UNIT II

KINEMATICS - VELOCITY
AND ACCELERATION
DIAGRAMS
SYLLUBUS:
Displacement, velocity and acceleration - analysis in simple mechanisms - Graphical Method
velocity and acceleration polygons - Kinematic analysis by Complex Algebra methods-Vector
Approach, Computer applications in the Kinematic analysis of simple mechanisms-Coincident points-
Coriolis Acceleration .

Content:

• Describe a mechanism.
• Define relative and absolute velocity.
• Define relative and absolute acceleration.
• Define radial and tangential velocity.
• Define radial and tangential acceleration.
• Describe a four bar chain.
• Solve the velocity and acceleration of points within a mechanism.
• Use mathematical and graphical methods.
• Construct velocity and acceleration diagrams.
• Define the Coriolis Acceleration.
• Solve problems involving sliding links.

It is assumed that the student is already familiar with the following concepts.
v Vector diagrams.
v Simple harmonic motion.
v Angular and linear motion.
v Inertia force.
v Appropriate level of mathematics.

All these above may be found in the pre-requisite tutorials.

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INTRODUCTION

A mechanism is used to produce mechanical transformations in a machine. This

transformation could be any of the following.

v It may convert one speed to another speed.

v It may convert one force to another force.
v It may convert one torque to another torque.
v It may convert force into torque.
v It may convert one angular motion to another angular motion.
v It may convert angular motion into linear motion.
v It may convert linear motion into angular motion.

A good example is a crank, connecting rod and piston mechanism.

Figure 1

v If the crank is turned, angular motion is converted into linear motion of the piston
and input torque is transformed into force on the piston.
v If the piston is forced to move, the linear motion is converted into rotary
motion and the force into torque. The piston is a sliding joint and this is called
PRISMATIC in some fields of engineering such as robotics.
v The pin joints allow rotation of one part relative to another. These are also called
REVOLUTE joints in other areas of engineering.

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 Consider the next mechanism used in shaping machines and also known as the
Whitworth quick- return mechanism.

Figure 2
v The input is connected to a motor turning at constant speed. This makes the
rocking arm move back and forth and the head (that carries the cutting tool)
reciprocates back and forth.
v Depending on the lengths of the various parts, the motion of the head can be
made to move forwards at a fairly constant cutting speed but the return stroke
is quick.
v Note that the pin and slider must be able to slide in the slot or the mechanism
would jam. This causes problems in the solution because of the sliding link and
this is covered later under Coriolis acceleration.
v The main point is that the motion produced is anything but simple harmonic motion
and at any time the various parts of the mechanism have a displacement, velocity
and acceleration.
v The acceleration gives rise to inertia forces and this puts stress on the parts in
addition to the stress produced by the transmission of power.
v For example the acceleration of a piston in an internal combustion engine can
be enormous and the connecting rod is subjected to high stresses as a result of
the inertia as well as due to the power transmission.
v You will find in these studies that the various parts are referred to as links and it
can be shown that all mechanisms are made up of a series of four links.
v The basic four bar link is shown below. When the input link rotates the output link
may for example swing back and forth. Note that the fourth link is the frame of
the machine and it is rigid and unable to move.
v With experience you should be able to identify the four bar chains in a mechanism.
All the links shown are rigid links which means they may push or pull. It is possible
to have links made of chain or rope which can only pull.

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 Figure 3

2. DISPLACEMENT, VELOCITY AND ACCELERATION

v All parts of a mechanism have displacement, velocity and acceleration. In the

tutorial on free vibration, a mechanism called the Scotch Yoke was examined
in order to explain sinusoidal or harmonic motion.
v The wheel turns at a constant speed and the yoke moves up and down.

Figure 4

It was shown that the displacement ‘x’, velocity ‘v’ and acceleration ‘a’ of point p was
given as follows. Angle = t
Displacement x = R sin ( t).
Velocity v = dx/dt = R cos( t)
Acceleration a = dv/dt = - 2Rsin( t)

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 v The values can be calculated for any angle or moment of time. The acceleration
could then be used to calculate the inertia force needed to accelerate and
decelerate the link.
v Clearly it is the maximum values that are needed. Other mechanisms can be
analyzed mathematically in the same way but it is more difficult.
v The starting point is to derive the equation for displacement with respect to angle
or time and then differentiate twice to get the acceleration.
v Without the aid of a computer to do this, the mathematics is normally much too
difficult and a graphical method should be used as shown later.

