ME 262 Theory of Machines I
(3, 0, 3)
F. W. ADAM
MECHANICAL ENGINEERING DEPARTMENT
KNUST
Kinematic analysis in mechanisms
particle motion in the three coordinate
systems
i. Cartesian/rectangular coordinate system
ii. Cylindrical coordinate system
iii. Spherical coordinate system
Transport Theorem
• Consider a vector A observed
from a moving coordinate
system xyz, which is rotating
with angular velocity Ω
Figure a
Figure b
5
General Motion
Position Vector
rB  rA  rB / A
Velocity
d
rB   d rA   d rB / A 
dt
dt
dt
vB  vA  vB / A xyz   x rB/A
Acceleration
d
v B   d v A   d v B / A xyz  d  x rB/A 
dt
dt
dt
dt
 x r  2 x v    x   x r 
aB  a A  a B/A xyz  
B/A
B/A xyz
B/A
For 2D motions, the above equation reduces to
 x r  2 x v    2 r
a B  a A  a B/A xyz  
B/A
B/A xyz
B/A
6
Example 1
Crank AB of the engine system shown has a constant
clockwise angular velocity of 2000 rpm. For the crank
position shown, determine the angular velocity as well
as the angular acceleration of the connecting rod BD
and the linear velocity and acceleration of point D.
solution
Since the crank rotates about A with constant 𝜔𝐴𝐵 = 2000 𝑟𝑝𝑚 =
209.4 𝑟𝑎𝑑/𝑠, The velocity of B can be calculated from
𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴 + Ω𝐴𝐵 x 𝑟𝐵/𝐴
𝑣𝐵/𝐴 = 𝑣𝐴 = 0
𝑣𝐵 = Ω𝐴𝐵 x 𝑟𝐵/𝐴 = 209.4 𝒌 x 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
𝑣𝐵 = − 4.04 𝒊 − 4.81 𝒋 𝑚𝑠 −1
𝑣𝐷 = 𝑣𝐵 + 𝑣𝐷/𝐵 + Ω𝐷𝐵 x 𝑟𝐷/𝐵
𝑣𝐷/𝐵 = 0
𝑣𝐷 𝒊 = −4.04 𝒊 + 4.81 𝒋 + Ω𝐷𝐵 𝒌 x 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
Expanding and comparing coefficients
𝑣𝐷 = −4.04 − 0.019Ω𝐷𝐵
0 = 4.81 + 0.023 Ω𝐷𝐵
Ω𝐷𝐵 = −6.28 𝑟𝑎𝑑/𝑠; 𝑣𝐷 = −209.13 𝑚𝑠 −1
Solution
Since 𝛼𝐴𝐵 = 0. The acceleration of B can be calculated from
𝑎𝐵 = 𝑎𝐴 + 𝑎𝐵/𝐴 + 2Ω𝐴𝐵 x 𝑣𝐵/𝐴 + 𝛼𝐴𝐵 x 𝑟𝐴𝐵 − Ω𝐴𝐵 2 𝑟𝐵/𝐴
𝑣𝐵/𝐴 = 𝑎𝐴 = 𝑎𝐵/𝐴 = 𝛼𝐴𝐵 = 0
𝑎𝐵 = −Ω𝐴𝐵 2 𝑟𝐵/𝐴 = −209.42 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
𝑎𝐵 = − 1007.63 𝒊 + 845.56 𝒋 𝑚𝑠 −2
𝑎𝐷 = 𝑎𝐵 + 𝑎𝐷/𝐵 + 2Ω𝐷𝐵 x 𝑣𝐷/𝐵 + 𝛼𝐷𝐵 x 𝑟𝐷/𝐵 − Ω𝐷𝐵 2 𝑟𝐷/𝐵
𝑣𝐷/𝐵 = 𝑎𝐷/𝐵 = 0
𝑎𝐷
= − 1007.63 𝒊 + 845.56 𝒋 + 𝛼𝐷𝐵 𝒌 x 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
− 6.282 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
Expanding and comparing coefficients
𝑎𝐷 𝒊
= −1007.63 − 0.019𝛼𝐷𝐵 − 0.91 𝒊
+ −845.56 +0.0236𝛼𝐷𝐵 −0.75 𝒋
0 = −846.33 + 0.023 𝛼𝐷𝐵
𝑎 = −1037.84 − 0.019𝛼
Example 2
Determine the velocity and acceleration
of the collar C at this instant
Solution
𝑣𝐷 = 𝑣𝐵 + 𝑣𝐷/𝐵 + Ω𝐷𝐵 x 𝑟𝐷/𝐵
𝑣𝐷 = 3𝒋 + 5 𝒌 x 0.2 𝒋
= −𝒊 + 3 𝒋 𝑚/𝑠
𝑎𝐷
= 𝑎𝐵 + 𝑎𝐷/𝐵 + 2Ω𝐷𝐵 x 𝑣𝐷/𝐵
+ 𝛼𝐷𝐵 x 𝑟𝐷/𝐵 − Ω𝐷𝐵 2 𝑟𝐷/𝐵
