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# Engineering Mechanics Part II: Dynamics. Lectures 1 - 3

## 1. Engineering Mechanics Part II: Dynamics Dr. Bahaa Saleh

## 2.

Engineering MechanicsDynamics

Kinematics

2

Statics

Kinetics

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## 3.

Course Supplemental MaterialsTextbook - Engineering Mechanics:

Dynamics, R. C. Hibbeler, 8th Edition,

Pearson Prentice Hall, 1998.

References: Engineering Mechanics:

Dynamics , J . L. Meriam and L. G. Kraige ,

6th Edition, John Wiley & Sons, Inc., 2008.

Lectures Notes prepared by instructors.

## 4. Course Grading System

• 20%• 20%

• 20%

• 40%

Attendance, participation,

Quizzes and assignments

1st Midterm Exam

2nd Midterm Exam

Final Exam

## 5. Course Topics

• Chapter 1: Introduction to dynamics• Chapter 2: Kinematics of a Particle:

Topic # 1: Particle motion along a straight line

Topic # 2: Particle motion along a curved path

Topic # 3: Dependent motion of connected particles

Topic # 4: Relative motion of two particles

• Chapter 3: Kinetics of a Particle:

Topic # 1: Force and Acceleration

Topic # 2: Work and energy

Topic # 3: Impulse and momentum

## 6. Course Topics – Cont.

• Chapter 4: Planer Kinematics of a RigidBody.

• Chapter 5: Planar Kinetics of a Rigid

Body: Force and Acceleration.

• Chapter 6: Introduction to Mechanical

Vibration.

6

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## 7. Chapter 1: Introduction to dynamics

Engineering MechanicsDynamics

Kinematics

7

Statics

Kinetics

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## 8. Definitions

• Statics: concerned with theequilibrium of a body that is

either at rest or moves with

constant velocity.

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## 9. Definitions – Cont.

Dynamics1- Kinematics: study of the motion of

particles/rigid

bodies

(relate

displacement, velocity, acceleration,

and time, without reference to the

cause of the motion).

• 2- Kinetics: study of the forces acting

on the particles/rigid bodies and the

motions resulting from these forces.

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## 10. Definitions – Cont.

• Rigid Body• Particle

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## 11. Review of Vectors and Scalars

• A Scalar quantity has magnitude only.• A Vector quantity has both magnitude and

direction.

## 12.

• Scalars (e.g)– Distance

– Mass

– Temperature

– Pure numbers

– Time

– Pressure

– Area

– Volume

• Vectors (e.g.)

– Displacement

– Velocity

– Acceleration

– Force

## 13. Vectors

• Can be represented by an arrow (called the “vector”).• Length of a vector represents its magnitude.

• Symbols for vectors:

–

(e.g. force) F , F , or F (bold type), or F

F

2F

## 14. Chapter 2: Kinematics of a Particle: Topic # 1: Particle motion along a straight line (Rectilinear Motion)

## 15. Definition

Rectilinear motion: A particle movingalong a horizontal/vertical/inclined straight

line.

## 16. Position of the particle (horizontal)

• Since the particle is moving, so theposition is changing with time (t):

• OP = Position = S = S (t)

## 17. Displacement of the particle (horizontal)

• Displacement (∆s) : The displacementof the particle is the change in its

position.

## 18. Displacement of the particle (horizontal)

1- ∆S is positive since the particle's finalposition is to the right of its initial position,

i.e., s` > s.

2- If the final position to the left of its initial

position, ∆S would be negative.

## 19. Velocity of the particle (horizontal)

• Velocity (v) : If the particle displacement ∆s duringtime interval ∆t, the average velocity of the

particle during this time interval is (displacement

per unit time)

• The magnitude of the velocity is known as the

speed, and it is generally expressed in units of m/s

v av

s

t

## 20. Velocity of the particle (horizontal)

• Instantaneous velocity :s

V lim

t 0 t

ds

v

dt

• So (v) is a function of time (t):

v v(t)

## 21. Acceleration of the particle (horizontal)

• Acceleration : The rate ofchange in velocity {(m/s)/s}

V V V

• Average acceleration :

aavg

V

t

• Instantaneous acceleration :

v dv d 2 s

a lim

2

t 0 t

dt dt

• If v ‘ > v

• If v ‘ < v

“ Acceleration “

“ Deceleration”

## 22. Acceleration of the particle (horizontal)

• Acceleration (a) : is the rate of change ofvelocity with respect to time.

2

dv d s

a

2

dt dt

## 23. Solved Examples

Example 1• A particle moves along a straight line such that its

position is defined by s = (t3 – 3 t2 + 2 ) m.

Determine the velocity of the particle when t = 4 s.

ds

2

v

3t 6t

dt

At t = 4 s,

the velocity (v) = 3 (4)(4) – 6(4) = 24 m/s

## 24.

Example 2• A particle moves along a straight line such that its

position is defined by s = (t3 – 3 t2 + 2 ) m.

Determine the acceleration of the particle when

t = 4 s.

ds

v

3t 2 6t

dt

dv

a

6t 6

dt

• At t = 4

a(4) = 6(4) - 6 = 18 m/s2

## 25. Relation involving s, v, and a No time t

Position sVelocity

Acceleration

ds

v

dt

ds

dt

v

dv

a

dt

dv

dt

a

a ds v dv

ds dv

v a

## 26. Motion with uniform/constant acceleration a

dva

dt

v

t

vo

0

dv

a

dt

dv a dt

v v 0 at

v v 0 at

## 27. Motion with uniform/constant acceleration a

dsv

v0 a t

dt

s

t

so

0

ds

(

v

a

t

)

dt

0

1 2

s s0 v 0 t a t

2

## 28. Motion with uniform/constant acceleration a

v dv a dsv

s

v0

s0

v

d

v

a

ds

1 2 1 2

v v 0 ac ( s s 0 )

2

2

v v 2 a ( s s0 )

2

2

0

## 29. Summary

• Time dependent acceleration• Constant acceleration

s s (t )

v v0 a t

ds

v

dt

2

dv d s

a

2

dt dt

a ds v dv

1 2

s s0 v 0 t a t

2

v v 2 a ( s s0 )

2

2

0

## 30.

Example 3• A car moves in a straight line such that for a short time its

velocity is defined by v = (3t^2 + 2t) m/s, where t is in

seconds. Determine its position and acceleration when t =

3 s. (When t = 0, s = o).

When t = 3 s

s 3 3 36 m

3

2

a 6 * 3 2 20 m s

2

## 31. Chapter 2: Kinematics of a Particle: Topic # 2: Particle Motion along a Curved Path

## 32.

Cartesian (Rectangular) CoordinatesTo describe the plane motion of a particle, we use the

Cartesian (Rectangular) Coordinates (x-y).

Y

Path

x

P

r

y

Y

vP

vy

path

P

vx

X

Y

path

X

aP

Ay

P

Ax

X

## 33. 12Projectile Motion

• Projectile: any body that is given an initialvelocity and then follows a path determined by

the effects of gravitational acceleration and air

resistance.

• Trajectory – path followed by a projectile

## 34.

Cartesian Coordinates ofProjectile Motion

Max. Height

y

V0

g

V0 sin

o

V0 cos

x

B

Range

X

## 35.

Horizontal and vertical components of velocityare independent.

Vertical velocity decreases at a constant rate

due to the influence of gravity.

## 36.

Cartesian Coordinates ofProjectile Motion

• Assumptions:

(1) free-fall acceleration

(2) neglect air resistance

• Choosing the y direction as positive upward:

ax = 0;

ay = - g (a constant)

y

v0

• Take x0= y0 = 0 at t = 0

• Initial velocity v0 makes an

x

angle with the horizontal

v 0x v 0 cosθ

v 0y v 0sin θ

## 37. Horizontal Motion of Projectile

• Acceleration in X-direction: ax= 0• Integrate the acceleration yields:

( ) v x v 0 cosθ constant

• Integrate the velocity yields:

( ) x v 0 t cosθ

## 38. Vertical Motion of Projectile

• ay = ac= -g = -9.81 m/s2• Integrate the acceleration yields:

