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www.soran.edu.iq 2
week1 General PhysicsContents A. Mechanics
1. Physics and measurement2 .Motion and dimensions
3 .Vectors4 .Motion in two dimensions
5 .Laws of motion6 .Circular motion
7 .Energy8 .Potential energy
9 .Linear momentum and collision10 .Rotation
11 .Angular momentum
B. Properties of Matter12 .Static and Elasticity
13 .Universal gravitation 14 .Fluid mechanics
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1. Physics and measurement1.1 SI prefixes for power of ten
1.2 The Greek Alphabet1.3 Standard Abbreviations and Symbols for Units
1.4 Mathematical Symbols and their meaning1.5The Fundamental SI units
1.6 Dimensional Analysis
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1.5 The Fundamental SI units
The Fundamental SI units
quantity unit abbreviation
Mass kilogram kg
Length meter m
Time second s
Temperature kilvin K
Electric current ampere A
Luminous intensity candela cd
Amount of substance mole mol
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Motion in one dimension
•week2 2.1 Position, velocity, and speed2.2 Instantaneous velocity and speed2.3 Acceleration2.4 One-Dimensional motion with constant acceleration2.5 Freely Falling Objects
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The displacement of a particle is defined as its change in position in some time interval.
∆x = xf – xi …………. (2.1)
Instantaneous velocityThe instantaneous velocity is vx equals the limiting value of the ratio
∆x / ∆t as ∆t approaches zero: vx = lim (∆x / ∆t) = dx /dt ……….. (2.4)
∆t---------0
The instantaneous velocity can be positive, negative, and zero.Instantaneous speed
Instantaneous speed of a particle is defined as the magnitude of its instantaneous velocity
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One-Dimensional motion with constant acceleration
•week3A simple type of one-dimensional motion is that in which the acceleration is constant.In this case the average acceleration āx over any time interval is numerically equal to the instantaneous acceleration ax at any instant within the interval, and the velocity changes at the same rate through the motion (āx = ax ).If ti = 0 and tf any later time t, we find that
ax = (vxf – vxi) / (t – 0)or
vxf = vxi + ax t (for constant ax) ……… (2.7)
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3 .Vectors
•Week43.1 Coordinate systems
3.2 Vector and scalar quantities3.3 Some properties of vectors
3.4 Components of a vector and unit vectors3.5 Vector product (multiplication)
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Coordinate systems
(a ) Cartesiancoordinate
This system of coordinate is represented by two or three dimensions, i.e., plane or space.
In two dimensions (see the figure), the vector from the origin O = (0,0) to the point A = (2,3) is simply
written as a = (2,3)
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3.2 Vector and scalar quantities
1 .The scalar and vector quantitiesThe scalar quantity is that quantity
determines by magnitude only, such as, temperature T and energy E.
The vector quantity is that quantity determines by magnitude and direction, such as, displacement x, velocity v, and
force F. A vector is a geometric entity characterized
by a magnitude (in mathematics a number, in physics a number times a unit) and a
direction .In rigorous mathematical treatments, a
vector is defined as a directed line segment, or arrow, in a Euclidean space.
•Week5
The Addition and subtraction of vectors A and B are two vectors, their sum is R , i.e., R = A + B
Vectors can be subtracted, i.e., if A and B are two vectors then their subtraction is
C = A – B = A + (-B)
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3.4 Components of a vector and unit vectors
Components of a vectorThe components of the vector A in two dimensions are the vectors
Ax and Ay, in such a way that : A = Ax + Ay
Ax = A cos θ
Ay = A sin θ A = [(Ax)2 + (Ay)2]½
tan θ = Ay / Ax
θ = tan-1 (Ay / Ax)
Week6
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Unit vectorsAnother way to express a vector in three dimensions is to introduce the three standard basis vectors:
e1 = (1,0,0) , e2 = (0,1,0) , e3 = (0,1,0)
These have the intuitive interpretation as vectors of unit length pointing up the x, y, and z axis of a Cartesian coordinate system, respectively, and they are sometimes referred to as versors of those axes.
In terms of these, any vector in three dimensions space can be expressed in the form:
( a,b,c = )a(1,0,0) + b(0,1,0) + c(0,0,1)
= ae1 + be2 + ce3
These three special vectors are often instead denoted i, j, k , the versors of the three dimensional space (or ) , in which the hat symbol (^) typically denotes unit vectors
(vectors with unit length) .The notation ei is compatible with the index notation and the summation convention commonly used in higher level mathematics, physics, and engineering.
