١

Tikrit University

College of Eng. Subject: Engineering Mechanics (Dynamics)

Mech. Eng. Dept. Syllabus

Class: 2nd Year

1. Principle of dynamic /Introduction.

2. Kinematic/ Plane rectilinear motion.

3. Kinematic/Plane curvilinear motion.

4. Rectangular coordinate of motion.

5. Normal-tangential coordinate of motion.

6. Polar coordinate of motion.

7. Relative Motion (Translating axes).

8. Kinematic/ Plane Rigid bodies.

9. Kinetics/Plane of Rigid bodies.

10. Work and energy.

11. Kinetic energy of particle.

12. Kinetic energy of rigid body.

13. Impulse Momentum Equations.

Text Book: Engineering Mechanics Dynamics J. L. MERIAM (6th Edition). References: Engineering Mechanics Dynamics twelfth edition R.C. Hibbeler.

Four tutorial sheets, Two for each Term.

Four Terms Exams., Two for each Term.

Marks Distribution: Quiz 10%, Terms Exams 30%, Final Exam 60%

٢

1. Principle of dynamic /Introduction

There are two parts of the mechanics which they are: Statics (equilibrium of bodies) and Dynamics (motion of bodies). The motion of bodies will be notes under forces action the force itself is invisible but the action of which is seen, so for dynamics and when the particles moves from one place to another, in this case another parameter will found which is the time. Here the dynamics will consider which deals with the accelerated motion of body. The subject of dynamic will be presented in two parts: kinematics, which treats only the geomteric aspects of mtion, and kinetics, which is the analysis of the forces causing the motion. Applications:

Strength of structures and machines (robots, cars, airplanes)

Vibrations (engine vibrations, bridges, wheels)

Fluid mechanics (airplanes, fluid machinery)

Electrical machines and apparatus (motors, transducers).

Basic concept: The basic concepts can be summarized as: Space: Is the region occupied by bodies. The position of which can be

determined by means of linear and angular measurements with reference of imaginary axes assumed to have no translation or rotation in space.

Force: Is the vector action of one body on another. Time: Is a measure of succession of events. Mass: Is the quantitative measure of the inertia or resistance to change in

motion of the body. Particale: Is a body with negligible dimensions. When the body on space

and the analysis is for the forces act on it in this case it treats as a particle.

Newton's Laws: Newton's three laws are very important in the dynamics, and can be summarize as:

Law 1: A particle remains at rest or continues to move with uniform velocity (in straight line with a constant speed) if there is no unbalanced force acting on it.

Law 2: The acceleration of particle is proportional to the resultant force acting on it and is in the direction of this force.

٣

Law �: The force of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.

Newton's second law forms the basis for most of the analysis in dynamics. For a particle of mass (m) subjected to a resultant force F, the law may be written as: F � ma …… (1-1) Where a is the resultant acceleration of the particle. There is an another acceleration which call the gravitational acceleration Known as � is the attractive of the earth to the bodies its values are take simply as (���1 m�s2) in SI unit and (�2�2 �t��s2) in US unit, these two values are often used in mechanics. The weight (� ) is different than the mass, which can be express as: � � m� ……. (1-2) The unit of the weight is the same of the unit of force (N) or �� m�s2� The four fundamental quantities of mechanics, and their units and symbol for the two systems, are summarized in the following table. Quantity Dimensional

Symbol SI Units U.S Customary Units

UNIT SYMBOL UNIT

SYMBOL

Mass M kilogram Kg slug ----- Length L meter m foot ft. Time T second s second Sec. Force F newton N pound Ib

2� �inematic� rectilinear motion:

Kinematics of a particle that moves along a rectilinear or straight line path with no rotation (Rectilinear kinematics). The kinematics of a particle is characterized by specifying, at any given instant, the particle's position, velocity, and acceleration. Position: The straight-line path of a particle will be defined using a single coordinate axis s, figure1.The origin 0 on the path is a fixed point, and from this point the position coordinate s is used to specify the location of the particle at any given

٤

instant. The magnitude of s is the distance from 0 to the particle, usually measured in meters (m) or feet(ft.), at time t� t the particle P has moves in a straight line to place and its coordinate becomes s � s the change in position coordinate during the interval t is called the displacement ∆s of the particle. The displacement would be negative if the particle moved in the negative s- direction.

