DET
Mechanical Engineering
Dynamics
(Higher)
8084
.
November 2000
HIGHER STILL
DET
Mechanical Engineering
Dynamics
(Higher)
Support Materials


CONTENTS
Teacher/lecturer notes:
Learning outcomes covered in the unit
Delivery
Details of possible starting points
Assessment
Resource materials
Student materials:
Integrative learning approaches
Tutorial examples for outcome 1
Tutorial examples for outcome 2 and 3
Practical example for outcome 4
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TEACHER AND LECTURER NOTES
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LEARNING OUTCOMES COVERED IN THIS UNIT
The outcomes and PCs have been listed in sequence for the unit. After each PC a note
has been included to indicate the extent of knowledge, understanding and application
which would require to be imparted and practiced.
Outcome 1: Solve problems using Newton's Second Law applied to linear and
angular motion
Performance Criteria
a)
The elements of motion are accurately described in accordance with
established theory.
Initial and final states are identified and the correct letters are used to describe
each element. Both linear and angular motion systems should be studied. The
radian and RPM should be discussed.
b)
The interrelationship between the elements of motion are analysed in
accordance with established theory.
Solve problems in linear and angular motion using Newton’s Second Law of
Motion. Explain the relationships between displacement, velocity and
acceleration for linear and angular systems. Use either a velocity/time
diagram approach or the equations of motion to relate the elements of motion
to a range or applications. When confidence in the separate systems has been
gained then combined systems that are limited to one linear and one angular
component can be introduced.
c)
Problems relating to systems affected by uniform acceleration are solved
correctly.
Apply Newton's law relating Force and Torque to acceleration and introduce
polar moment of Inertia. Show the determination of applied force or torque
required to produce stated changes in motion to linear, angular and combine
systems.
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Outcome 2: Apply work power transfer theory to linear and angular systems
Performance Criteria
a)
Work and power quantities are defined correctly in accordance with the
established theory.
Explain the work and power requirements again for linear, angular and
combined systems. Then use the same range of systems involving linear,
angular and combined forms of motion to determine work and power
requirements.
b)
Work and power calculations are performed correctly.
Show how problems are solved that are associated with work and power
c)
Problems related to systems affected by constant and variable applied forces
are solved accurately.
Applications that are studied should include average and instantaneous power
requirements.
Outcome 3: Analyse the kinetics of motion using the energy balance approach
Performance Criteria
a)
The influences affecting kinetic energy are defined correctly in accordance
with the established theory.
Revise potential, kinetic and strain energy formulae
b)
Energy balance principles are applied to motion systems correctly.
Develop energy balance equations. Apply the equations initially to linear
systems, then use polar moment of Inertia and develop energy balance
equations to angular systems. Finally develop energy balance equations to
combined systems
c)
Problems relating to systems affected by uniform acceleration are solved
accurately.
Again the same range of systems as used in the first two outcomes should be
covered. The energy balance approach to the analysis and calculations
involved in the solution of a range of problems should be used. These can
build for linear problems to angular motion problems and then when
confidence is established in the separate systems combined systems can be
introduced.
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Outcome 4: Analyse uniform circular motion force systems
a)
Centripetal acceleration is defined correctly in terms of the requirement for
circular motion in accordance with the established theory.
Develop the theory and equation for centripetal force. The size and scale of
centripetal force can be explained using the idea of a mass whirling on the end
of a piece of string this can then be scaled up to the moons motion around the
Earth.
b)
An experiment to verify the theory of centripetal acceleration is performed
satisfactorily.
Solve problems containing centripetal force and acceleration elements. At this
stage many examples of circular motion can be found but care should be take
so that they can be easily assimilated by the students.
c)
Mechanisms affected by centripetal acceleration are described correctly.
Perform an experiment to confirm the equation for centripetal force and write
a report to describe the experimental procedure. The report should contain an
explanation of how the standard apparatus was used, the theory being verified,
experimental results, an analysis of the results, discussion of the results and a
valid conclusion. Then a selection of various diagrams could be drawn for the
basic centripetal acceleration situations such as cars going around curves and
centrifugal clutches.
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DELIVERY
Initially it must be stressed that this unit contains many new concepts. For students to
acquire and use these concepts successfully time must be allowed for them not only to
become familiar with new words and the letters that represent them but also the
difficult underlying ideas. It is likely that most students attempting this unit will have
already achieved a Standard Grade Mathematics and experience will have been
attained in the mathematical manipulation of formulae required for the unit. Students
should have worked at either Structures Intermediate 2 or Technological Studies
Intermediate 2 so that they can represent devices and force systems. They should also
be able to make diagrammatic representations from text descriptions. They should
also have a working knowledge of work and power calculations along with the basic
ideas behind Newton’s Second Law. Students should also have been exposed to the
various energies that a body can poses although they may not have come across both
angular forms of kinetic and potential energy.
Outcome 1 should be introduced initially with equal effort given to graphical and
analytical approaches. The emphasis should be on understanding the concept of
displacement/time graphs lead one to velocity and that velocity/time graphs lead to
acceleration. The opportunity should be taken to showing magnitude, direction, sense
and position of application of the forces that cause changes to motion. The
application of the equations of motion and Newton’s Second Law will then follow.
Care should be taken to ensure that the students are able to use the appropriate letters
which represent the variables. Time should be allowed for them to become familiar
with the letters, equations and their manipulation in the respective formulae. When
linear motion has been successfully completed the similarity between linear and
angular motion can be demonstrated. Great care should be taken here to ensure that
the use a radians is understood and that students can convert RPM and radian
quantities. A series of questions should be designed to give the students a clear
progression through linear only, then angular only and finally on to simple combined
systems.
Outcome 2 requires that students should have a good working knowledge of the basic
work done and power calculations from the Intermediate 2 level work. The same
examples from Outcome 1 can be revisited and work and power calculations for them
can be added to the earlier solutions.
Outcome 3 offers a different approach to solving these types of problems. Great care
is needed, however, at the point were angular quantities are discussed as these often
prove to be very difficult to conceptualise. Polar moment of Inertia is particularly
difficult for the students. The same set of tutorial questions can be extended even
further, using this alternative approach to solve the same quantities as in the first
outcome.
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The move to circular motion in Outcome 4 can only be attempted after the students
are confident with the work in the first three outcomes. Centripetal force is another
very difficult idea to conceptualise. This is often introduced by asking the students to
imagine that they are travelling in a car around a fast bend and saying which way they
are ‘pushed’ by this. Then they can be lead into the idea that they ‘push’ into the
centre of the corner so that they can still sit upright. This could then lead onto a
diagram that would be converted into a conventional representation and then the
appropriate letters and symbols added. From here the equations can be introduced.