3. VELOCITY DIAGRAMS

This section involves the construction of diagrams which needs to be done accurately
and to a suitable scale. Students should use a drawing board, ruler, compass,
protractor and triangles and possess the necessary drawing skills.

ABSOLUTE AND RELATIVE VELOCITY

An absolute velocity is the velocity of a point measured from a fixed point (normally the
ground or anything rigidly attached to the ground and not moving). Relative velocity is the
velocity of a point measured relative to another that may itself be moving.

TANGENTIAL VELOCITY

Consider a link A B pinned at A and revolving about A at angular velocity . Point B

moves in a circle relative to point A but its velocity is always tangential and hence at

90o to the link. A convenient method of denoting this tangential velocity is (vB)A
meaning the velocity of B relative to A. This method is not always suitable.

Figure 5

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RADIAL VELOCITY

v Consider a sliding link C that can slide on link AB. The direction can only be radial
relative to point A as shown.
v If the link AB rotates about A at the same time then link C will have radial and
tangential velocities.

Figure 6

v Note that both the tangential and radial velocities are denoted the same so
the tags radial and tangential are added.
v The sliding link has two relative velocities, the radial and the tangential. They are
normal to each other and the true velocity relative to A is the vector sum of both
added as shown.
v Note that lower case letters are used on the vector diagrams. The two
vectors are denoted by c1 and c2. The velocity of link C relative to point A is the
vector a c2.

Figure 7

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CRANK, CONNECTING ROD AND PISTON

Consider this mechanism again. Let’s freeze the motion (snap shot) at the position
shown. The diagram is called a space diagram.

Figure 8

v Every point on every link has a velocity through space. First we label the centre
of rotation, often this is the letter O. Point A can only move in a tangential direction
so the velocity of A relative to O is also its absolute velocity and the vector is
normal to the crank and it is designated (vA)O. (Note the rotation is
anticlockwise).
v Now suppose that you are sat at point A and everything else moves relative
to you. Looking towards B, it would appear the B is rotating relative to you (in
reality it is you that is rotating) so it has a tangential velocity denoted (VB) A.
v The direction is not always obvious except that it is normal to the link.Consider the
fixed link OC. Since both points are fixed there is no velocity between them so
(vC) O = 0.
v Next consider that you at point C looking at point B. Point B is a sliding link and
will move in a straight line in the direction fixed by the slider guides and this is
velocity (vB) C. It follows that the velocity of B seen from O is the same as that
seen from C so (vB) C = (vB) O.

v The absolute velocity of B is (vB) C = (vB) O and this must be the vector sum of
(VA) O and (vB) A and the three vectors must form a closed triangle as shown. The

velocity of the piston must be in the direction in which it slides (conveniently

horizontal here). This is a velocity diagram.

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 Figure 9
v First calculate the tangential velocity (vA)O from v = x radius = x OA

v Draw the vector o - a in the correct direction (note lower case letters).
v We know that the velocity of B relative to A is to be added so the next vector ab starts
at point a. At point a draw a line in the direction normal to the connecting rod but of
unknown length.
v We know that the velocity of B relative and absolute to O is horizontal so the vector ob
must start at a. Draw a horizontal line (in this case) through o to intersect with the other
line. This is point b. The vectors ab and ob may be measured or calculated. Usually
it is the velocity of the slider that is required.
v In a design problem, this velocity would be evaluated for many different positions of the
crank shaft and the velocity of the piston determined for each position.
v Remember that the slider direction is not always horizontal and the direction of o - b
must be the direction of sliding.

WORKED EXAMPLE No.1

The mechanism shown has a crank 50 mm radius which rotates at 2000 rev/min.
Determine the velocity of the piston for the position shown. Also determine the angular
velocity of link AB about A.

Figure 10

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SOLUTION
v Note the diagrams are not drawn to scale. The student should do this using a
suitable scale for example 1 cm = 1 m/s.

v This is important so that the direction at 90o to the link AB can be transferred to
the velocity diagram.
v Angular speed of the crank = 2 N/60 = 2 x 2000/60 = 209.4 rad/s
(vA)O = x radius = 209.4 x 0.05 = 10.47 m/s.