𝑎𝐷
= 2𝒋 + 10 𝒌 x 2 𝒋 − 6 𝒌 x 0.2𝒋
− 25(0.2 𝒋)
−2
Example 3
Determine the angular motion of
rod DE at this instant. The collar at
C is pin connected to AB and slides
over rod DE.
Solution
Assignment
1. If link CD has an angular velocity of = 6 rad/s, determine
the velocity and acceleration of point E on link BC and the
angular velocity and angular acceleration of link AB at the
instant shown.
Assignment
2. At the instant shown rod AB has an angular velocity 𝜔𝐴𝐵 = 4
rad/s and an angular acceleration 𝜔𝐶𝐷 = 2 𝑟𝑎𝑑/𝑠 2 . Determine
the angular velocity and angular acceleration of rod CD at this
instant. The collar at C is pin connected to CD and slides freely
along AB.
Assignment
3.The"quick-return" mechanism
consists of a crank AB, slider
block B, and slotted link CD. If
the crank has the angular
motion shown, determine the
angular motion of the slotted
link at this instant.
Assignment
4. Ball C moves with a speed of 3
m/s, which is increasing at a
constant rate of 1.5 m/𝑠 2 , both
measured relative to the circular
plate and directed as shown. At the
same instant the plate rotates with
the angular velocity and angular
acceleration shown. Determine the
velocity and acceleration of the
ball at this instant.
Velocity Analysis using Instant Centre
Method
• An instant (or instantaneous) centre (IC) is a point where there is
no relative velocity between two links of a mechanism at that
instant.
• Instant centres are useful for calculation of linear and angular
velocities, and mechanical advantage of mechanisms.
𝜔
Example
Solution
The crankshaft AB turns with
a clockwise angular velocity
of 10 rad/s,. Determine the
velocity of the piston at the
instant shown.
𝑣𝐵 = 𝜔𝑟 = 10 × 0.25 = 2.5 𝑓𝑡/𝑠
Instant Centres in Mechanisms
,
In any mechanism, obvious instant centres are points where
the links join. An instant centre may not coincide with on any joint.
Such an instant centre needs to be determined using Kennedy’s
theorem. The number of instant centres may be determined by
pairing all the links. Thus,
nn  1
N C 
;n2
2
n
2
where
N = number of instant centres and
n = number of links
For the four-bar
mechanism shown
n=4
N=4(4-1)/2=6
Kennedy’s Theorem
Kennedy’s theorem deals with three instant centres of three links
in a mechanism. It states that the three instant centres of three
links moving relative to another must lie on a straight line.
Mechanical Advantage of Mechanism
For a conservative mechanism
Pin  Tinin  Toutout  Pout
 Tout   in 