Fi��re 1

�elocit� an� �cceleration:

The particle when move from one place to another along a displacement s it take a time t, in an average velocity

If we take smaller and smaller values of ∆t, the magnitude of ∆s becomes smaller and smaller. Consequently, the instantaneous velocity is a vector defined as . or

Similarly for the acceleration

. or

From equations (2-1 &2-2) a new useful relation between displacement, velocity, and acceleration. By eliminating dt can be obtained the following:

�onstant acceleration � : At velocity constant the acceleration will be Zero. But when the acceleration is constant the following equations for displacement, velocity can be deriving as:

٥

from equation (2-2)

Or

(a) From equation (2-1)

Substitute the values of the velocity from equation (a) and integrate the above equation then

or

(b) And from equation (2-3)

After integration and rearrangement, the teams it can be deduce:

N�T�: T�e a�o�e e��ations �a� �� an� c� are �se� w�en t�e acceleration is constant � �NL�

٦

��ample:

The car in �i��re 1 moves in a straight line such that for short time its velocity is defined by v� ��t2 �2t� m�s, where it is in seconds. Determine its position and acceleration when t�� s, the initial conditions are t � �� s � �

S�L�T��N:

Since v=f (t) the position can be determined from

Where

s = t3 +t2

When t=3 s

S= (3)3 + (3)2 =36 m Ans.

Since v=f (t) the acceleration can be determined from

Where

When t=3 Ans.

٧

��ample: A rocket is fired vertically up from rest. If it is designed to maintain a constant upward acceleration of 1���, calculate the time t required for it to reach an altitude of ���m and its velocity at that position.

S�L�T��N: Since the acceleration is constant then

And the rocked start from rest it mean ( )

So, = Ans.

And the velocity at that �� �m or ����� m altitude can be calclated from

Where = =940 m/s Ans. Or

� 1.5(9.81) (63.9) = 940 m/s Ans.

٨

T��T B��� P��BL�MS: 2/3 The velocity of a particle which moves along s-axis is given by v = 2 - 4t + 5t3/2, where t is in seconds and v is in meters per second. Evaluate the position s,

velocity v, and acceleration a when t = 3 s. The particle is at the position s0 =3 m when t =0. Ans. s = 22.2 m, v = 15.98 m/s a = 8.99 m/s2 2/5

The acceleration of a particle is given by a = 2t -10, where a is in meters per second squared and t is seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is so = - 4, and the initial velocity vo = 3 m/s. Ans. v =3 – 10t + t2 (m/s) s = -4 + 3t -5t2 + 1/3 t3 (m)

2/8 The velocity of a particle moving in a straight line is decreasing at the rate of 3 m/s per meter of displacement at an instant when the velocity is 10 m/s. Determine the acceleration a if the particle at instant. Ans. a = - 30 m/s2

٩

�� �inematic� ��r�ilinear motion:

It is clear that the curvilinear motion is that the particle path is a curve so more than one axes is necessary to describe the movement of which or a plane with two axes. The types of axes which they are useful for solving the curvilinear motion can be summarized as:

a. The Cartesian for two dimensional are (x, y) coordinates.

b. The normal and tangential (n, t) coordinates.

c. The Polar (r,) coordinates.

If the motion of the particle a long a plane curve as shown in the figure (1-3) at time t the particle is at position �, which is located by the position vector r measured from some convenient fixed origin �. If both the magnitude and direction of r are known at time t� then the position of the particle is completely specified. At time t� t, the particle is at � locate by the position vector r� r. As shown that a vector addition not scalar. Where the displacement of the particle during time t is a vector r. and the distance is

the scalar length s measured.

Figure (1-3)

�elocit� an� �cceleration:

The particle when move along the path from � to �. the displacement ∆s it take a time ∆t, in an average velocity

If we take smaller and smaller values of ∆t, the magnitude of ∆s becomes smaller and smaller. Consequently, the instantaneous velocity is a vector defined as . or

١٠

The velocity of the particle at position � is always a vector tangent to the path , the velocity of the new position of the particle is the tangential vector as shown in figure(2-3).

Figure (2-3)

Similarly for the acceleration

. or

The acceleration vector is neither normal to path nor tangent to the path, and the relation between the vectors velocities of the particle as shown in figure (3-3) can be written as:

=

��1 �ectan��lar coor�inates ��� ��

١١

Cartesian

A curvilinear motion is of a particle defined by �� � ��� – 1�t� m�s and �� �1�� – �t2� m� It is also known that ��� when t�� plot the path of the particle and determine its velocity and acceleration, and the angle when the position � �� is reached.