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4 .Motion in two dimensionsWeek7
4.1 The position, Velocity, and Acceleration Vectors
4.2 Two-dimensional motion with constant acceleration
4.3 Projectile motion4.4 Uniform circular motion
4.5 Tangential and radial acceleration
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4.1 The position, velocity, and acceleration vectorsWe describe here the motion of a particle in two dimensions, i.e., motion in xy-plane.
The description of the position of a particle is by position vector r.
The displacement vector ∆r is defined as the difference between its final position vector and its initial position vector:
Displacement vector ∆r ≡ rf – ri ………………….. (4.1)
The direction of ∆r is shown in the Figure below.
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Two-dimensional motion with constant acceleration Week8
4.2 Two-dimensional motion with constant accelerationThe position vector for a particle moving in the xy-plane can be written asr = xi + yj ………………. (4.6)
where x, y and r change with time as the particle moves. From the Equations 4.3 and 4.6 the velocity of the particle can be obtained as
v =dr/dt = dx/dt i +dy/dt j
= vx i + vy j ………….. (4.7)
Because the acceleration is constant, its components ax and ay are also constants.
Therefore, the equations of kinematics to the x and y components of the velocity vector can be applied.
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4.4 Uniform circular motion
Week94.4 Uniform circular motionThe movement of an object in a circular path with constant speed v is called uniform circular motion.Even though an objects move at constant speed in circular path, it still has acceleration.The acceleration depends on the change in the velocity vector.The acceleration depends on the change in the magnitude of the velocity and / or by a change in the direction of the velocity.The change in the direction of the velocity occurs for an object moving with constant speed in a circular path.The velocity vector is always tangent to the path of the object and perpendicular to the radius of the circular path.
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Laws of motion (Particle’s dynamics) ]
•Week105.1 The concept of the Force An object accelerates due to an external force.The object accelerates only if the net force acting on it is not equal zero.The net force acting on an object is defined the vector sum of all forces acting on the object.The net force is the total force, the resultant force, or the unbalanced force.If the net force exerted on an object is zero, the acceleration is zero and its velocity remains constant. Definition of equilibriumWhen the velocity of an object is constant (including when the object is at rest), the object is said to be in equilibrium.
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5.2 Newton’s first law
“In the absence of external force, when viewed from an inertial reference frame, an object at rest remains at rest and an
object in motion continues in motion with constant velocity .” That is to say, when no force acts on an object, the acceleration of the object is zero.
From the first law, we conclude that any isolated object (one that does not interact with its environment) is either at rest or moving with constant velocity.
Definition of inertiaThe tendency of an object to resist any attempt to change its velocity is called inertia
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5.3 Mass Definition of mass
“Mass is an inherent property of an object and is independent of the object’s surrounding and of the method used to measure it.”
Mass is a scalar quantity Mass and weight are two different quantities
The weight of an object is equal to the magnitude of the gravitational force exerted on the object and varies with location.
For example, a person who weight 900 N on the Earth weights only 150 N on the Moon.
The mass of an object is the same everywhere: an object having a mass of 2 kg on the Earth also has a mass of 2 kg on the Moon.
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5.5 The Gravitational force and weight
Week11The attraction force exerted by the Earth on an object is called the gravitational force Fg. This force is directed toward the center of the Earth, and its magnitude is called the weight of the object.Freely falling object experience an acceleration g acting toward the center of the Earth.Applying Newton’s second ΣF = ma to the freely falling object of mass m, with a = g and ΣF = Fg, we obtain
Fg = mg …………………… (5.4)Thus, the weight of an object, being defined as the magnitude of Fg , is equal to mg, and g = 9.8 m/s2.
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5.6 Newton’s third lawNewton’s third law states that:
If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1:
F12 = - F21 …………… (5.5)
The force that object 1 exerts on object 2 is called the action force and the force of object 2 on object 1 the reaction force.Either force can be labeled the action or reaction force.
In general, “The action force is equal in magnitude to the reaction force and opposite in direction”.
The action and reaction forces act on different objects
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5.7 Some applications of Newton’s laws
Week12Objects in EquilibriumIf the acceleration of an object is zero; the particle is in equilibrium. If apply the second law to the object, noting that a = 0 , we see that because there are no forces in the x direction, ΣFx = 0 The condition ΣFy = may = 0 gives
ΣFy = T – Fg = 0 or T = Fg
The forces T and Fg are not an action-reaction pair because they act on the same object.The reaction force to T is T’, the downward force exerted by the object on the chain.The ceiling exerts on the chain a force T” that is equal in magnitude to the magnitude of T’ and points in the opposite direction.
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Work and Energy
Week13 6.1 Work done by a constant forceIf an object undergoes a displacement Δr under the action of a constant force
F, the work done, W , by the force is ,W = F Δr cos θ ………. (6.1)From Equation 6.1, the work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application, i.e., θ = 90o ,then W = 0 because cos 90o = 0.