  

 Tin   out 
By definition, the mechanical advantage
(MA) is the ratio of the magnitude of output
force, to the magnitude of input force,
MA 
Fout
Fin
 Tout


rout 



 Tin



r
in


 Tout
MA  
 T
 in
 rin



 rout




 in
MA  

 out
 rin


r
 out




Relative Velocity Method
The crank of a slider crank mechanism rotates clockwise at a
constant speed of 300 rpm. The crank is 150 mm and the
connecting rod is 600 mm long. Determine:
1. Linear velocity and acceleration of the midpoint of the connecting
rod, and
2. Angular velocity and angular acceleration of the connecting rod, at a
crank angle of 45° from inner dead centre position.
solution
Velocity analysis
𝑣𝐵 = 𝜔𝑟𝐵 = 0.15 × 10𝜋 = 4.713 𝑚/𝑠
From the velocity diagram
𝑣𝐴𝐵 = 3.4 𝑚/𝑠
𝑣𝐴 = 4 𝑚/𝑠
𝑣𝐷 = 4.1 𝑚/𝑠
Acceleration analysis
2
2
𝑣
4.713
𝐵
𝑎𝐵 𝑟 =
=
= 148.1 𝑚/𝑠 2
𝑂𝐵
0.15
2
2
𝑣
3.4
𝐴𝐵
𝑎𝐴𝐵 𝑟 =
=
= 19.3 𝑚/𝑠 2
𝐴𝐵
0.6
From the acceleration diagram
𝑎𝐴𝐵 𝑡 1032
𝛼𝐴𝐵 =
=
= 171.67 𝑟𝑎𝑑/𝑠 2
𝐴𝐵
0.6
Example 2
In the mechanism shown, the slide QD is driven with a uniform
angular velocity of 10 rad/s and the block C has a mass of 1.5 kg.
Draw the velocity and acceleration diagrams for the mechanism in
the position shown. OQ = 1 m; OC = 2 m.
solution
Velocity analysis
𝑣𝑐 = 28.8 𝑚/𝑠
From the velocity diagram
𝑣𝑐𝑐′ = 7.5 𝑚/𝑠
Acceleration analysis
𝑣𝑐 2
𝑟
𝑎𝑐 =
= 414 𝑚/𝑠 2
𝑂𝐶
𝑎𝐶 ′ 𝑄 𝑟 = 288 𝑚/𝑠 2
𝑎𝐶 ′ 𝐶 𝐶 = 2𝑣𝑐𝑐′ 𝜔𝐶 ′ 𝑄
𝑎𝐶 ′ 𝐶 𝐶 = 150 m/𝑠 2
From the acceleration diagram
𝑎𝐶 = 417 𝑚/𝑠 2
ROTARY MOTION TRANSMISSION
Gears
In the transmission of motion or power between two
shafts. The effect of slipping is to reduce the velocity
ratio of the system. In precision machines, in which a
definite velocity ratio is of importance (as in watch
mechanism), the only positive drive is by means of gears
or toothed wheels. A gear drive is also provided, when
the distance between the driver and the follower is very
small.
A gear or cogwheel is a rotating machine part having cut
teeth, or cogs, which mesh with another toothed part in
order to transmit torque, in most cases with teeth on the
one gear being of identical shape, and often also with
that shape on the other gear.
Advantages and disadvantages of the
gear drive
The following are the advantages and disadvantages of the
gear drive as compared to belt, rope and chain drives:
Advantages
1. It transmits exact velocity ratio.
2. It may be used to transmit large power.
3. It has high efficiency.
4. It has reliable service.
5. It has compact layout.
Disadvantages
1. The manufacture of gears require special tools and
equipment.
2. The error in cutting teeth may cause vibrations and
Types of Gears
There are many different types of gears. For the purposes of this
section we will focus on different gear geometry.
• Spur gears are by far the most common
type of gear. Spur gears have teeth that
run perpendicular to the face of the gear.
• Helical gears are very similar to spur gears
except the teeth are not perpendicular to the