First draw vector oa. (Diagram a)

v Next add a line in the direction ab (diagram b)
v Finally add the line in the direction of ob to find point b and measure ob to get
the velocity.
(Diagram C).

Figure 11a Figure 11b Figure 11c

v The velocity of B relative to O is 7 m/s.

v The tangential velocity of B relative to A is the vector ab and this gives 9.2 m/s.
v The angular velocity of B about A is found by dividing by the radius (length of AB).
v for AB is then 9.2/0.09 = 102.2 rad/s. (note this is relative to A and not an
absolute angular velocity)

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4.BAR CHAIN

v The input link rotates at a constant angular velocity 1. The relative velocity of each
point relative to the other end of the link is shown.
v Each velocity vector is at right angles to the link. The output angular velocity is 2 and
this will not be constant. The points A and D are fixed so they will appear as the
same point on the velocity diagram.
v The methodology is the same as before and best shown with another example.

Figure 12

WORKED EXAMPLE No. 2

Find the angular velocity of the output link when the input rotates at a constant speed of
500 rev/min. The diagram is not to scale.

Figure 13

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SOLUTION

First calculate 1.
1 = 2 x 500/60 = 52.36 rad/s.
Next calculate the velocity of point B relative to A.
(VB)A = 1 x AB = 52.36 x 1 = 52.36 m/s.
Draw this as a vector to an appropriate scale.

Figure 14a

Next draw the direction of velocity C relative to B at right angles to the link BC
passing through point b on the velocity diagram.

Figure 14 b Figure 14 c

v Next draw the direction of the velocity of C relative to D at right angles to link
DC passing through point a (which is the same as point d). Point c is where the
two lines intersect,
v Determine velocity cd by measurement or any other method. The velocity of point C
relative to D and is 43.5 m/s.
v Convert this into angular velocity by dividing the length of the link DC into it.
2 = 43.5/0.7 = 62 rad/s.

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4. ACCELERATION DIAGRAMS

v It is important to determine the acceleration of links because acceleration produces

inertia forces in the link which stress the component parts of the mechanism.
v Accelerations may be relative or absolute in the same way as described for velocity.
v We shall consider two forms of acceleration, tangential and radial. Centripetal
acceleration is an example of radial.

CENTRIPETAL ACCELERATION

v A point rotating about a centre at radius R has a tangential velocity v and angular
velocity and it is continually accelerating towards the centre even though it
never moves any closer. This is centripetal acceleration and it is caused by the
constant change in direction.
v It follows that the end of any rotating link will have a centripetal acceleration towards
the opposite end.

The relevant equations are: v= R a= 2 R or a = v2/R.

v The construction of the vector for radial acceleration causes confusion so the rules
must be strictly followed. Consider the link AB. The velocity of B relative to A is
tangential (vB)A.
v The centripetal acceleration of B relative to A is in a radial direction so a suitable
notation might be aR. It is calculated using aR = x AB or aR = v2/AB.

Note the direction is towards the centre of rotation but the vector starts at a and ends
at b1. It is very important to get this the right way round otherwise the complete diagram will be
wrong.

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 Figure 15

TANGENTIAL ACCELERATION

Tangential acceleration only occurs if the link has an angular acceleration rad/s2.
Consider a link AB with an angular acceleration about A.

Figure 16

v Point B will have both radial and tangential acceleration relative to point A. The true
acceleration of point B relative to A is the vector sum of them. This will require an extra
point. We will use b1 and b on the vector diagram as shown.

v Point B is accelerating around a circular path and its direction is tangential (at right
angles to the link). It is designated aT and calculated using aT = x AB.
v The vector starts at b1 and ends at b. The choice of letters and notation are arbitrary
but must be logical to aid and relate to the construction of the diagram.

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 Figure 17
WORKED EXAMPLE No.3

A piston, connecting rod and crank mechanism is shown in the diagram. The crank
rotates at a constant velocity of 300 rad/s. Find the acceleration of the piston and the
angular acceleration of the link BC. The diagram is not drawn to scale.

Figure 18
SOLUTION:

First calculate the tangential velocity of B relative to A.

(vB)A = x radius = 300 x 0.05 = 15 m/s.

Next draw the velocity diagram and determine the velocity of C relative to B.

