Introduction to Dynamics

Introduction To Dynamics

Edexcel HN Unit:
Engineering Science (NQF L4)

Author: Leicester College
Date created:
Date revised: 2009

Abstract;
The basic concepts associated with some elements of dynamic engineering systems and
calculations are discussed. The theory associated with Linear and angular motion, energy and
simple harmonic motion are outlined with sample calculations for each topic being given.

Contents
Introduction To Dynamics.................................................................................................................................1
Dynamic Systems..............................................................................................................................................2
Angular motion - Equations of Angular motion...........................................................................................3
Energy and Rotation..........................................................................................................................................5
Simple Harmonic Motion – Pendulum..............................................................................................................7
Credits................................................................................................................................................................9

These files support the Edexcel HN unit – Engineering Science (Mechanical)

File name Unit Key words
outcome
Stress 1.1 Stress, strain, statics, young’s modulus
introduction
BM, shear force 1.1 Shear force, bending moment, stress
diagrams
Selecting beams 1.2 Beams, columns, struts, slenderness ratio

Torsion 1.3 Torsion, stiffness, twisting
introduction
Dynamics 2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation
introduction

For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications

© Leicester College 2009 This work is licensed under a HYPERLINK
"http://creativecommons.org/licenses/by/2.0/uk/" n _parentCreative Commons Attribution 2.0 License.

Introduction to Dynamics

Dynamic Systems

These notes give some sample examples of Dynamic Mechanics. (Outcome 2)

Linear Motion

Equations of Linear Motion

v = u + at
s = (u + v ) t Where u = initial velocity (m/s)
2 v = final velocity (m/s)
2 2
v = u + 2as s = distance (m)
s = ut + ½ at2 t = time (s)
a = acceleration (m/s2)

Example 1

A vehicle starting from rest accelerates at a rate of 3m/s2 for 8 seconds. It then continues at this
velocity for 20 seconds before coming to rest again in 5 seconds. Determine the maximum
velocity, rate of deceleration and total distance travelled.
First – Draw a v/t graph of this motion;

V (m/s)

3m/s2

1 2 3

8 28 33 t(s)

We know u = 0, t1 = 8s, t2 = 20s, t3 = 5s and a = 3m/s2

To find v; v = u + at1 = 0 + 3 x 8 = 24 m/s

To find s1 s = (u + v ) t1 = (0 + 24) 8 = 96m
2 2
To find s2 s = (u + v ) t2 = (24 + 24) 20
2 2 = 480m

To find s3 s = (u + v ) t3 = (24 + 0) 5 = 60m
2 2

Stotal = s1 + s2 + s3 = 636m

Page PAGE 9 of NUMPAGES *Arabic 9


To find deceleration;

v = u + at where v = 0

a = - u/t = -24/5 = 4.8 m/s2

Angular motion - Equations of Angular motion
θ = (ω1 + ω2 ) Where θ = Angular displacement(rads)
2 ω1= Initial Ang. Velocity (rad/s)
ω2 = ω1 + άt ω2 = Final Ang. velocity (rad/s)
ά = Angular acceleration (rad/s2)
ω22= ω12 + 2άθ

θ = ω1t + ½ άt2

Conversion of angular to linear to angular motion v = ωr and a = άr

Example

An electric motor starting from rest is accelerated to 500rpm in 3 seconds. It continues at this
velocity for 20 seconds before coming to rest in 4 seconds. Determine the rate of acceleration and
deceleration, the maximum angular velocity and the total number of revolutions during this motion.
500rpm to Rad/sec - 500 x 2π / 60 = 52.36 rad/s

First – Draw a v/t graph of this motion;

ω
(rad/s)

1 2 3

3 23 27 t(sec)

Angular acceleration - ω2 = ω1 + άt and ω1 = 0

So ά = ω2 / t = 52.36 / 3 = 17.45 rad/s2

Revs in period 1 - θ = (ω1 + ω2 )t = (0 + 52.36) 3
2 2
= 78.54 Rads



Revs in period 2 - θ = (ω1 + ω2 )t = (52.36 + 52.36) 20
2 2
= 1047.2 Rads

Revs in period 3 - θ = (ω1 + ω2 )t = (52.35 + 0) 4
2 2
= 104.7 Rads

Total radians = 1230.46 Convert to revs – 1 / 2π

= 195.83 Revolutions

Deceleration

As acceleration; ω2 = 0 and t = 4s

ά = ω1 / t = 52.36 / 4
= 13.09 rad/s2



Energy and Rotation
Principle of conservation of energy – Energy cannot be created or destroyed but only
converted from one for to another.

There are two types of MECHANICAL energy. For linear motion they are;
• Potential energy – Pe = mgh (The energy of position)
• Kinetic energy - Ke = ½ mv2 (the energy of motion)

When an object rotates there is no PE and Ke = ½ I ω2

Where I is the ‘moment of Inertia’ – this is a measure of an object’s resistance to change of
motion or state.