face. The teeth are at an angle to the face giving
helical gears more tooth contact in the same
area.
Helical gears can also be used on non-parallel
shafts to transmit motion. Helical gears tend to
run quieter and smoother than spur gears due to
the increased number of teeth in constant
contact at any one period of time.
Types of Gears
• Herringbone gears resemble two helical gears
that have been placed side by side. They are
often referred to as "double helicals". One
benefit of herringbone gears is that it helps to
avoid issues related to side thrust created with
the use of helical gears.
• Bevel gears are used mostly in situations
that require power to be transmitted at right
angles (or applications that are not parallel).
Bevel gears can have different angles of
application but tend to be 90°.
Types of Gears
• Worm gears are used to transmit power at
90° and where high reductions are
required. The worm resembles a thread
that rides in concaved or helical teeth.
• A rack is basically a straight gear used to
transmit power and motion in a linear
movement.
Types of Gears
• Internal gears typically resemble inverted spur gears
but are occasionally cut as helical gears.
• Face gears transmit power at (usually)
right angles in a circular motion. Face
gears are not very common in industrial
application.
Types of Gears
• Sprockets are used to run chains or belts.
They are typically used in conveyor
systems.
• In epicyclic gearing one or more of the
gear axes moves. Examples are sun and
planet gearing (see below) and
mechanical differentials.
Types of Gears
• Hypoid gears resemble spiral bevel
gears except the shaft axes do not
intersect. The pitch surfaces appear
conical but, to compensate for the
offset shaft, are in fact hyperboloids of
revolution. Hypoid gears are almost
always designed to operate with shafts
at 90 degrees.
• Spiral Bevel Gears
Classification of Gears
The gears or toothed wheels may be classified as follows:
1. According to the position ofaxes of the shafts. The axes of the
two shafts between which the motion is to be transmitted, may be
(a) Parallel, (b) Intersecting, and (c) Non-intersecting and nonparallel.
2. According to the peripheral velocity of the gears. The gears,
according to the peripheral velocity of the gears may be classified
as : (a) Low velocity, (b) Medium velocity, and (c) High velocity.
The gears having velocity less than 3 mls are termed as low velocity
gears and gears having velocity between 3 and 15 m/s are known
as medium velocity gears. If the velocity of gears is more than 15
m/s, then these are called high speed gears.
Classification of Gears
3. According to the type of gearing. The gears, according to the
type of gearing may be classified as :
(a) External gearing, (b) Internal gearing, and (c) Rack and
pinion.
4. According to position of teeth on the gear surface. The teeth on
the gear surface may be (a) straight, (b) inclined, and (c) curved.