The assessment for this section is based around practical work therefore easing the
students’ way into this difficult area.
Examples of useful equipment, tutorial questions and possible sources of further
useful materials are indicated later in this pack.
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STARTING POINTS FROM WHICH TO COMMENCE DELIVERY
This unit does not build upon a previous Higher Still unit. The entry point will
depend upon the student’s backgrounds. A general guideline would be that the
majority of students should have studied some of these areas in a school based science
class. This would not have included any mention of rotary motion, however in most
cases students would have covered the basic ideas behind linear motion. This usually
would have been looking at motion in a straight line often using distance time graphs
done with a ticker tape timer. The students would have also covered the concepts of
different types of energy and their interconversion. In some cases this could have
included the basic Kinetic and Potential Energy equations.
Using this as a starting point means that some students may appear to be able to use
the correct words but the underlying concepts are very little understood. It is
therefore important that the time should be made available for the candidates to fully
grasp the underlying principles of linear systems before any movement toward the
angular systems is attempted.
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ASSESSMENT
Assessment packs for this unit are available from the National Assessment Bank and
should be used in conjunction with the Mechanical Engineering Arrangement
Document and Subject Guide.
The assessment packs suggests how and when assessment tasks should be
administered, the recommended duration of each assessment and the conditions under
which assessment should take place.
(Part of the section – organisation and conditions for assessment has been reproduced
from the first assessment pack).
Using instrument(s) of assessment
There are four assessments in this unit each one can be administered as an end of
topic test. However, the assessment for the fourth learning outcomes in designed to
be of a practical nature.
Assessment
It is possible to combine the assessment of the first three outcomes into one
instrument to be attempted once learning on the three outcomes has reached an
appropriate stage. However, the approach that has been taken is to use end of section
assessments through the first three outcomes. The fourth outcome instrument of
assessment requires the student to devise and perform an experiment using a standard
piece of apparatus and to evidence this by writing and submitting a report, then
describe a simple system affected by centripetal acceleration such as a car moving
around a curve.
INSTRUMENT OF
ASSESSMENT
OUTCOME & PC
COVERED
TYPE OF INSTRUMENT
1
1 a, b, c
Structured question
2
2 a, b, c
Structured question
3
3 a, b, c
Structured question
4
4 a, b, c
Report
+
Structured question
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Suggested timing and duration of assessments
The recommended forty hours should be divided into approximately
OUTCOME
TEACHING TIME
ASSESSMENT TIME
Outcome 1
14 hours 30 mins
30 mins
Outcome 2
4 hours 30 mins
30 mins
Outcome 3
9 hours 30 mins
30 mins
Outcome 4
7 hours 30 mins
2 hours
+
30 mins
With assessments administered as end of topic tasks. The 2 hours for Outcome 4
allows for the experiment and report requirement. Contents of the outcomes are set
out in the tables below.
OUTCOME 1
KNOWLEDGE AND UNDERSTANDING TO BE ASSESSED
Newton's law applied to linear and angular motion
1. Motion relationships between displacement, velocity and acceleration for linear angular and
combined systems.
2. Newton's law relating Force and Torque to acceleration I and introduce polar moment of Inertia.
CONTEXTS, APPLICATIONS, ILLUSTRATIONS
1. Use either a Velocity/Time diagram approach or the equations of motion to relate the elements of
motion to a range or applications.
2. Determination of applied force or Torque required to produce stated changes in motion to, linear,
angular and combine systems. Application can be limited to one combined system for assessment
purposes.
INSTRUMENT OF ASSESSMENT 1A
This assessment covers all PCs.
Where PC (a) asks for all the elements of motion to be accurately defined is implied because the
student has to have this knowledge to be able to do this question.
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OUTCOME 2
KNOWLEDGE AND UNDERSTANDING TO BE ASSESSED
Work and power determination
1. Work and power requirements again for linear, angular and combined systems.
2. Solve problems associated with work and power.
CONTEXTS, APPLICATIONS, ILLUSTRATIONS
The same range of systems involving linear, angular and combined forms of motion is used to
determine work and power requirements. Applications should include average and instantaneous
power requirements.
INSTRUMENT OF ASSESSMENT 2A
This assessment covers all P C’s.
OUTCOME 3
KNOWLEDGE AND UNDERSTANDING TO BE ASSESSED
Energy balance applied to linear and angular motion
1. Revise potential, kinetic and strain energy formulae.
2. Develop energy balance equations to linear systems.
3. Use polar moment of Inertia and develop energy balance equations to angular systems.
4. Develop energy balance equations to combined systems.
CONTEXTS, APPLICATIONS, ILLUSTRATIONS
Again the same range of systems as used in the first two outcomes should be covered.
The energy balance approach to the analysis and calculations involved in the solution of a range of
problems can be evidenced by one application involving a combined linear/angular system. The forms
of energy can be limited to those of Potential and Kinetic Energies.
INSTRUMENT OF ASSESSMENT 3A
This assessment covers all PCs.
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OUTCOME 4
KNOWLEDGE AND UNDERSTANDING TO BE ASSESSED
Centripetal Acceleration
1. Develop the theory and equation for centripetal force.
2. Solve problems containing centripetal force and acceleration elements.
3. Perform an experiment to confirm the equation for centripetal force and write a report to describe
the experimental procedure.
4. Describe mechanisms that are affected by centripetal acceleration.
CONTEXTS, APPLICATIONS, ILLUSTRATIONS
The report should contain how the standard apparatus was used, the theory being verified,
experimental results, analysis of results, discussion of results and a valid conclusion.
INSTRUMENT OF ASSESSMENT 4A
This assessment covers all PCs.
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Reassessment
Time is allowed within units for assessment and reassessment of outcomes. Where a
student has not attained the standard necessary to pass a particular outcome or
outcomes, they should have the opportunity to be reassessed. Reassessment should
focus on the outcome(s) concerned and, as a general rule, should be offered on a
maximum of two occasions following further work concentrated on areas of
difficulty. Evidence from the original unit assessment assists teachers and lecturers to
identify why an individual student has failed to achieve a particular outcome and to
plan focused support for learning.
General guidelines for reassessment
 students may be given the opportunity to orally correct minor errors and these oral
responses recorded on the original documentation
 where part of a complete assessment instrument is required to be repeated this
should be done by the student as soon as possible following the original
unsuccessful attempt and again the new evidence should be attached to the original
documentation.
The reassessment could take the form of questions with a similar standard. Those in
the other two assessment packs would be suitable.
The conditions under which assessment takes place
Arrangement documents refer to assessment being carried out under controlled
conditions to ensure reliability and credibility. For the purposes of internal
assessment, this means that assessment evidence should be compiled under
supervision to ensure that it is the students' own work. A teacher, lecturer, invigilator
or other responsible person, for example, a workplace provider, may carry out
supervision.
All three assessments should be 'closed book' and carried out under controlled
conditions. Students should be allowed to use data booklets and non programmable
calculators. Those responsible for supervision of the assessments should ensure that
unmarked copies of the date booklets are available for student use.
During presentation of the learning materials for each outcome students should be
informed of the form and nature of each assessment also that it is the students
knowledge and understanding that is being assessed. Before a particular assessment
is attempted students must have experienced the answering of similar types of
question via dedicated learning materials, homework or tutorial sessions. The student
during delivery of the unit should acquire this experience.



data books should be available during all tests
non programmable calculators may be used
the three instruments of assessment should be unseen and of the closed book type.
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Examples of instruments of assessment are also given in the assessment pack along
with solutions, possible methods or recording student performance, advice on
reassessment and student guidance.
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RESOURCE MATERIALS
Useful software
Software is available to solve many of the types of problems encountered throughout
the unit and could be used to consolidate learning. The temptation to use software as
the main vehicle of learning should be resisted as the routine calculations such as
resolution of forces and application of the principle of moments are generally
completed automatically without any interaction with the user. Software is however,
very useful for remedial work were students can work independently.
The sources of this type of software are continually changing and each new example
should be looked at carefully before it is introduced.
Sources of information and course work materials
This unit although written to satisfy the requirements of the Higher Mechanical
Engineering Course, contains many elements which have already been covered by
earlier SQA modules. Materials already developed for these earlier modules can
provide a source of useful examples and applications.
The Subject Guide Engineering 2: Mechanical Engineering – Appendix 4 maps the
content of this unit with the relevant SQA modules. Some of the centres will present
this unit as part of an Mechanical Engineering Higher course, intend to produce
comprehensive teaching packs for group and individual study purposes as a long term
development. These packs will be made available for purchase by centres that may
wish to use this service.
The course work materials contained in this pack were developed from first principles
but similar material is available from most textbooks on Mechanical Engineering
Science or Applied Mechanics. The unit was developed from the module dynamics
and therefore a great deal of the information may already be available in note form
from this source.
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STUDENT MATERIALS
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INTEGRATIVE LEARNING APPROACHES
Tutorial examples
The following tutorials are intended to be used directly after related engineering
principles have been described and discussed. The layout of the tutorial sheets may
require further development of these principles to prepare the students to efficiently
attempt to solve the problems contained in the exam. Students should be encouraged
to set out their answers in logical and repeatable steps. For example, the simple
question below shows a suitable layout.
Example question
Newton’s second law
A vehicle of mass 850kg is moving at 28kmh-1. If its speed is reduced to 8kmh-1 in a
distance of 55m, what is the average braking force?
Example answer
Step 1: all units to SI form:
Initial velocity, u = 28 kmh-1 =
28 x 100
60 x 60
= 7.78ms-1
Final Velocity, v = 8 kmh-1 =
8 x 1000
60 x 60
= 2.22ms-1
Step 2: work out the deceleration from the equation
v2 = u2 +2as
Rearranging we get:
a
=
v2 - u2
2xs
(2.22)2 – (7.78)2
2 x 55
=
4.93 – 90.53
110
=
-0.505ms-2
Deceleration = a =-0.505 ms-2
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Step 3: calculate the braking force using:
F = ma
F = 850 x 0.505
=
429.25
Average braking force
=
429.25 N
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TUTORIAL EXAMPLES FOR OUTCOME 1
Outcome 1: Solve problems using Newton's Second Law applied to linear and
angular motion
Performance Criteria
a)
the elements of motion are accurately described in accordance with established
theory;
b)
the interrelationship between the elements of motion are analysed in
accordance with established theory;
c)
problems relating to systems affected by uniform acceleration are solved
correctly.
The first part of this tutorial should lead through all three criteria for linear motion.
The second part should go through all three criteria for angular motion. The third part
of the tutorial should go through all three criteria for combined systems.
The examples shown are not intended to be entire tutorial sheets but examples of the
build up from easy questions through to harder ones in each section.
The solutions rely on the student building up a systematic method for the solution of
problems. This system could include writing down all the information contained
within the question, converting were necessary to SI units. Drawing a diagram of the
system of changes and writing down the equations they intend to use. The initial and
final states should be identified and the correct engineering letters used to describe
each element. The radian and RPM should be discussed when moving from linear to
angular systems and then some time spent converting RPM to radians and radians to
degrees and vice versa.
The first problems should be using Newton’s Second Law of Motion applied to linear
motion (and then angular). This could be achieved by looking at the relationships
between displacement, velocity and acceleration for linear systems and then angular
systems. Using both a velocity/time diagram approach and the equations of motion to
relate the elements of motion to a range or applications. Newton's law relating force
and torque to acceleration and the introduction of polar moment of inertia is an
advanced concept. Only when confidence in the separate systems has been gained
then combined systems should be tackled and but these should be limited to one linear
and one angular component. The analysis should go deep enough to allow the
determination of applied force or torque required to produce stated changes in motion
to linear, angular and combine systems.
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1. A train starts from rest and accelerates uniformly to a velocity of 50km h-1 in 15
seconds. Determine:
a)
the acceleration of the train;
b)
the distance traveled by the train during this time.
Answer one.
Known information:
 u = 0 ms-1
 v = 40 km h-1  13.89 ms-1
 t = 15 seconds
 a = ? using
a)
vu
13.89  0

