the body at different distances from Thus:

the axis of rotation will have different value of >
which is proportional to >

3) In topic 6 we found Extra note for better understanding
that a point moving in
circular path

undergoes centripetal
(radial) acceleration.

1) Figure 1 As a rigid body rotates 2) Figure 2 As a rigid body rotates 4) Resultant linear
about fixed axis through point O, about a fixed axis through point acceleration of point
the point P has a tangential O, the point P experiences P (Figure 2 is given
velocity that is always tangent tangential (linear) acceleration , by
to the circular path of radius r. and centripetal (linear)
acceleration . (i) Magnitude
From : From :
 

(ii) direction



8.2 Equilibrium of a uniform rigid body
8.2.1 Torque,

Tendency of a force to rotate object about an axis.
Definition OR turning effect produced by a force.

OR cross product between distance, r and force, F

where : force applied Point of F
: distance from point of rotation to point rotation
Equation 
where force is applied r
: angle between and

o SI unit : Nm : ; clockwise :
Important o Vector quantity
concept(s)) o Anticlockwise

How to Tips : Guna jangka lukis OR DyFingers turn method F

identify Thumb : Fix at +
direction point of rotation
Index finger : Move
of − Pen : Move according according to the
torque, to the direction of direction of Force, F to
Force, F to get the get the direction of
leg : fix on the direction of
point of rotation

⃗

Step(s) to calculate torque,

Point of rotation   180  60  120

F  50N r F  50N r  F  50N r F  50N
60 60
0.4m 0.4m 0.4m 60 0.4m

Step 1 : Choose the point of Step 2 : Draw r from point of Step 3 : identify the angle θ Step 4 : Determine Step 5: Calculate
rotation. rotation to point where F which is the angle between r direction of rotation
applied. and F. using DyTurn =
Fingers method
= .
= + . Nm

Example (TLO 8.2(b)) o Ability of a force create torque and cause a rotation depends on
three factors:
Find the magnitude and direction of 1. magnitude F of the force; ( ↑ ↑)
the torque each force produces about 2. distance r from the point of application of F to the point of
pivot point A. rotation; ( ↑ ↑)
3. The angle at which the force is applied. ( maximum if Ʇ to
F1 F3 F4 and = 0 if || with )

F5 Even though same force is used, torque applied to the bolt increases as the distance from
your hand to the bolt increases. In Fig. (b) below, you produce more torque by placing your
5.0 m 5.0 m hand on the end of the wrench handle compare to (a)

Point of rotation F2

Solution

 1  0Nm * r  0
 2  60(5) sin 90  300Nm
 3  80(5) sin143  240.73Nm

 4  70(1) sin 60  60.62Nm
 5  0Nm * r  0

8.2.2 Static equilibrium of a uniform rigid body

Conditions for equilibrium of a
uniform rigid body

First condition for equilibrium Second condition for
(translational/linear equilibrium): equilibrium
The net force acting on the body
(rotational equilibrium):
is zero: The net torque about any
point on the body is zero:

NON FACE TO FACE SLT

Problem Solving Strategy: Equilibrium of Rigid Body

1. Draw a free body diagram (we are dealing with
extended objects, the position of forces are
important, uniform body  drawn at center )

2. Pick any point as rotation point and label distance .
Tips : choose the point where most unknown forces pass
through as rotation axis

3. Apply
4. Choose a convenient x- and y-axis and resolve forces

acting into their x- and y-components
5. Apply
6. Solve equations in 3 and 5 to determines unknowns

Example (TLO 8.2(b)) Solution B
D
A traffic light hangs from a structure as show (a) ① FBD
in figure above. The uniform aluminum pole 37°
AB is 7.5 m long has a mass of 8.0 kg. The AB : 7.5 m
mass of the traffic light is 12.0 kg. Determine AE : 3.75 m E
the AD : 6.31 m 37°
(a) tension in the horizontal cable CD,
(b) force exerted by the pivot A on the WT = 12(9.81) A
= 117.72 N
aluminum pole.
WP = 8(9.81)
= 78.48 N