Gear Trains
Sometimes, two or more gears are made to mesh with each
other to transmit power from one shaft to another. Such a
combination is called gear train or train of toothed wheels. The
nature of the train used depends upon the velocity ratio required
and the relative position of the axes of shafts. A gear train may
consist of spur, bevel or spiral gears.
Types of Gear Trains
The following are the different types of gear trains, depending
upon the arrangement of wheels:
1. Simple gear train
2. Compound gear train
3. Reverted gear train and
4. Epicyclic gear train.
Gear Trains
In a simple gear train there is only one
gear on each shaft, as shown. The gears
are represented by their pitch circles.
When there are more than one gear on
a shaft, as shown in the Figure, it is
called a compound train of gear.
Gear Trains
When the axes of the first
gear (i.e. first driver) and the last
gear (i.e. last driven or follower)
are co-axial, then the gear train
is known as reverted gear train
as shown in Fig. 13.4.
We have already discussed that in an
epicyclic gear train, the axes of the
shafts, over which the gears are
mounted, may move relative to a fixed
axis.
Gear Trains
𝑽 = 𝝎𝑨 𝑹𝑨 = −𝝎𝑩 𝑹𝑩
𝝎𝑨
𝑹𝑩
𝑫𝑩
𝑪𝑩
𝑵𝑩
=−
=−
=−
=−
𝝎𝑩
𝑹𝑨
𝑫𝑨
𝑪𝑨
𝑵𝑨
Where
𝜔 is the angular velocity
R is the radius
D is the diameter
C is the circumference
N is the number of teeth
Example 1
The gearing of a machine tool is
shown below. The motor shaft is
connected to gear A and rotates at
975 rpm. The gearwheels B, C, D and E
are fixed to parallel shafts rotating
together. The final gear F is fixed on
the output shaft. What is the speed of
gear F ? The number of teeth on each
gear are as given below:
Gear
A B C D E F
No. of teeth 20 50 25 75 26 65
𝝎𝑭
𝑵𝑨
𝑵𝑪
𝑵𝑬
=−
.−
.−
𝝎𝑨
𝑵𝑩 𝑵𝑫
𝑵𝑭
𝝎𝑭 = −𝝎𝑨
𝑵𝑨 𝑵𝑪 𝑵𝑬
.
.
𝑵𝑩 𝑵𝑫 𝑵𝑭
𝟐𝟎 𝟐𝟓 𝟐𝟔
𝝎𝑭 = −𝟗𝟕𝟓
.
.
𝟓𝟎 𝟕𝟓 𝟔𝟓
𝝎𝑭 = −𝟓𝟐 𝒓𝒑𝒎
Example 2
In an epicyclic gear train, an arm
carries two gears A and B having 36
and 45 teeth respectively. If the
arm rotates at 150 rpm. in the
anticlockwise direction about the
centre of the gear A which is fixed,
determine the speed of gear B. If
the gear A instead of being fixed,
makes 300 rpm. in the clockwise
direction, what will be the speed of
gear B ?
𝝎𝑨/𝑪 = 𝝎𝑨 − 𝝎𝑪 ; 𝝎𝑩/𝑪 = 𝝎𝑩 − 𝝎𝑪
𝝎𝑨/𝑪
𝑵𝑩 𝝎𝑨 − 𝝎𝑪
=−
=
𝝎𝑩/𝑪
𝑵𝑨 𝝎𝑩 − 𝝎𝑪
𝑵𝑨 = 𝟑𝟔; 𝑵𝑩 = 𝟒𝟓
a. 𝝎𝑪 = 𝟏𝟓𝟎 𝒓𝒑𝒎; 𝝎𝑨 = 𝟎; 𝝎𝑩 =?
𝑵𝑩 𝝎𝑨 − 𝝎𝑪
𝟒𝟓
𝟎 − 𝟏𝟓𝟎
−
=
⇒−
=
𝑵𝑨 𝝎𝑩 − 𝝎𝑪
𝟑𝟔 𝝎𝑩 − 𝟏𝟓𝟎
𝝎𝑩 = 𝟐𝟕𝟎 𝒓𝒑𝒎
b. 𝝎𝑪 = 𝟏𝟓𝟎 𝒓𝒑𝒎; 𝝎𝑨 = −𝟑𝟎𝟎; 𝝎𝑩 =?
𝟒𝟓 −𝟑𝟎𝟎 − 𝟏𝟓𝟎
−
=
𝟑𝟔
𝝎𝑩 − 𝟏𝟓𝟎
𝝎𝑩 = 𝟓𝟏𝟎 𝒓𝒑𝒎
Example 3
The Figure below shows a reverted planetary
train. Gear 2 is fastened to its shaft and is
driven at 250 rpm in a clockwise direction.
Gears 4 and 5 are planet gears which are
joined but are free to turn on the shaft
carried by the arm. Gear 6 is stationary. Find
the speed and direction of rotation of the
arm.
We must first decide which
gears to designate as the
first and last members of
the train. Since the speeds