 0.926ms 2
t
15
Acceleration of the train equals 0.926 ms-2 in the direction of the motion.
a
b)
13.89  15
(u  v)t

 104.175m
2
2
The distance travelled by the train in this time is 104.18m.
s
Using a velocity/time graph can equally well solve this problem.
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2. A fault in the braking system causes a lift, in a tall building, to fall. If the final
velocity of the lift is 62 ms-1. Determine:
a)
the height from which the lift fell;
b)
the time taken for the lift to fall this distance.
Known information:
 u = 0 ms-1
 v = 62 ms-1
 a = g = 9.81 ms-2
v2
=
u2 + 2as 
s
=
v2 – u2
2xa
622 – 0
2 x 9.81
=
=
195.92m
The height from which the lift fell = 195.92m.
The time taken to fall this distance can be found from:
v
=
u + at 
t
=
v–u
a
=
62 – 0
9.81
=
6.32s
The time taken to fall this distance = 6.32 seconds.
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3. Imagine that it is possible for a child, propelling a scooter in a play area, to
accelerate at the rate of 0.53 ms-2. Determine:
a)
how far the child will have travelled 5 seconds after reaching the velocity of
2.65 ms-1;
b)
how fast will the child then be travelling;
c)
how far did the child travel between the velocities of 2.65 ms-1 and 3.2 ms-1.
Known information:
 u = 2.65 ms-1
1
 v=?
s  ut  at 2  s  2.65  5  0.5  0.53  52  19.875m
2
 a=?
 s=?
 t=5s
The distance travelled for the five seconds after reaching 2.65 ms-1 is 19.88m.
Using the same information we get:

v  u  at  v  2.65  0.53  5  5.30 ms-1
The final velocity attained 5 seconds after reaching the velocity of 2.65 ms-1 is 5.3
ms-1
For part c) we can add the final velocity v = 3.2 ms-1
v 2  u 2  2as  s 
v2  u2
3.2 2  2.652

 3.035m
2a
2  0.53
The distance travelled between the two velocities is 3.04m.
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4. A vehicle of mass 8.4 tonne is moving along a level road at 36km h-1. If the
vehicle’s velocity is uniformly increased to 57km h-1 over a distance of 240m,
what is the average accelerating force?
Known information:

Mass of vehicle
m = 8.4 tonne
=
8400 kg

Initial velocity
u = 36 km h-1
=
36 x 103 = 10ms-1
60 x 60

Final velocity
v = 57 km h-1
=

57  10 3
 15.83 ms 1
60  60
Displacement
s = 240m
Firstly we must find the acceleration from:
v 2  u 2  2as  a 
v2  u2
15.832  10 2

 0.314 ms 2
2s
2  240
Acceleration is 0.314 m s-2
Now using F + ma
F = 8400 x 0.314 = 2637.6 N
Therefore the accelerating force is 2.64kN.
The next set of questions is for angular motion.
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5. A pulley has a diameter of 300mm and it rotates at 240 revs/min. It drives a belt
that passes round the rim. Calculate the linear velocity of the belt in metres per
second.
Known information:
Angular velocity in rad s-1