S�L�T��N:

١٢

1. To plot the path on (x, y) coordinate y is already available so

Vx= by integrating dx the values of X can be found so

X = 50t -8t2

From x and y

2. To determine the velocity and the acceleration:

vy= Vy= = (100 – 4t2)= -8t m/s

ay = ay= (-8t)= -8 m/s2

ax = ax= m/s2

when y=0

0=100 – 4t2 t = 5 s

So, vx=50 – 16(5) =-30 m/s

vy = -8(5) = - 40 m/s

= 50 m/s

= 17.89 m/s2

= = 53.13 o

١٣

The velocity and acceleration can be written in a vector form as follows

��1�1 Pro�ectile Motion:

For a special case when a particle is used as extruded and for the following assumptions:

a. No aerodynamic drag.

b. Neglecting the earth rotation.

c. The altitude change is small enough, the acceleration due to

gravity can be treated as constant but in a mines sign (ay= -g), ax = 0,

Then the rectangular coordinate can be useful for the trajectory analysis, and all equation of constant acceleration can be used as following:

Where

, and Where is the projectile angle as shown below.

١٤

Example:

A rocket has expended all its fuel when it reaches position A, where has a velocity of � at an angle with respect to the horizontal. It then begins unpowered flight and attains a maximum added height h at position B after traveling a horizontal distance s from A. Determine the expressions for h and s, the time t of flight from A to B, and the equation of path. For the interval concerned, assume a flat earth with a constant gravitational acceleration g and neglect any atmospheric resistance.

Solution:

…. (a)

…(b)

Position B is reached when , so t can be evaluated as:

١٥

Substitution of this value for the time into the expression for y gives the maximum added altitude

and

Note that s is maximum when

The equation of the path can be obtained by substituting the value of t from equation (a) in y expression equation (b), so

TEXT BOOK PROBLEMS:

2/62 A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to and . Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 2 s.

Ans. v=14.42 mm/s , a = 7.2 mm/s2, 0 , 0

2/66 The x- and y- motions of guides A and B with right-angle slots control the curvilinear motion of the connecting pin P, which slides in both slots. For a short interval the motions are governed by and , where x and y are in millimeters and t in seconds. Calculate the magnitudes of the velocity v and acceleration a of the pin for t = 2 s. sketch the direction of the path and indicate its curvature for this instant.

Ans. v = 2.24 mm/s, a = 2.06 mm/s2

١٦

2/�� A rocket is released at point A from a jet aircraft flying horizontally at 1000 km/h at an altitude of 800 m. If the rocket thrust remains horizontal and gives the rocket a horizontal acceleration of 0.5g , determine the angle from the horizontal to the line of sight to the target.

Ans.

١٧

��2 ������ ��� ���������� ����������� �����:

If a particle moves along a curvilinear motion is movement can be described using the normal and the tangential axes ����� position of which are on the particle place. The normal and tangential axes are usually perpendicular to each other.

As shown in the figure that the particle when moves from � to B to � the positive direction of the normal � at any position is taken toward the center of the curvature of the path.

Velocity and acceleration:

The velocity direction is always tangent to the path and the acceleration have two parts which they the normal acceleration �� its direction toward the center of curvature and tangential acceleration �� with a direction tangential to the path. To calculate the velocity and acceleration magnitude a unit vectors �� and �� are to the normal and tangential directions respectively.

When the particle moves from A to with time increment dt it have a distance of �� so the magnitude of the velocity can be written as:

From the figure (a) below

١٨

where is the radius of curvature.

�

The velocity vector is

And the acceleration vector is

To determine From the above figure (b). the arc det can be writtwn as

d and the unit

The direction of is given by ��, so � dividing by dt we get:

or

……(c)

From equation a, b, and c the accelertation can be written as:

١٩

Where

= ρ

�

Where

and and

If the equaion of the carvilinear of motion of the partical is known, then the radius of carvature can be calculated from:

� � �����

T�� �������� M�����:

The velocity and acceleration in the spatial case when the path is a circle where the radius of curvature is replaced by the radius of the circle r and the angle is replaced by as shown in the figure, so

����: all the equations of constant acceleration are valid only replace the acceleration ac by at .

٢٠

Example:

To anticipate the dip and hump in the road, the driver of a car applies her brakes to produce a uniform deceleration. Her speed is 100 km/h at the bottom A of the dip and 50km/h at the top C of the hump, which is 120 m along the road from A. If the passengers experience a total acceleration of 3 m/s2 at A and if the radius of curvature of the hump at C is 150 m, calculate (a) the radius of curvature ρ at A, (b) the acceleration at the inflection point B, and (c) the total acceleration at C.

Solution The velocities in meter/second are:

Since a uniform (constant) deceleration:

At A:

At B: since the radius of curvature is infinite at the inflection point, so

٢١

At C:

The acceleration in unit vectores can be written as:

And the magnitude of the acceleration is

= =2.73 m/s2

E������: When the skier reaches point � along the parabolic path in figure below, he has a speed of 6 �/� which is increasing at 2 �/�2� Determine the direction of his velocity and the direction and magnitude of his acceleration at this instant.

S�������