② pick a point as rotation point

③ apply ∑ =

A  0

R WP WT T  0

0  78.48(3.75)sin53117.72(7.5)sin53T (6.31)sin37  0
3.7475T  940.15
T  247.57N

NON FACE TO FACE SLT

Solution ⑥ solve for the unknown

(b) ④ choose axes and resolve forces

B

D
E
∴ Force exerted by the pivot A on the aluminum pole

A R  Rx2  Ry 2
R  (247.57)2  (196.2)2  315.89N

⑤ Apply ∑ ⃗ = 0, ∑ ⃗ = 0  Fy 0 above +

 Fx 0 Ry WP WT  0

Rx  T  0 Ry  78.48 117.72
Rx  247.57N
Ry  196.2N

8.3 Rotational dynamics

8.3.1 Moment of inertia,

o Moment of Inertia represents an

object’s resistance to change in

rotational motion.

o depends on (1) mass (2) distribution of
mass and (3) rotation axis chosen.

• Mass close to axis • Mass farther from axis
• Smaller moment of inertia • Larger moment of inertia
• Easy to start rotating. • Harder to start rotating

o The larger moment of inertia requiring
more torque to change the body's

(a) (b) rotational speed.

Moment of inertia, of a particle Example (TLO 8.3(a))

Rotation Rigid rods of
axis negligible mass
lying along the y
o Scalar quantity ; axis connect three
o SI unit : kg m2 particles. The
system rotates
Moment of inertia, of particle system about the x axis.
Find the moment of
Moment of Inertia, is defined as the sum of the inertia about the x
products of mass, m of each particle and the axis.
square of its respective distance, r from the
rotation axis. Solution
=
 



 
 



kg m2
 

Moment of inertia of various uniform rigid bodies

Shape Diagram Equation Shape Diagram Equation

Hoop or ring Thin rod with
or thin
rotation axis
cylindrical through one
shell CM end

Solid Solid CM
cylinder or CM Sphere

disk

Thin rod with CM Hollow CM
rotation axis Sphere
or
through Thin spherical
center
shell

8.3.2 Relation between torque, and angular acceleration,

Net torque Newton second law for
Rotational motion
o The net torque is the sum of the
torques due to the applied forces: o A rigid object that experience a net

  torque ( ) about the axis of

  rotation undergoes an angular

acceleration, .

where is the moment of inertia about
the axis of rotation.

A net torque is the cause of angular
acceleration

o The relationship is analogous to NON FACE TO FACE SLT
(Newton’s Second Law in linear motion)
Problem solving Strategy:
o Rearrange : Rotational Dynamics
i. The angular acceleration is directly
proportional to net torque 1) As always, draw a clear and complete diagram
ii. The angular acceleration is inversely 2) Choose the object or objects that will be the
proportional to the moment of inertia
of the object. system to be studied.
3) Draw a free body diagram for the object under

consideration (showing all (and only) the forces
acting on that object and exactly where they act,
so you can determine the torque due to each.
4) Identify the axis of rotation and determine the
torques about it. Assign the correct sign for each
torque.
5) Apply = (* = ∑   )
6) Also apply ∑ = , and other laws or principle
as needed.
7) Solve the resulting equation(s) for the
unknown(s).

Example (TLO 8.3(c)) NON FACE TO FACE SLT
Solution
To get some exercise without going anywhere, you set Given : = 18 N
your bike on a stand with the rear wheel free to turn. As
you pedal, the chanin applies a force of = 18 N to the = 7.0 cm = 0.07 m
rear sprocket wheel at a distance of = 7.0 cm from the Wheel is hoop  =
rotation axis of the wheel. Consider the wheel to be a = 35 cm = 0.35 m
hoop of radius = 35 cm and mass = 2.4 kg. What is = 2.4 kg
the angular velocity of the wheel after 5.0 s?

rad s‒2

= +
= 0 + (4.286)(5)
= 21.43 rad s‒1

8.4 Conservation of angular momentum

8.4.1 Angular momentum,

Angular Momentum of a particle o the angular momentum of the particle
about the axis of rotation is given by :
For a particle P of mass m travels in a
circle of radius with tangential linear
velocity as shown in figure below,

o Magnitude of angular momentum :

o Known that : & θ = 90°, thus:

o since :

Angular Momentum of a rigid body o SI unit for : kg m2 s−1

Angular momentum of a body rotating o Angular momentum is a vector
about a fixed axis is the product of the
body’s moment of inertia and its quantity
angular velocity with respect to that
axis:

where Angular velocity and
: angular momentum angular momentum
: moment of inertia (kg m2) vectors point along
: angular velocity (rad s−1) the rotation axis in

the direction
determined by the

right hand rule.