W = F Δr cos 90o
If the applied force F is in same direction as the displacement Δr, then θ = 0 and cos θ = 1. In this case, Equation 6.1 gives W = F Δr
Work is scalar quantity, and its units are force multiplied by length .The SI unit of work is the newton . meter (N.m), this unit is called joule (J) .
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6.2 Work done by a varying force
Consider a particle being displaced along the x axis under the action of a force, Fx , that varies with position, x.We can express the work done by Fx as the particle moves from xi to xf as
W = ∫ Fx dx………. (6.2)
This equation reduces to Equation 6.1 when the component Fx = F cos θ is constant
If more than one force acts on a particle, the total work done on the system is the work done by the net force.
If the net force in the x direction is Σ Fx then the total work, or net work, done as the particle moves from
xi to xf is Σ W = Wnet = ∫ (∑ Fx )dx ………….. (6.3)
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6.3 Kinetic energy and the Work–Kinetic energy Theorem
Week14 6.3 Kinetic energy and the Work–Kinetic energy TheoremConsider a system consisting of a single object.The figure below shows a block of mass m moving through a displacement directed to the right under the action of a net force ΣF. From Newton’s second law that the block moves with an acceleration a.If the block moves through a displacement Δr = Δxi = (xf - xi)i the work done by the net force ΣF is
)..........(....................dx)F(Wf
i
x
x86
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6.5 Potential energy of a system
Week15We introduced the concept of kinetic energy associated with motion of objects. Now we introduce potential energy, the energy associated with the configuration of a system of objects that exert forces on each other.
Potential energyConsider a system consists of an object and the Earth, interacting via the gravitational force.
We do some work on the system by lifting the object slowly through a height Δy = yb – ya .
While the object was at the highest point, the energy of the system had the potential to become kinetic energy, but did not do so until the book was allowed to fall .
Thus, the energy storage mechanism before releasing the object is called potential energy.
In this case, the energy is gravitational potential energy.
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7 .Linear momentum and collision Week16
The momentum of an object is related to both its mass and its velocity.The concept of momentum leads us to second conservation law, that of conservation of momentum.We introduce a new quantity that describes motion, linear momentum .
Consider two particles interact with each other.According to Newton’s third law, F12 = – F21 and then
F12 + F21 = 0
m1a1 + m2a2 = 0
and
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7.2 Impulse and Momentum Week17
Assume that a single force F acts on a particle and that this force may vary with time.According to Newton’s second law, F = dp/dt, or
dp = Fdt …………………… (7.7)If the momentum of the particle changes from pi at time ti to pf at time tf , then
The quantity is called the Impulse of the force F acting on a particle over the time interval Δt = tf –ti Impulse is a vector defined by
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)...(..........dtdf
i
f
i
t
tif 87 Fppppp
p
)...(..........dtf
i
t
t97 FI
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7.3 Collisions in One Dimension
Week18We use the law of conservation of linear momentum to describe
what happens when two particles collide .There are two types of collision, elastic and inelastic.
An elastic collision between two objects is one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision.
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Week19 Fluid statics: study of fluids at restDifferent from fluid dynamics in that it concerns pressure forces perpendicular to a plane (referred to as hydrostatic pressure)If you pick any one point in a static fluid, that point is going to have a specific pressure intensity associated with it:
P = F/A whereP = pressure in Pascals (Pa, lb/ft3) or Newtons (N, kg/m3)F = normal forces acting on an area (lbs or kgs)A = area over which the force is acting (ft2 or m2)
Fluid Statics
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Bernoulli’s EquationWeek20
Z1 + (P1/) + (V12/2g) = Z2 + (P2/) + (V2
2/2g)Wow! Z = pressure head, V2/2g = velocity head (heard of these?), 2g = (2)(32.2) for Eng. SystemIf we’re trying to figure out how quickly a tank will drain, we use this equation in a simplified form: Z = V2/2g
Example: If the vertical distance between the top of the water in a tank and the centerline of it’s discharge pipe is 14 ft, what is the initial discharge velocity of the water leaving the tank? Ans. = 30 ft/s
Can you think of any applications for this?
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Stagnation Point: Bernoulli EquationStagnation point: the point on a stationary body in every flow where V= 0
Stagnation Streamline: The streamline that terminates at the stagnation point.
Symmetric:
Axisymmetric:
If there are no elevation effects, the stagnation pressure is largest pressure obtainable along a streamline: all kinetic energy goes into a pressure rise:
2
2Vp
streamlineaontconspzVp T tan2
1 2
Total Pressure with Elevation:
Stagnation Flow I:
Stagnation Flow II: