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Mod1 Chp.1- CIE 262-ppt e913bfa55c8e52fc3e6d9a626a89dcbb

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Mulungushi University
Engineering Mechanics
Dynamics- CIE 262
Mr. Kalolo Mukanga J.R
Dynamics- CIE 262 _ 2022
1
Course Grading System
Continuous Assessment (CA)
- Assignments/Quizzes
- Tests
- Laboratory experiments
- Laboratory Tests
Final Theory Examination
Total:
40%
5%
20%
10%
10%
60%
100%
Prescribed Text book
Text-book: Beer & Jehnson (2007), Vector Mechanics For
Engineers 8th edition, McGraw-Hill
Recommended Text book: J.L. Meriam and I,G, Kraige,
Engineering Mechanics (Dynamic), John Wiley & Sons inc.
Vol 2, SI Version, Sixth Edition, ISBN 978-0-471-78703-7
Lecturers Notes
Course Contents
Module 1: Particle Dynamics
Chp1. Kinematics of Particles
1.1
Rectilinear kinematics- 1D Motion
1.2
Curvilinear kinematics
Chp2. Kinetics of Particles
2.1 Kinetics of particles_newtons 2nd law
2.2 Kinetics of particles_energy method
2.3 Kinetics of particles_momentum method
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Course Contents
Module 1: Particle Dynamics
Chp3. System of Particles
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Course Contents
Module 2: Rigid Body Dynamics
Chp4. Review of Plane Kinematics of Rigid Bodies
1.1
Introduction
1.2
Translation (rectilinear and curvilinear)
1.3
Fixed-axis rotation
1.4
General plane motion
1.5
Absolute and relative motions, motion relative to rotating axes
Chp5. Plane Kinetics of Rigid Bodies
2.1 Equations of motion (force and moment) for translation
2.2 Fixed-axis rotation and general motion
2.3 Work and energy method, impulse and momentum method
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Course Contents- Cont.
Chp6. Vibration and Time Response
3.1 Equation of motion and time response for: undamped and damped free vibration (free-body
diagrams and work-energy method) for single-degree-of-freedom systems
3.2 Undamped and damped forced vibration of single DOF systems (external harmonic vibration, base
excitation, vibration measuring instruments)
3.3 Equivalent single DOF system of continuous elements/systems using energy equivalence and
equation of motion
Chp7. Introduction to 3D Dynamics of Rigid Bodies
4.1 Translation, fixed-axis and fixed-point rotations
4.2 General motion, inertia as a tensor quantity
4.3 Euler's equations as applied to rate of change of momentum
4.4 Coordinate system where inertia tensor is invariant
4.5 Gyroscopic motion
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Chp1. Kinematics of Particles
Module 1: Particle Dynamics
1.1
Introduction
Engineering Mechanics
Dynamics
Kinematics
Statics
Kinetics
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Chp1. Kinematics of Particles
Position
𝑟⃗ = 𝑥 𝑖 ; 𝑥 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒, 𝑖 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟
Displacement
∆𝑥⃗ = 𝑥⃗ − 𝑥⃗ ; 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟𝑚
∆𝑥 = 𝑥 − 𝑥 ; 𝑠𝑐𝑎𝑙𝑎𝑟 𝑓𝑜𝑟𝑚
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Chp1. Kinematics of Particles
Instantaneous Velocity and acceleration
Instantaneous Velocity
𝑣⃗ = lim
→
∆𝑥⃗ 𝑑𝑥⃗
=
; 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟𝑚
∆𝑡 𝑑𝑡
𝑣=
𝑑𝑥
; 𝑠𝑐𝑎𝑙𝑎𝑟 𝑓𝑜𝑟𝑚
𝑑𝑡
𝑎⃗ =
𝑑𝑣
; 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟𝑚
𝑑𝑡
𝑎=
𝑑𝑣
; 𝑠𝑐𝑎𝑙𝑎𝑟 𝑓𝑜𝑟𝑚
𝑑𝑡
Instantaneous acceleration
We can equally define the critical point as the point where the velocity is zero hence indicating a change of
direction.
∆𝑥⃗ 𝑑𝑥⃗
𝑣⃗ = lim
=
=0
→ ∆𝑡
𝑑𝑡
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Chp1. Kinematics of Particles
Relationship between Position, Velocity & Acceleration
We have defined that acceleration and velocity are given by:
𝑎=
𝑑𝑣
(1)
𝑑𝑡
𝑣=
𝑑𝑥
𝑑𝑡
(2)
𝑑𝑡 =
𝑑𝑥
𝑣
by making 𝑡 subject in (2), we obtain:
Substituting in (1):
𝑎𝑑𝑥 = 𝑣𝑑𝑣
Dynamics- CIE 262 _ 2022
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Chp1. Kinematics of Particles
Application
1. Consider a car idealised by a particle moving with a displacement given by:
𝑥 𝑡 = 𝑡 − 15𝑡 + 50𝑡 − 25 𝑚
Find the total distance travelled in 10sec.
Dynamics- CIE 262 _ 2022
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Chp1. Kinematics of Particles
Acceleration as a function
When acceleration is as a function of time, position or velocity, below are the formulas and procedure to follow:
Of Time:
𝑎=
𝑎(𝑡)
Of Position:
𝑑𝑣
𝑑𝑡
𝑣=
𝑣(𝑡))
𝑣=
𝑎𝑑𝑥 = 𝑣𝑑𝑣
𝑎(𝑥))
Of Velocity:
𝑣(𝑠))
𝑎=
𝑎(𝑣))
𝑑𝑥
𝑑𝑡
𝑥(𝑡))
𝑑𝑥
𝑑𝑡
𝑡(𝑥))
𝑑𝑣
𝑑𝑡
𝑣=
𝑡(𝑣))
Dynamics- CIE 262 _ 2022
𝑣(𝑡))
𝑑𝑥
𝑑𝑡
𝑥(𝑡))
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Chp1. Kinematics of Particles
Application
1. The acceleration of a particle attached to a spring is defined as
𝑎 𝑡 = 𝐶 sin(𝜔𝑡)
Where
𝐶 = −1.5𝑚/𝑠 and 𝜔 = 2.5 𝑟𝑎𝑑
At time 𝑡 = 0, the particle has velocity 𝑣 = 0.6𝑚/𝑠 and position,𝑥 = 0.
i. Find the expression of the position, velocity and acceleration as a function of time.
ii. Determine the velocity and position at 𝑡 = 0.5 𝑠𝑒𝑐.
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Chp1. Kinematics of Particles
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Chp1. Kinematics of Particles
Constant Acceleration
Let the acceleration be constant with value a. then
𝑑𝑣
=𝑎 →𝑣=
𝑑𝑡
where k is the constant of integration.
If 𝑣 = 𝑣 at 𝑡 = 0 then 𝑘 = 𝑣 so
𝑎 𝑑𝑡 = 𝑘 + 𝑎𝑡
𝒗 = 𝒗𝟎 + 𝒂𝒕 (1)
However, 𝑣 =
so
𝑑𝑥
= 𝑣 + 𝑎𝑡
𝑑𝑡
𝑥=
(𝑣 +𝑎𝑡)𝑑𝑡 = 𝑐 + 𝑣 𝑡 +
𝑎𝑡
2
Where c is the constant of integration
if 𝑥 = 𝑥 at 𝑡 = 0 then 𝑐 = 𝑥 and so
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Chp1. Kinematics of Particles
𝒙 = 𝒙𝟎 + 𝒗𝟎 𝒕 +
𝒂𝒕𝟐
𝟐
→ 𝒔 = 𝒗𝟎 𝒕 +
𝒂𝒕𝟐
𝟐
(2)
𝒙 − 𝒙𝟎 ≡ 𝒔 ≡ 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
From (1) we have 𝑣 − 𝑣 = 𝑎𝑡, so
𝑣−𝑣 𝑡 =
Substituting in (2) gives
𝑥 =𝑥 +𝑣 𝑡+
𝒙 = 𝒙𝟎 +
We need one final equation, from (1) 𝑡 =
𝟏
𝟐
1
𝑣−𝑣 𝑡
2
𝒗 + 𝒗𝟎 𝒕 → 𝒔 =
𝟏
𝟐
𝒗 + 𝒗𝟎 𝒕
(3)
, putting in (3)
1
𝑣−𝑢
𝑣+𝑣
2
𝑎
= 𝑣+𝑣 𝑣−𝑣 =𝑣 −𝑣
𝑥−𝑥 =
2𝑎 𝑥 − 𝑥
𝒗𝟐 = 𝒗𝟎 𝟐 + 𝟐𝒂 𝒙 − 𝒙𝟎 → 𝒗𝟐 = 𝒗𝟎 𝟐 + 𝟐𝒂𝒔 (4)
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Chp1. Kinematics of Particles
From (1) 𝑣 = 𝑣 − 𝑎𝑡
, we can obtain an alternative of (2) getting (2a)
𝒔 = 𝒗𝒕 −
𝒂𝒕𝟐
𝟐
(2a)
In these equations the 𝑡 is the time, s is the displacement, u is the velocity at 𝑡 = 0, v is
the velocity at time 𝑡 and 𝑎 is the acceleration.
Notes
1. Each equation omits one of the five: 𝑡, 𝑠, 𝑣 , 𝑣, 𝑎.
2. If the acceleration is negative, it is usually called the deceleration.
3. 𝑎 and 𝑣 (and 𝑥 ) are constants.
4. It may be necessary to use more than one of these equations
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Chp1. Kinematics of Particles
Application
1. A car travels from rest at a constant deceleration is 3𝑚/𝑠 and initial speed is 20 𝑚/𝑠.How far will it travel at
𝑡 = 3𝑠𝑒𝑐?
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Chp1. Kinematics of Particles
Absolute dependant motion
This is defined as motion of a particle that depends on another eg. Pulley and cranes.
Approach to solve such problems:
I. Select reference or datum for each particle ( fixed datum and perpendicular to motion).
II. Define positive distance or motion from reference.
III. Develop an equation for the length of the cable.
IV. Take time derivative.
V. Solve with other kinematic relationships.
Example
Position of a particle may depend on position of one or more other particles.
Position of block B depends on position of block A. Since rope is of constant
length, it follows that the sum of lengths of segments must be constant.
𝑥 + 𝑥 = 𝑐𝑠𝑡 → 𝑣 + 𝑣 = 0 → 𝑎 + 𝑎 = 0
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Chp1. Kinematics of Particles
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