8.4.2 Conservation of Angular momentum

Principle of conservation of angular momentum states the total angular
momentum of a system remains constant if the net external torque acting

on it is zero, that is, if the system is isolated.

constant or

Because the magnitude of the system is , we can express the principle of
conservation of angular momentum as

Initial @ before final @ after

Conservation of angular momentum requires that the product of and must

remain constant. Thus, a change in for an isolated system requires a
change in .

Examples that demonstrate conservation of angular momentum

A skater doing spin on ice, illustrating conservation of As the students walks towards the centre of
angular momentum Fig. (a) with her arms outstretched is the rotating platform, moment of inertia of the
large so is small; Fig (b) when she pulls her arm toward system decreases, the angular velocity of the
her body, is small so is larger, she spin faster. system increases because the angular
momentum of the system remain constant

Example (TLO 8.4(b)) NON FACE TO FACE SLT

A 25 kg child with an initial speed
of 2.5 m s−1 along the path
tangent to the rim of a merry go
round with a radius of 2.0 m and
jumps on. The merry go round,
which is initially at rest has a
moment of inertia of 500 kg m2.
Find the final angular velocity of
the child and merry go round.

Solution

Beginning of dive : she pulls
arms/legs closer
Intention: is reduced 
increase  made more rotation

End of dive : layout position
Purpose: increases 
increase, slow rotation rate 
less “water-splash”
rad s−1

Topic 9
Simple Harmonic Motion (SHM)

9.1 Kinematics of Simple Harmonic Motion
9.2 Graphs of Simple Harmonic Motion
9.3 Period of Simple Harmonic Motion

MODE Face to face Non Face to face SLT
Lecture SLT 1
Tutorial 8
1

8

Prepared by ChongYL/KMJ/session 20192020

LEARNING OUTCOMES

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

9.1 Kinematics of simple harmonic motion e) Apply velocity, acceleration, kinetic energy
a) Explain SHM and potential energy for SHM.

(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) (Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

b) Solve problems related to SHM displacement 9.2 Graphs of simple harmonic motion
equation, =
a) Discuss the following graphs:
c) Derive eq(Tuuatotiroianl s: C: 3&C4, CLO3,PLO4, CTPS3, MQF LOD6) i. Displacement - time
i. Velocity, = = ±   − ii. Velocity – time
iii. Acceleration – time
ii. Acceleration, = = = − iv. Energy - displacement

iii. Kinetic energy, = ( − ) & potential (Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
energy, =
9.3 Period of simple harmonic motion
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6) a) Use expression for period of SHM, T for

d) Emphasis the relationship between total SHM simple pendulum and single spring.
energy and amplitude.
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

We have learnt : Periodic motion

 Linear motion the back & forth motion of
 Circular motion
an object about a fixed point
 Rotational motion called the equilibrium position.

In this topic, we are going
to study a special kind

of periodic motion
which occurs in
mechanical systems. This

motion is called,
simple
harmonic

Example: motion
1. oscillations of a block attached to a spring (SHM).
2. movement of child on a swing

3. motion of a pendulum

4. motion of the piston in a car engine

5. vibrations of atoms in a solid

9.1 Kinematics of Simple Harmonic Motion

9.1.1 Simple Harmonic Motion (SHM)

SHM is defined as periodic motion without o The angular frequency, is always
loss of energy in which the acceleration of a constant thus
body is directly proportional its
displacement from the equilibrium position o The negative sign in the equation
and is directed towards the equilibrium position but in indicates that the direction of the
opposite direction of the displacement . acceleration, is always opposite to
the direction of the displacement, .
OR mathematically,

{Hallmark of SHM}

where
= acceleration of the body
= angular frequency
= displacement from the equilibrium position.

o Examples of SHM system

(1) Simple Pendulum (3) Vertical spring oscillation

(2) Frictionless horizontal spring
oscillation

Figure (b)

Figure (a) Figure (c)

Equilibrium o A point where the acceleration and force exerted on the body
position undergoing oscillation is zero.