( ) v v0 ac t

( )

v y (v0 ) y gt

v y v 0 sinθ gt

• Integrate the velocity yields:

( )

gt 2

y v 0 t sinθ

2

## 39.

• ax = 0;ay = - g (a constant)

• Integration of these acceleration yields

v x v0 cosθ constant

v y v 0 sin θ g t

x v 0 t cosθ

(1)

y v 0 t sin θ g t /2

2

(2)

• Elimination of time t from Eqs. 1 & 2 yields

• Equation of the path of projectile

y x tan θ (g x sec θ / 2 v )

2

2

2

o

(3)

## 40.

Maximum Height of ProjectileMax. Height

y

V0

o

x

X

## 41.

Maximum Height of ProjectileAt the peak of its trajectory, vy = 0.

v y v 0y a t v o sin gt 0

v 0 sin

Time t1 to reach the peak t1

g

g

1 2

Substituting into: y v 0 sin t gt

2

v 0y

h y max

v 0sinθ 1 v 0 sinθ

g

v 0sinθ

g 2 g

v 0sinθ

2

h y max

g

1 v 0sinθ

2

g

2

2

## 42.

Maximum Height of Projectilev 0 sin

1 v 0 sin

2

g

2

h y max

h y max

2

g

(v o sin )

v sin v

2g

2g

2g

2

2

o

2

2

0y

v 0 sinθ

v 0 sinθ

x v 0x t v 0x

v 0 cosθ

g

g

2

v o sin 2θ

sin2θ

x

sinθ cosθ

2g

2

## 43.

Maximum Height of Projectile andthe corresponding time and X

h y max

v sin

2g

2

o

2

o

v sin 2θ

x

2g

v 0 sin

t1

g

2

## 44.

The Horizontal Range of Projectiley

V0

V0

o

Range OB

B

X

## 45.

The Horizontal Range of ProjectileThe range (OB) where y = 0.

1 2

y v 0 sin t gt

2

Time for the range OB

2 v 0 sin

tB

g

For the rang OB substitute into:

x v 0 t B cosθ

2

o

2v o

v

X OB v o cosθ

sinθ

2 sinθ cosθ

g

g

2

2

vo

vo

X OB

sin2(90 - θ)

sin2 θ

g

g

## 46.

The Horizontal Range of ProjectileFrom the Rang equation it is clear that an

angle of firing with the horizontal gives the

same range OB as an angle of firing (90 - )

with the horizontal or as an angle of with

vertical.

## 47.

Maximum Range OB* of Projectiley

V0

o

Range OB

B

Max. Range OB*

B*

X

## 48.

Maximum Range OB* of ProjectileTo calculate max. Range (OB*) and its angle

2

o

v

X OB

sin2(90 - θ)

g

sin2(90 - θ) 1 sin(2 )

θ 45

2

vo

OB

g

*

o

## 49.

Projection Angle• The optimal angle of projection is dependent

on the goal of the activity.

• For maximal height the optimal angle is 90o.

• For maximal horizontal distance the optimal

angle is 45o.

## 50.

Projection angle = 10 degrees10 degrees

## 51.

Projection angle = 45 degrees10 degrees

30 degrees

40 degrees

45 degrees

## 52.

Projection angle = 60 degrees10 degrees

30 degrees

40 degrees

45 degrees

60 degrees

80 degrees

## 53.

Projection angle = 75 degrees10 degrees

30 degrees

40 degrees

45 degrees

60 degrees

75 degrees

80 degrees

So angle that maximizes Range

( optimal) = 45 degrees

## 54.

Example: A ball traveling at 25 m/s drive off ofthe edge of a cliff 50 m high. Where do they

land?

25 m/s

Horizontally

x = x0 + (v0)x t

Initial Conditions

vx = 25

m/

s

vy0 = 0 m/s

a =- 9.8

m/ 2

s

Vertically

v = v0-gt

y = y0 + v0t + 1/2gt2 ….

t=0

v2 = v02 - 2g(y-y0)….

y0 = 0 m

x = 25 *3.19 = 79.8 m

-50 = 0+0+1/2(-9.8)t2 … t = 3.19 s

y =- 50 m

x0 =0 m

79.8 m