How we determine the value of I depends on the physical shape of the object;

For a solid disk or cylinder I = mr2/2
For a concentrated mass I = mr2.
Where m = mass (kg) and r = radius (m)

In order to change the state of uniform motion or rest of a rotating object about a central point we
need to apply a TORQUE.
T = Fr = I ά

Example1
A cylinder, 150mm diameter, with a mass of 8.5 kg, rolls along a flat surface at a constant velocity
of 1.6 m/s. Determine the total KE of the cylinder (ie both linear and rotational KE)

In this case I = mr2/2 = (8.5 x 0.0752)/2
= 0.0239 kgm2

Ke linear = ½ mv2 = 0.5 x 8.5 x 1.62 = 10.88J

From previous examples v = ω r
therefore ω = v/r = 1.6/0.075 = 21.33 rads/sec

Ke rotational = ½ I ω2 = ½ x 0.0239 x 21.332
= 5.438J

Total KE = Ke Linear + Ke rotational =

10.88 + 5.438 = 16.32J

Example 2

A motor vehicle starting from rest free wheels down a slope of incline 1 in 8. Neglecting resistance
to motion determine its velocity after it has traveled 200m down the slope.

Pe lost = Ke gained (From conservation of energy)



Mgh = ½ mv2 The slope is 1/8 = 0.125.

Becomes gh = ½ v2 Therefore h = 0.125 x 200 = 25m
So v2 = 2gh and v = √(2 x 9.81 x 25) = 490J

Example 3

A solid cylinder 0.1m diameter and 0.05m long is allowed to roll down a 15 degree slope. What is
its linear velocity after it has traveled 0.8 m down the slope. Assume a density of cylinder material
of 6800 kg/m3.

Solid cylinder

150

Volume of cylinder = π r2l = π x 0.052 x 0.05
= 3.927 x 10-4 m3

Mass = density x volume
= 6800 x 3.927 x 10-4 = 2.67 kg

Pe = mgh where h = 0.8 x sin 15 = 0.207m

= 2.67 x 9.81 x 0.207 = 5.422J

Following the principle of conservation of energy;

Pe lost = Ke gained = Ke linear + Ke rotational

Ke linear = ½ mv2 = 0.5 x 2.67 x v2 = 1.335v2

Ke rotational = ½ I ω2 where I = ½ mr2 = ½ x 2.67 x 0.052 = 3.3375 x 10-3

And ω = (v/r)2

So Ke rotational = ½ I(v/r)2
= ½ x 3.3375 x 10-3 x v2/2.5x10-3

= 0.6667v2

According to the principle of conservation of energy:

Pe = Ke lin. + Ke rot

5.422 = (1.335 + 0.6675)v2
v = √(5.422/2.0045) = 1.648 m/s



Simple Harmonic Motion – Pendulum
For a pendulum
The maximum linear velocity occurs at A

The maximum linear acceleration B
occurs at B
v= ωr
a = ω2r B
Where ω = 2f A
And f = 1 √(g/l)
2π

Example 1

Determine the natural frequency, maximum linear velocity and acceleration of a simple pendulum
with a cord length of 0.5m.

f = 1 √(g/l) = 1 √(g/l) = 1/6.284 x √(9.81/0.5)
2π 2π
= 0.705 Hz.

Where ω = 2πf 2πx 0.705 = 4.43 rad/s

v= ωr = 4.43 x 0.5 = 2.215 m/s

a = ω2r = 4.432 x 0.5 = 9.81 m/s2

Example 2

A simple pendulum has a periodic time of 2s. What is the length of the cord ?

Periodic time t = 1/f so f = 0.5Hz

f = 1 √(g/l) Transpose for l
2π

L= g
(2 π f )2 = 0..994m

Simple Harmonic Motion – Springs

For a spring there are two displacements to consider

Initially, an unloaded spring has no displacement. When a mass is placed on the spring it will
deflect and remain static.
This deflection is due to the stiffness of the spring – k



Where k = f/x = mg/x
x is the extension of the spring due to load f
m is the mass
and g is acceleration due to gravity

The second displacement (A – amplitude) is when the spring is extended with the spring loaded
and released causing an oscillating motion. This gives us SHM about the point of deflection due to
pre-loading of the spring.

In this case;

f = 1 √(k/m) and k = mg/x so
2π
f = 1 √(g/x) Note m cancels
2π

The maximum linear velocity, v= ωA
and the maximum linear acceleration a = ω2A

Example

A spring extends by 25mm when a pre load of 1 kg is applied. It is then extended by a further
30mm and released. Determine the natural frequency of oscillation, the maximum velocity and
acceleration for this situation.

k = f/x = mg/x = 1 x 9.81/ 0.025 = 392.4 N/m

f = 1 √(g/x) = 1 √(9.81 /0.025) = 3.15Hz
2π 2π

As before ω = 2π f = 2π x 3.15 = 19.81 rad/s

maximum linear velocity,
v= ωA = 19.81 x 0.030 = 0.594 m/s

maximum linear acceleration
a = ω2A 19.812 x 0.030 = 11.774 m/s2

Note in this example that x is used to determine frequency and A is used to determine v and a.



Credits

This resource was created Leicester College and released as an open educational resource
through the Open Engineering Resources project of the Higher Education Academy Engineering
Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the
JISC/HE Academy UKOER programme.

© 2009 Leicester College

This work is licensed under a Creative Commons Attribution 2.0 License.

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Licence. All reproductions must comply with the terms of that licence.

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that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher.

The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College.


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