Linear velocity from the expression
240  2
 8 rad s 1
60
V=r
Note: r should be in m and is half the diameter.
v = 0.15 x 8 = 3.77 ms-1
The linear velocity of the belt = 3.77 m s-1.
6. Starting from rest a road vehicle having wheels 0.7 m diameter reaches a velocity
of 15 m s-1 in 20 seconds. Calculate the angular acceleration of the wheels.
Linear acceleration of the wheels
a
From the expression:
v  u 15  0

 0.75 ms 2
t
20
a=r

a 0.75

 2.14 ms 2
r
.35
The angular acceleration  = 2.14 rad s-2.
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7. A wheel rotating at 9.6 rads-1 accelerates at a constant 0.5ms-2 for half a minute.
Determine:
a)
the angular velocity after the half minute;
b)
the angular displacement after the half minute.
From the equation:
 2  1  t  9.6  0.5  30  24.6 rad s 1
The angular velocity after half a minute is 24.6 rad s-1
From the equation:
1
2
   1t  t 2  9.6  30  0.5  0.5  30 2  513 rad
The angular displacement after half a minute is 513 rad.
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8. A flywheel starting from rest increases its speed to 300 RPM at a uniform rate. If
it turns through 80 revolutions while its speed increases from 240 RPM to 300
RPM. Find:
a)
the angular acceleration in rads-2;
b)
the time taken to reach maximum speed from rest.
The total number of revolutions made in reaching maximum speed from rest.
Known information:
Therefore:
rad s-2
The angular acceleration is 0.3534 rad s-2.
The time taken to reach the maximum speed from rest is given by:
s
The time take to go from rest to 300 RPM is 88.9 seconds.
The total angular displacement can be found from:
rad
To convert from radians to revs 2 radians = 1 rev
Total revs = 1396.4 / 2 = 222.25
Total number of revolutions of the flywheel is 222.25.
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9. A flywheel has a mass of 1400kg and a radius of gyration of 1.2m. When it is
running at 300 RPM the driving torque is removed and the flywheel comes to rest
in 80 secs. Calculate:
a)
the frictional torque;
b)
the work done against friction per revolution, assuming that the friction is
constant through out the motion.
Known information:
angular deceleration  = /8 rad s-2
Frictional or decelerating TORQUE
T = I = Nm
The frictional Torque is 791.68 Nm.
The work done per rev against friction
W.d.
=T
= 791.68 x 1 x 2 
W.d.
= 4974 J
The work done per rev against friction is 4974 J.
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Combined systems
10. A load is raised vertically by a cable wound round a drum.
From rest, the drum accelerates uniformly to a maximum
angular velocity of 15 rad s-1 in 8 seconds. The friction
torque at the drum shaft bearings is constant at 25O Nm.
Determine:
the magnitude of the uniform angular acceleration of
the drum;
a)
the tension in the cable during acceleration;
b)
the total torque required at the drum shaft during
acceleration.
To find the angular acceleration of the drum:
rad s-2
Angular acceleration of the drum = 1.875 rad s-2.
To find the tension in the cable we have to convert the angular to linear quantities.
From the expression
a=r
a = 1.4 x 1.875
a = 2.625 ms-2
Linear acceleration of the load is 2.625 m s-2.
The tension in the cable whilst accelerating upwards is the combination of the tension
to overcome gravity plus the tension to accelerate the mass upwards.
F = mg + ma
F = 400 ( 9.81 + 2.625)
F = m (g+a)
F = 4974 N
The tension in the cable during the acceleration is 4974 N.
DET: Mechanical Engineering: Dynamics (Higher)
32
The total Torque at the shaft is a combination of three factors firstly the torque to
balance the tension in the cable secondly the torque to overcome the drum inertia and
thirdly the torque required to overcome frictional resistance at the bearings.
Torque required
at the shaft
during acceleration
= torque to balance
tension in the
cable
= ( f x r)
= (1974 x 1.4)
+ torque to
overcome
drum inertia
+ (Ix)
+ (700 x 1.875) +
250
+ frictional
torque at
the bearings
+ friction torque
= 8526.1 Nm
Total torque required at the drum shaft during the acceleration = 8526 Nm.
DET: Mechanical Engineering: Dynamics (Higher)
33
TUTORIAL EXAMPLES FOR OUTCOME 2 AND 3
Outcome 2: Apply work power transfer theory to linear and angular systems
Performance Criteria
a)
work and power quantities are defined correctly in accordance with the
established theory;
b)
work and power calculations are performed correctly;
c)
problems related to systems affected by constant and variable applied forces
are solved accurately.
Apply work and power requirements again for linear, angular and combined systems.
The same range of systems involving linear, angular and combined forms of motion
can be used again but this time to determine work and power requirements. It should
be noted that applications should be limited to only include average and instantaneous
power requirements.
Note that there are many variations on this type of question and a full tutorial would
contain examples of simple linear and then simple angular work and power
calculation. These are not shown here but many examples can be found in
engineering mechanics or engineering science text books. The student should practise
the type of combined calculation shown, as the combination of the linear and angular
quantities often proves difficult to master.
11. A mass of 50kg hangs vertically from a light cord, the other end of which is
wound around the rim of a drum of 600mm diameter. The moment of inertia of
the drum is 27kg m2. Initially the system is at rest, and then a constant torque of
200Nm is applied to the drum shaft such that the load is raised.
Determine:
a)
the work done by the torque in raising the
load 30 m;
b)
the angular acceleration of the drum;
c)
the time taken to raise the load 30 m;
d)
the power developed at the drum shaft after
this time.
Linear displacement of the load:
Angular displacement of the load:
= 30m
==s/r
= 30 /0.3
 = 100 rad
Work done by the torque:
=Tx 
= 200 x 100 = 20 000
= 20 kJ
Work done:
DET: Mechanical Engineering: Dynamics (Higher)
34
Total torque required
To accelerate drum
= Torque to overcome
drum inertia
= I + Fr
+ torque to balance
tension in the cable
(where F is the tension in the cable)
To find the tension in the cable for the mass to accelerate upwards:
 F = m (g+a)
F = mg + ma
 F = m ( g + r)
Using this expression to replace F in the total torque to accelerate the drum equation:
Total Torque
Total Torque
Total Torque - mgr
Total Torque - mgr
=
=
=
=
I
I
I
( I+ mr2 )
Total Torque - mgr
( I+ mr2 )
=