of gears 2 and 6 are given,
either may be used as the
first. The choice makes no
difference in the results, but
once the decision is made,
it cannot be changed. We
shall choose gear 2 as the
first; then gear 6 will be the
last. Thus
𝝎𝟔/𝟑
𝑵𝟐 𝑵𝟓 𝝎𝟔 − 𝝎𝟑
=−
.−
=
𝝎𝟐/𝟑
𝑵𝟒 𝑵𝟔 𝝎𝟐 − 𝝎𝟑
𝟐𝟎 𝟏𝟔
𝟎 − 𝝎𝟑
−
.−
=
𝟑𝟎 𝟑𝟒 −𝟐𝟓𝟎 − 𝝎𝟑
𝝎𝟑 = 𝟏𝟏𝟒 𝒓𝒑𝒎
Example 4
In the bevel-gear train shown the input is to
gear 2, and the output from gear 6, which is
connected to the output shaft. The arm 3 turns
freely on the output shaft and carries the
planets 4 and 5. Gear 7 is fixed to the frame.
What is the output speed if gear 2 rotates at
2000 rpm?
The problem is solved in two steps. In
the first step we consider the train to
be made up of gears 2, 4, and 7 and
calculate the velocity of the arm. Thus
𝝎𝟕/𝟑
𝑵𝟐 𝑵𝟒 𝝎𝟕 − 𝝎𝟑
=−
.
=
𝝎𝟐/𝟑
𝑵𝟒 𝑵𝟕 𝝎𝟐 − 𝝎𝟑
−
𝟐𝟎 𝟓𝟔
𝟎 − 𝝎𝟑
.
=
𝟓𝟔 𝟕𝟔 𝟐𝟎𝟎𝟎 − 𝝎𝟑
𝝎𝟑 = 𝟒𝟏𝟔. 𝟕 𝒓𝒑𝒎
Now consider the train as composed of
gears 2, 4, 5, and 6.. The train value
𝝎𝟔/𝟑
𝑵𝟐 𝑵𝟓 𝝎𝟔 − 𝝎𝟑
=−
.
=
𝝎𝟐/𝟑
𝑵𝟒 𝑵𝟔 𝝎𝟐 − 𝝎𝟑
−
𝟐𝟎 𝟐𝟒
𝝎𝟔 − 𝟒𝟏𝟔. 𝟕
.
=
𝟓𝟔 𝟑𝟓 𝟐𝟎𝟎𝟎 − 𝟒𝟏𝟔. 𝟕
𝝎𝟔 = 𝟐𝟖. 𝟗𝟏 𝒓𝒑𝒎
Tabular Analysis of Planetary Trains
It involves analyzing two easily described parts of the
motion separately, and then adding the results together.
The analysis is carried out in the following steps:
(1) Lock all gears rigidly to the rotating arm and rotate
the arm with rigidly attached gears by the number of
revolutions proportional to the angular velocity of the
arm.
(2) Tabulate the resulting turns of each gear.
(3) Motion relative to arm: Separate the gears from the
arm while holding the arm fixed, and rotate the rest
of the gears back so that the total rotation (step 1 +
step2) of one or more of the gears matches their
given rotations.
(4) Add the turns of each gear in the table.
Example 2
In an epicyclic gear train, an arm
carries two gears A and B having 36
and 45 teeth respectively. If the
arm rotates at 150 rpm. in the
anticlockwise direction about the
centre of the gear A which is fixed,
determine the speed of gear B. If
the gear A instead of being fixed,
makes 300 rpm. in the clockwise
direction, what will be the speed of
gear B ?
a. X+150=0; x=-150
𝟑𝟔
= 𝟐𝟕𝟎 𝒓𝒑𝒎
𝟒𝟓
b. . X+150=-300; x=-450
𝟑𝟔
𝝎𝑩 = 𝟏𝟓𝟎 − −𝟒𝟓𝟎 .
= 𝟓𝟏𝟎 𝒓𝒑𝒎
𝟒𝟓
Gear B
𝝎𝑩 = 𝟏𝟓𝟎 − −𝟏𝟓𝟎 .
Step number
Arm C
Gear A
1. Gears locked
+150
+150
+150
2. Arms fixed
0
+x
-x. 𝑵𝑨
𝑵
𝑩
3. Results
+150
150+x=0
𝑵
150-x. 𝑵𝑨
𝑩
Assignment 1
Find the speed and direction of rotation of gear 8 in the figure.
What is the speed ratio of the train.
Assignment 2
In the epicyclic gear shown in the figure, gear B has 120 teeth
externally and 100 teeth internally. The driver A has 20 teeth
and the arm E is connected to the driven shaft. Gear D has 60
teeth. If A revolves at 100 rpm ccw and D at 27 rpm ccw, find
the speed of the arm E.
Assignment 3
In the epicyclic gear shown in the figure, shaft A is
stationary. If gear 2 rotates at 800 rpm cw. What are
the speed and direction of rotation of shaft B.