Figure 19
From the velocity diagram (vC)B = 7.8 m/s
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v Next calculate all accelerations possible and construct the acceleration diagram to
find the acceleration of the piston.
v The tangential acceleration of B relative to A is zero in this case since the link has no
angular acceleration ( = 0).
v The centripetal acceleration of B relative to A

aR = 2x AB = 3002 x 0.05 = 4500 m/s2.

The tangential acceleration of C relative to B is unknown.

The centripetal acceleration of C to B
aR = v2/BC = 7.82 /0.17 = 357.9 m/s2.
The stage by stage construction of the acceleration diagram is as follows.

Figure 20a Figure 20b Figure 20c

v First draw the centripetal acceleration of link AB (Fig.a). There is no tangential

acceleration so designate it ab. Note the direction is the same as the direction of
the link towards the centre of rotation but is starts at a and ends at b.
v Next add the centripetal acceleration of link BC (Figure b). Since there are two
accelerations for point C designate the point c1. Note the direction is the same

as the direction of the link towards the centre of rotation.

v Next add the tangential acceleration of point C relative to B (Figure c). Designate
it c1 c. Note the direction is at right angles to the previous vector and the length is

unknown. Call the line a c line.

v Next draw the acceleration of the piston (figure d) which is constrained to be in
the horizontal direction. This vector starts at a and must intersect the c line.
Designate this point c.

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 Figure 20d

v The acceleration of the piston is vector ac so (aC) B = 1505 m/s2. The tangential

acceleration of C relative to B is c1 c = 4000 m/s2.

v At the position shown the connecting rod has an angular velocity and acceleration
about its end even though the crank moves at constant speed.
v The angular acceleration of BC is the tangential acceleration divided by the length
BC.
2
(BC) = 4000 / 0.17 = 23529 rad/s .

WORKED EXAMPLE No.4

The diagrams shows a “rocking lever” mechanism in which steady rotation of the
wheel produces an oscillating motion of the lever OA. Both the wheel and the lever are
mounted in fixed centers. The wheel rotates clockwise at a uniform angular velocity ( )
of 100 rad/s. For the configuration shown, determine the following.

Figure 21
(i) The angular velocity of the link AB and the absolute velocity of point A.
(ii) The centrifugal accelerations of BC, AB and OA.
(iii)The magnitude and direction of the acceleration of point A.
The lengths of the links are as follows.
BC = 25 mm AB = 100 mm OA = 50 mm OC = 90 mm

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SOLUTION

The solution is best done graphically. First draw a line diagram of the mechanism to
scale. It should look like this.

Figure 22

Next calculate the velocity of point B relative to C and construct the velocity diagram.

Figure 23

(vB)C = x radius = 100 x 0.025 = 2.5 m/s

Scale the following velocities from the diagram.
(v A)O = 1.85 m/s {answer (i)} (vA)B = 3.75 m/s
Angular velocity = tangential velocity/radius
For link AB, = 3.75/0.1 = 37.5 rad/s. {answer (i)} Next calculate all the accelerations
possible.

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 v Radial acceleration of BC = 2 x BC = 1002 x 0.025 = 250 m/s2.

v Radial acceleration of AB = v2/AB = 3.752/0.1 = 140.6 m/ s2.

v Check same answer from 2 x AB = 37.52 x 0.1 = 140.6 m/ s2.

v Radial Acceleration of OA is v2/OA = 1.852/0.05 = 68.45 m/ s2.

Construction of the acceleration diagram gives the result shown.

Figure 24

The acceleration of point A is the vector o- a shown as a dotted line. Scaling this

we get 560 m/s2.

WORKED EXAMPLE No. 5

Find the angular acceleration of the link CD for the case shown.

Figure 25

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SOLUTION

First calculate or scale the length CB and find it to be 136 mm.

Next find the velocities and construct the velocity diagram. Start with link AB as this
has a known constant angular velocity.
(vB)A = x radius = 480 x 0.08 = 38.4 m/s

Figure 26

1. Next calculate all the accelerations possible.

2. The centripetal acceleration of B to A is 38.42/0.08 = 18 432 m/s2
3. The centripetal acceleration of C to D is 152/0.16 = 1406 m/s2
4. The centripetal acceleration of C to B is 312/0.136 = 7066 m/s2.
5. We cannot calculate any tangential acceleration at this stage.
6. The stage by stage construction of the acceleration diagram follows.
7. First draw the centripetal acceleration of B to A (Figure a). There is no tangential to
add on).

Figure 27a Figure 27b Figure 27c

8. Next add the centripetal acceleration of C to B (figure b)

9. Next draw the direction of the tangential acceleration of C to B of unknown
length at right angles to the previous vector (figure c). Designate it as a c line.
10. We cannot proceed from this point unless we realize that points a and d are the
same (there is no velocity or acceleration of D relative to A). Add the
centripetal acceleration of C to D (figure d). This is 1406 m/s2 in the direction of
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 link CD. Designate it d c2.

Figure 27d Figure 27e

Finally draw the tangential acceleration of C to D at right angles to the previous

vector to intersect the c line (figure e).

From the diagram determine c2 c to be 24 000 m/s2. This is the tangential acceleration of
C to D. The angular acceleration of the link DC is then:

2
(CD) = 24000/0.16 = 150 000 rad/s in a clockwise direction.

Note that although the link AB rotates at constant speed, the link CD has angular
acceleration.

WORKED EXAMPLE No. 6

The same arrangement exists as shown for example 5 except that the link AB is
decelerating at 8000 rad/s2 (i.e. in an anticlockwise direction). Determine the acceleration
of the link CD.

SOLUTION

The problem is essentially the same as example 5 except that a tangential

acceleration now exists for point B relative to point A. This is found from

aT = x AB = 80000 x 0.08 = 6400 m/s2

v The direction is for an anticlockwise tangent. This is vector b1 b which is at right angles
to a b1 in the appropriate direction. The new acceleration diagram looks like this.

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 Figure 28
Scaling off the tangential acceleration c2 c we get 19 300 m/s2. Converting this into the

angular acceleration we get

= 19 300/0.16 = 120 625 rad/s2 in a clockwise direction.

5. INERTIA FORCE

One of the reasons for finding the acceleration of links is to calculate the inertia force
needed to accelerate or decelerate it. This is based on Newton’s second law.

Force = mass x acceleration F=Ma

Torque = moment of inertia x angular acceleration T=I

WORKED EXAMPLE No.7

A horizontal single cylinder reciprocating engine has a crank OC of radius 40 mm

and a connecting rod PC 140 mm long as shown.
The crank rotates at 3000 rev/min clockwise. For the configuration shown,
determine the velocity and acceleration of the piston.

The sliding piston has a mass of 0.5 kg and a diameter of 80 mm. The gas pressure
acting on it is 1.2 MPa at the moment shown. Calculate the effective turning moment
acting on the crank. Assume that the connecting rod and crank has negligible inertia and
friction.

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 FIGURE: 29
SOLUTION
Draw the space diagram to scale.

Figure 30

The moment arm should be scaled and found to be 34 mm (measured at right angles
to the connecting rod PC.
Calculate the velocity of C relative to O.
= 2 N/60 = 2 x 3000/60 = 314.16 rad/s
(VC)O = x radius = 314.16 x 0.04 = 12.57 m/s
Draw the velocity diagram.

Figure 31
From the velocity diagram we find the velocity of the piston is 11 m/s.
Next calculate all the accelerations
possible. Point C only has a radial
acceleration towards O
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Radial acceleration of C is v2/radius = 12.572/0.04 = 314.16 m/s2
Point P has radial and tangential acceleration relative
to C. Tangential acceleration is unknown.

Radial acceleration = (v P)C2/CP = 92/0.14 = 578.57 m/s2

Now draw the acceleration diagram and it comes out like this.

Figure 32

The acceleration of the piston is 770 m/s2.

Now we can solve the forces.

Pressure force = p x area = 1.2 x 106 x x 0.082/4 = 6032 N and this acts left to right.
Inertia force acting on the piston = M a = 0.5 x 770 = 385 N and this must be provided
by the pressure force so the difference is the force exerted on the connecting rod. Net
Force = 6032 – 385 = 5647 N.

The connecting rod makes an angle of 11o to the line of the force (angle scaled from
space diagram). This must be resolved to find the force acting along the line of the
connecting rod.

Figure 33

The force in the connecting rod is 5637 cos 11o = 5543 N.

This acts at a radius of 34 mm from the centre of the crank so the torque provided by the
crank is
T = 5545 x 0.034 = 188.5 N m.

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