200 – 50 x 9.81 x 0.3
(27 + 50 x 0.32 )
=


=
+
+
+
m ( g + r)r
mgr + mr2
mr2
1.67 rad s-2
The angular acceleration of the drum is 1.67 rad s-2.
The time taken to raise the load can be calculated from:
1
2
  1t  t 2  t 
t 
2

as
1 = 0
2  100
= 10.91 s
1.68
The time taken to raise the load 30m is 10.91 seconds.
We need 2 in this case
Power = T 2
2
=T
= 1 + t
= 0 + 1.68 x 10.91
= 18.33 rad s-1
= 200 x 18.33
= 3666 kW
Power
2
The power developed at the drum shaft is 3.66 k W.
DET: Mechanical Engineering: Dynamics (Higher)
35
Outcome 3: Analyse the kinetics of motion using the energy balance approach
Performance Criteria
a)
the influences affecting kinetic energy are defined correctly in accordance
with the established theory;
b)
energy balance principles are applied to motion systems correctly;
c)
problems relating to systems affected by uniform acceleration are solved
accurately.
Initially students should start this section by revising potential, kinetic and formulae.
They would then go on to develop energy balance equations. This would initially be
applied to linear systems, then use polar moment of inertia could be introduced and
this would lead to the development of the energy balance equations for angular
systems. Finally when the student had developed enough skill in the separate
calculations of linear and angular system they could be carefully lead into the
development of the energy balance equations as applied to the combined systems.
Again the same range of systems as used in the first two outcomes should be covered.
The following examples show the energy balance approach to the analysis and the
calculations involved in the solution of a range of problems. These build from linear
problems to angular motion problems and then separate systems are combined into
systems involving both linear and angular components.
12. A mass of 30 kg is allowed to fall through a height of 12 metres. What is its
velocity at this lower point?
Gain in Kinetic Energy
½ m v2
½ v2
v
=
=
=
=
=
=
=
loss in Potential Energy
mgh
gh
2gh
2 x 9.81 x 12
15.34 m s-1
The velocity at the bottom of the fall is 15.34 m s-1.
DET: Mechanical Engineering: Dynamics (Higher)
36
13. The potential energy of a mass is increased by 20kJ when it is lifted vertically
through a height of 25m. It is then released and allowed to fall freely. Neglecting
air resistance, find its kinetic energy and hence its velocity after it has fallen 10m.
The gain in PE when the mass is raised 25m.
20 000
20 000
9.81 x 25
m
= mgh
= 9.81 x m x 25
= m
= 81.55kg
The mass is 81.55kg.
The body falls freely under gravity (9.81ms-2) through 10m
Loss in PE
Loss in PE
= gain in KE
= mgh
= 81.55 x 9.81 x 10
= 8 000.05 kJ
Kinetic energy at bottom of 10m fall is 8 kJ.
KE = ½ mv2
V=
2  KE
2  8000

 14.007 m s-1
m
81.55
The velocity at the bottom of the fall of 10m is 14 m s-1.
DET: Mechanical Engineering: Dynamics (Higher)
37
14. A rotating shaft carries a load having a moment of inertia about the shaft axis of
48 kg m2 . Calculate, using an energy method, the torque required to accelerate the
shaft from rest to a speed of 10 revs s-1 in 12 revs. The bearing friction is equal to
a couple of 300 Nm.
Known information:



Initial angular velocity
Final angular velocity
Angular displacement
1 = 0 ( at rest)
2 = 10 rev s-1
 = 12 rev
=10 x 2
=12 x 2
= 20 rad s-1
= 24 rad
Now:
Work done on shaft
T
T x 24
T
= Gain in KE of shaft
= ½ I 22
= ½ I 22
= ½ x 48 x (20 )2
= 94748.2
= 1556.6 kNm
+ friction couple x angular disp.
+ friction couple x angular disp.
+ 300 x 
+ 300 x 24
+ 22619.5
The torque required to accelerate the shaft is 1.56kNm.
DET: Mechanical Engineering: Dynamics (Higher)
38
15. A shaft carries a flywheel of mass 4000kg and has radius of gyration of 800mm.
Determine:
a)
the torque required to run up the flywheel to a speed of 300 RPM in one
minute from rest against a resisting torque of 250Nm;
b)
the kinetic energy of the flywheel at 300 RPM;
c)
the number of revolutions made in coming to rest from 300 RPM against a
braking torque of 350Nm applied in addition to the 250Nm resisting torque.
max speed of flywheel
= 300 RPM
 max angular velocity
= 2
1
= 300 x 2 rad s-1
60
= 10  rad s-1
= 0 rad s-1 ( at rest)
Acceleration of the fly wheel:
  1 10  0
 2  1    t    2

 0.167 rads-2
t
60
Acceleration of the fly wheel = 0.167 rad s-2
I = mk2 = 4000 x 0.82 = 2560 kg m2
Total Torque
= Accelerating Torque
=I
= 2560 x .167 + 250
= 1343 = 250kNm
+ Frictional Torque
+ 250
The total accelerating torque required is 1.59kNm.
The KE of the flywheel at 300 RPM:
KE
=
½ I 22
= ½ x 2560 x (10 )2
= 1263309 MJ
Total KE of the flywheel at 300 RPM = 1.26MJ.
The total decelerating torque = 350 + 250 = 600 Nm
The total decelerating torque = I 
I
= 600
  = 600
2560
=
.2344 rad s-2
DET: Mechanical Engineering: Dynamics (Higher)
39
Angular displacement from:
 22  12  2   
Number of rev =
 22  12
0 2  (10 ) 2
 100 2


 2105.29
2
2  .2344
 .4688
2105.29
2
=
335
The total number of revolutions made during the slowing down section is 335.
DET: Mechanical Engineering: Dynamics (Higher)
40
Combined question
16. A diesel powered road roller has a body mass of 7.2 Tonne excluding the wheels.
The single front wheel has a mass of 1.25 Tonne which for the purposes of
rotation can be considered to be concentrated at its outside radius of 0.4 m. The
rear axle together with its pair of wheels has a mass of 2 Tonne which for the
purposes of rotation can be assumed concentrated at the wheel rim of radius 0.6m.
If the roller is moving at a constant speed of 16.2 km h-1. Calculate:
a)
the linear kinetic energy of the body alone;
b)
the rotational kinetic energy of the wheels and axles;
c)
the total kinetic energy of the road roller;
d)
the total change in kinetic energy if the speed is increased up to 24 km h-1.
Linear Kinetic Energy of the body:
v
16.2  10 3
 4.5 ms 1
60  60
1
1
KE  mv 2 
 7200  4.52  72900 kJ
2
2
Total linear Kinetic Energy of the body is 72.9kJ.
Rotational KE of the wheels = KE Front wheel (1) + KE rear wheels & axle(2)
I 1  m1k 2  1250  0.4 2  200
1 
I 2  m2 k 2  2000  0.62  720
v 4.5

 11.25 ms 1
r 0.4
2 
v 4.5

 7.5 ms 1
r 0.6
Rotational KE of the wheels:
1 I112 + 1 I222
2
2
 0.5x200x11.25
+
2
0.5x720x7.52
=
12656.25
+20250
The total rotational KE of the wheels is 32.9kJ.
DET: Mechanical Engineering: Dynamics (Higher)
41
The total KE of the road roller = lin KE of body + rot KE of wheels + lin KE of
wheels.
Linear KE of wheels:
Total mass of wheels
Total KE of wheels
Total KE of the roller
= 2 + 1.5 = 3.5 tonnes
= ½ m v2 = 0.5 x 3.5 x 4.52 = 32200
= 72900 +32900 +32200
= 138700 kJ
The total KE of the roller is 138.7kJ.
To find the increase in KE:
v 6.67
v 6.67
24  103
 16.685 ms 1  2  
 11.12 ms 1
v
 6.67 ms 1 1  
r 0.4
r 0.6
60  60
Total KE at 24 km h-1 =
1
1
1
2
mbody wheels v12  I frontwheel 2fw  I rearwheel rw
2
2
2
1
1
1
(7.2  1.25  2)  6.67 2   200  16.6852fw   720  11.12 2rw
2
2
2
=
232455
+
27822
+
44516
=
304793
=
Total kinetic energy at 24 km h-1 is 304.8 kJ
Change in KE
=
304.8 – 138.7
=
166.1 kJ
The total change in kinetic energy when the speed of the road roller is increased
from 16.2km h-1 to 24km h-1 is 166.1kJ.
DET: Mechanical Engineering: Dynamics (Higher)
42
PRACTICAL EXAMPLE FOR OUTCOME 4
Outcome 4: Analyse uniform circular motion force systems
Performance Criteria
a)
centripetal acceleration is defined correctly in terms of the requirement for
circular motion in accordance with the established theory;
b)
an experiment to verify the theory of centripetal acceleration is performed
satisfactorily;
c)
mechanisms affected by centripetal acceleration are described correctly.
Students now move from angular quantities into the full circular motion theory and
develop the equation for centripetal force. The size and scale of centripetal force can
be explained here using the idea of a mass whirling on the end of a piece of string and
then scaling this up to the moons’ motion around the Earth. The students must be
able to solve problems containing centripetal force and acceleration elements. At this
stage many examples of circular motion can be found but care should be taken so that
the students can easily assimilate them.
The Performance Criteria that asks for an experiment to confirm the equation for
centripetal force and a report can be performed on one of many apparatuses. The
report should contain an explanation of how the standard apparatus was used, the
theory being verified, experimental results, an analysis of the results, discussion of the
results and a valid conclusion. An example of a meritous report is shown here. Then
a selection of various diagrams could be drawn for the basic centripetal acceleration
situations such as cars going around curves, centrifugal clutches etc.
The following example show an exemplar report that could be used as a hand out to
students or it could be used as a bench mark to help assess whether a candidate has
reached grade A.
DET: Mechanical Engineering: Dynamics (Higher)
43
Centripetal Force Experiment
Theory
When the turntable is driven, the rotating masses
fixed to the horizontal arms generate centripetal
forces acting outwards. At the same time, masses
mounted on the central spindle generate a
gravitational force vertically downwards.
Neglecting friction (that can not be quantified
here), we equate the centripetal forces to the
gravitational forces acting on the system and at the
instant they balance the spindle masses will just
start to rise off their seat. At this point we can write
an equation saying that the centripetal force is
equal to the gravitational force. A graph of these
two quantities for the various masses should
produce a straight line. Other interesting graphs
would be:
Centripetal Force Vs
Speed
Centripetal Force Vs
Speed squared
LOG Centripetal Force Vs
LOG Speed
Should give a curve
Should give a straight
line
Should give a straight
line
Object
To prove for a mass rotating at a constant angular
velocity, that its centripetal force is proportional to
the square of its angular velocity multiplied by a
constant equal to mass times the radius.
DET: Mechanical Engineering: Dynamics (Higher)
44
Diagram
DET: Mechanical Engineering: Dynamics (Higher)
45
Method
The two masses on each arm were fixed at the same
distance from the centre, 100mm in this case, to
enable the full range of values to be assessed. A
mass M was placed on the central spindle and the
apparatus motor was switched on, the apparatus
started to rotate. Using the variable speed control
the speed of rotation was increased until the mass M
lifted off its seat. The tachometer was then read.
The speed was then allowed to fall until the mass M
returned to its seat and a second reading with the
tachometer was taken.
The experiment was repeated with incremental
increases of the mass M on the central spindle. The
results were then tabulated and the graphs were
plotted.
Results
experimental experimental experimental fixed
set
Rotational Rotational
practical
Speed 1
Speed 2
av. Speed Radius Balancing Rotating velocity in Velocity experimen centripetal
t
in RPM
in RPM
in RPM
mm
mass g Mass kg
Rads /
squared
number
force N
sec
21
20
20.5
100
20
0.4
2.15
4.61
1
0.1843
48
46
47
100
100
0.4
4.92
24.22
2
0.9690
76
73
74
100
250
0.4
7.75
60.05
3
2.4020
108
102
105
100
500
0.4
11.00
120.90
4
4.8361
132
126
128
100
750
0.4
13.40
179.67
5
7.1868
150
149
149
100
1000
0.4
15.60
243.46
6
9.7384
168
166
166
100
1250
0.4
17.38
302.18
7
12.0874
197
195
197
100
1750
0.4
20.63
425.59
8
17.0235
448
243
245
100
2750
0.4
25.66
658.25
9
26.3299
Equations used
rpm conversion

DET: Mechanical Engineering: Dynamics (Higher)
rpm  2
60
46
theoretical
centripetal
differen
force N
0.1962
0.9810
2.4525
4.9050
7.3575
9.8100
12.2625
17.1675
26.9775
0.011
0.012
0.050
0.068
0.170
0.071
0.175
0.144
0.647
centripetal force from velocity
CF  m   2  r
Theoretical centripetal
Force from mass
DET: Mechanical Engineering: Dynamics (Higher)
CF  m  g
47
Graphs
Actual Vs Calculated Centripetal Force
Calculated Force from mass
30.0000
25.0000
20.0000
15.0000
10.0000
5.0000
0.0000
0.0000
10.0000
5.0000
20.0000
15.0000
25.0000
30.0000
Practical Centripetal Force
Centripetal Force Vs Velocity
Squared
Centripetal Force Vs Speed
Velocity Squared
Velocity
30.00
20.00
10.00
0.00
0.0000
10.0000
20.0000
30.0000
800.0
600.0
400.0
200.0
0.0
0.0000
10.0000
20.0000
30.0000
Centripetal Force
Centripetal Force
LOG Centripetal Force VS LOG Velocity
1.6000
1.4000
LOG Velocity
1.2000
1.0000
0.8000
0.6000
0.4000
0.2000
-1.0000
-0.5000
0.0000
0.0000
0.5000
1.0000
1.5000
2.0000
LOG Centripetal Force
DET: Mechanical Engineering: Dynamics (Higher)
48
Conclusion
Centripetal Force Vs Velocity Squared
700.0
Velocity Squared
600.0
500.0
400.0
300.0
200.0
100.0
0.0
0.0000
5.0000
10.0000 15.0000 20.0000 25.0000 30.0000
Centripetal Force
From the graph of centripetal force against velocity
squared we can see that they form a straight line,
allowing for experimental errors, this proves that
these two functions are proportional to each other.
From the data and the graph we can deduce that
because the equation of a straight line is;
y  m x  c
and the intercept is through the point (0,0)
therefore the value of C must be zero leaving;
y  m x
or
CF   2  mass  radius
Therefore the constant or slope of the graph should
equal mass times the radius.
Slope
dy 26.33

 0.04
dx 658.25
m  r  0.4  01
.  0.04
This shows that the theoretical equation is proved to
be correct by this experiment.
DET: Mechanical Engineering: Dynamics (Higher)
49
Sources of Errors
The errors in the readings could have been caused
by;
1.the reading on the tachometer was fluctuating
and an instantaneous reading was estimated;
2.the bands on the rotating platform were not
exactly uniform;
3.friction in the mechanical components;
4.the motor rpm was difficult to regulate to give a
constant velocity.
Improvements
1.fit a digital tachometer instead of a analogue;
2.fit an electronic switch on the apparatus to stop
the tachometer;
3.repeat the experiment a number of times (time
did not allow this in this case).
The above is shown only as an example and many variations are possible and the
standard of this report is meant to show what a potential grade A student could
produce.
DET: Mechanical Engineering: Dynamics (Higher)
50
Examples of diagrams that show centripetal motion.
DET: Mechanical Engineering: Dynamics (Higher)
51
DET: Mechanical Engineering: Dynamics (Higher)
52