9.1.2 Displacement ( ) o Distance from the equilibrium position.

Term(s) o is defined as the maximum magnitude of the displacement from
used to Amplitude ( ) the equilibrium position.
describe
o SI unit is metre (m).

SHM o is defined as the time taken for one complete cycle.

Period ( ) o SI Unit : second (s).

Frequency ( ) o is defined as the number of complete cycles in one second.

o SI Unit : hertz (Hz) where 1 Hz = 1 cycle s1 = 1 s1

o Equation : =

Angular o Equation :
frequency ( ) o SI Unit : rad s−1

o Force that tends to bring a system back toward equilibrium.
o It is always directed towards the equilibrium position and whose

magnitude at any instant is directly proportional to the displacement

of the particle from the equilibrium position at that instant.

Term(s) where is known as force constant.

used to Restoring force
describe ( )

SHM

9.1.3 Kinematics of SHM Kinematic equation of
motion for an object in
SHM can be found
from a relationship

between SHM with
uniform circular
motion and

sinusoidal curve.

Object does SHM behave
very similar to objects that
move in circles. The radius
of the circle is symbolic of
the amplitude, A

Slide 8 As time passes, the oscillating object traces out a sinusoidal curve on the moving paper.
r1
MOE7 rebound_asia; 10/12/2002

The shadow of an object in uniform circular motion has same vertical motion as an object oscillating on a spring in SHM.

It follows that the equation of motion for the shadow of the object in circular motion is the same as the equation of motion for
the oscillating object on spring.

Thus the equation of motion for an object in SHM can be found from a relationship between SHM & Uniform circular motion.

Ministry of Education, Malaysia; 14/07/2004



SHM Circular motion Sinusoidal curve

amplitude x (m)
AT


x A x t (s)
θ


Equilibrium
position

–A



x but Note

θ can also be rewrite as

SHM displacement equation


o What if the SHM not start from at ? Example : At = 0 , = ⇒ = 90° =

Angle at time x (m)
is = +
A Extra note for better understanding

Initial position = 0
0
t (s)
of particle at

= 0

−


–A Figure (a) = + or =



Refer figure above, the SHM displacement Example : At = 0 , = − = 270° =

equation is given as :

In which therefore,

Given that , the equation become: = 0
0


o is known as which indicates −

the starting point in simple harmonic motion
Figure (b) = + or = −
where time

GRAPH IN TERMS OF TIME ( ) IN TERMS OF DISPLACEMENT
Note: Set Calculator in MODE RAD ( )

when do calculation

Graph displacement-time

(m) where:
x : displacement from the
equilibrium position at
time
A : amplitude
Displacement,  : angular frequency or
½ ¾ angular velocity
(s)
¼ ( ) : phase (unit in radian,

rad)
t : time
above eq. can rewrite as:
−



GRAPHS of SHM IN TERMS OF TIME ( ) IN TERMS OF DISPLACEMENT
Note: Set Calculator in MODE RAD ( )
Graph velocity-time
(m) when do calculation

From : ; = cos
= = cos ...(1)
From trigonometry identities
where
sin + cos = 1
= Therefore,

 = 1 − … (2)
Substitute eq. (2) into (1)
velocity, ¼ ½ The maximum velocity,
¾ (s) occurs when = (1 − sin )
, From : =
therefore,

− sin = ⋯ (3)

At amplitude, = 0
At equilibrium position, = (1 − )
is maximum.
−
= ( )

= ( − )

 

GRAPH of SHM IN TERMS OF TIME ( ) IN TERMS OF
Graph acceleration-time Note: Set Calculator in MODE DISPLACEMENT ( )

RAD when do calculation From equation
)

= =

(m)  Displacement equation
is given by
The maximum acceleration,
Acceleration, occurs when Substitute (2) into (1)
, therefore
¼ ½

− ¾ (s)

At amplitude :
acceleration, is maximum
At equilibrium :
acceleration, is zero

9.1.4 Energies of SHM

o Total mechanical energy of a SHM system consists of Kinetic energy, & Potential
energy, .

Graph Equation and concepts

Consider the oscillation of a spring as SHM

Graph potential energy-displacement

(J) Potential energy for the spring is given by

Potential Substitute into the equation of
energy,

potential energy:

− (m)


Graph Equation and concepts

Graph kinetic energy-displacement Recall :

(J) Knowing that :
Kinetic energy as a function of time,

Kinetic But remember,   , therefore
energy,

(m)

− where : mass
: angular frequency
: amplitude
: displacement from equilibrium

Graph Equation and concepts

Graph energy-displacement Without damping (no loss of energy from a
system ) , the total energy of the system
(J) remain constant.





Total Total mechanical energy of a simple
mechanical harmonic oscillator is directly proportional
energy, to the square of amplitude.

− There is an interchange of energy
(m) between potential and kinetic energies.

As potential energy decreases, kinetic energy

increases.

Equation(s) used for Kinematics of SHM Displacement, In terms of time ( ) In terms of
Velocity, displacement (
CAUTION : Set calculator in MODE
RAD when do calculation using the  
equations in this column

x  Asin t

Acceleration, 
Kinetic energy,
Potential energy,
Total energy,

Example (LO 9.1 b,e) NON FACE TO FACE SLT

The displacement of a 0.2 kg particle (2) Determine (i) maximum displacement
undergoing linear simple harmonic (ii) angular frequency (iii) period of the
motion is given by motion and (iv) frequency of the
motion
where and are measured in cm and
second respectively. Compare with s
(1) Write the SHM equation as cosine (i) = 5 cm
(ii) = 20 rad s−1
function (iii)

can also be rewrite as (iv)

.


where in cm , in s

(3) Determine the displacement, velocity NON FACE TO FACE SLT
and acceleration at time = 2 s
acceleration, at s

 



Displacement, at s cm s−2

Set calculator in mode RAD (radian) (4) What is the maximum velocity and
acceleration the particle can achieved ?
cm

velocity, at s cm s−1




Set calculator in mode RAD (radian) 19.74 cm s−2

cm s−1

(5) Determine the velocity and acceleration (6) Determine kinetic energy and
when the particle is 3.0 cm from the potential energy when its
equilibrium position. displacement is 3.0 cm.

When cm At cm = 0.03 m

 

 

cm s−1

84 cm s−2 NON FACE TO FACE SLT

9.2 Graphs of Simple Harmonic Motion Graph displacement- (m)
time

½ (s) At amplitude ( =
− )

Graph velocity- (m s−1)  = 0 m s−1;
time  = maximum

½ = magnitude in the
− opposite direction.

(s)
(m s−2)

Graph ½ = At equilibrium ( =
acceleration-  = maximum
time


(s)  = zero (0)

−

9.2 Graphs of Simple Harmonic Motion Graph energy-displacement o U is small when K is

At = ± the energy is all potential energy; the kinetic energy is large, and vice versa,
zero.
because the sum (total
At = 0 the energy is all kinetic energy;
the potential energy is zero. energy, ) must be

(J) Total energy is constant. constant.

o Total energy, is equal to the
maximum potential energy

stored in the spring when

because at these

points and thus there is no
(m)
kinetic energy.
Remarks:
o At the equilibrium position,
: Potential energy ; : Kinetic energy ; : Total energy where because ,

the total energy, all in the form

of kinetic energy.

9.3 Period of simple harmonic motion

9.3.1 Single spring oscillation o Consider the oscillation of a spring as
SHM as shown in Fig. (a) & (b).

o The only force acting on it is the
restoring force of the spring.

o Hooke’s law :
o Newton second law:

 
 

Fig. (a) vertical spring oscillation eq. (1)

o 5) From eq. (1) :
 executes linear SHM

− 0 o By comparing with :

o Thus :  

Fig. (b) Horizontal spring oscillation