· web viewclassical mechanics describes the relationship between the motion of an object and the...

Mechanics

Sometimes referred to as classical mechanics or Newtonian mechanics

is concerned with the effects of forces on material objects A major

development in the theory was provided by Isac Newton in 1687

The part of mechanics that describes motion without regard to its causes is

called kinematics

Here we will focus on one dimensional motion

Position

To describe the motion of an object one must be able to specify its position

at all time using some convinient coordinate system For example a particle

might be located at x=+5 m which means that it is 5 m in the positive direction

from the origin If it were at x=-5 m it would be just as far from the origin but

in the opposite direction

Displacement

A change from one position to another position is called displacement

∆ x=x2minusx1

For examples if the particles moves from +5 m to +12 m then

∆ x=(+12m)minus (+5m )=+7m

( 18 )

the + sign indicates that the motion is in the positive direction The

displacement is a vector quantity

Average velocity

Which is the ratio of the displacement that occurs during a particular

time interval to that interval A common unit of velocity is the meter per second

(ms)

iquestυgtiquest ∆ x∆ t

=x2minusx1

t 2minust 1

Average speed

Is a different way of describing ldquohow fastrdquo a particle moves and involves

the total distance covered independent of direction(ms)

iquest sgtiquest totaldistance∆ t

Average acceleration

When a particlersquos velocity changes the particle is said to accelerate For

motion along an axis the average acceleration over a time interval is

iquestagtiquest ∆υ∆t

=υ2minusυ1

t2minust1

Acceleration is vector quantity The common unit of acceleration is the meter

per second per second (mssup2)

In many types of motion the acceleration is either constant or approximately

so In that case the instantaneous acceleration and average acceleration are equal

( 19 )

a=iquestagtiquestυ2minusυ1

t 2minust 1

Also let υ1 =υ0 (the initial velocity) and υ2=υ (the velocity at arbitrary time)

with this notation we have

υ =υ0+at

In a similar manner we can have

x=x0+υ0 t+12a t 2

where x0is the position of the particle at initial time

Finally from this two equations we can obtain expression that does not contain

Time

υ2=υ02+2a (xminusx0 ) and if x0 = 0 then

υ2=υ02+2ax

These equations may be used to solve any problem in one-dimensional motion

with constant acceleration

All objects dropped near the surface of the earth in the absence of air

resistance fall toward the Earth with the same nearly constant acceleration

We denote the magnitude of free-fall acceleration as g The magnitude of free-

fall acceleration decreases with increasing altitude Furthermore slight

variations occur with latitude At the surface of the Earth the magnitude is

approximately

( 20 )

98 mssup2 The vector is directed downward toward the center of the Earth Free-

fall acceleration is an important example of straight-line motion with constant

acceleration When air resistance is negligible even a feather and an apple fall

with the same acceleration regardless of their masses

Example

υ =υ0+at = 7 + 08x2 =86 ms

Example

d = 220 m υ = 1097 kmsec = 1097 x103 msec a

υ2=υ02+2a (xminusx0 )

= 2a d (υ0 = 0 x0 = 0 )

a= υ2

2d=1097 x103

2 x220=27 x105asymp28 x104g ( g = 98 ms2 )

( 21 )

Exercises

The Concept of Force

Classical mechanics describes the relationship between the motion of an

object and the force acting on it There are conditions under which classical

mechanics does not apply Most often these conditions are encountered when

dealing with objects whose size is comparable to that of atoms or smaller andor

which move at speed of light If the speeds of the interacting objects are very

large we must replace Classical mechanics with Einsteinrsquos special theory of

relativity which hold at any speed including those near the speed of light

If the interacting bodies are on the scale of atomic structure we must replace

Classical mechanics with quantum mechanics Classical mechanics is a special

case of these two more comprehensive theories But still it is a very important

special case because it applies to the motion of objects ranging in size from the

very small to astronomical

The concept of force Force is an interaction that can cause deformation andor

change of motion of an object and it is a vector quantity

( 22 )

Fundamental forces are

gravitational

electromagnetic

weak nuclear ( between leptons )

strong nuclear ( repulsion and attraction )

Newtonrsquos second law

ldquoThe net force on a body is equal to the product of the bodyrsquos mass and the

acceleration of the bodyrdquo In equation form

sum F=ma sum F represents the vector sum of all external forces acting on

the object m is its mass and a is the acceleration of the object

sum F x=max sum F y=ma y sum F z=maz

The SI unit of force is the newton (1 N = 1 kgmssup2)

The magnitude of the gravitational force is called the weight of the object

W=Fg=mg

where g is the magnitude of the free-fall acceleration

( 23 )

There many forces

1 - The Normal Force is perpendicular to the surface

2 - The Frictional Force

3 - The Tension Force FT

4 - The Spring Force F = - kx ( Hooks law ) The minus sign indicates that the

spring force is always opposite in direction from the displacement of the free

end The constant k is called the spring constant

( 24 )

Example

ExercisesAn object has only one force acting on it Can it be at rest Can it have an

acceleration

If a single force acts on it the object must accelerate If an object accelerates

at least one force must act on it

An object has zero acceleration Does this mean that no forces act on it

If an object has no acceleration you cannot conclude that no forces act on it

In this case you can only say that the net force on the object is zero

Is it possible for an object to move if no net force acts on it

Motion can occur in the absence of a net force Newtonrsquos first law holds that an

object will continue to move with a constant speed and in a straight line if

there is no net force acting on it

( 25 )

What force causes an automobile to move

The force causing an automobile to move is the force of friction between the

tires and the roadway as the automobile attempts to push the roadway

backward

The force of the wind on the sails of a sailboat is 390 N north The water exerts

force of 180 N east If the boat has a mass of 270 kg what are the magnitude

and direction of its acceleration

Work

1048708 A force does no work on a object if the object does not move

1048708 The sign of the work depends on the angle θ between the force and

displacement

1048708 Work is a scalar quantity and its units is joule (1 J = 1 Nm)

( 26 )

Kinetic energy

( 27 )

Example

Linear momentum The linear momentum of an object of mass m moving

with a velocity υ is defined as the product of the mass

and velocity

Momentum is a vector quantity with its direction matching that of the velocity

Newton didnrsquot write the second law as F=m a but as

( 28 )

F= changeisinmomentumtimeinterval

=∆ p∆t

Where ∆ t is the time interval during which the momentum change ∆ p This

expression is equivalent to F = ma for an object of constant mass

Example

Centripetal force Consider a ball of mass m tied to a string of length r and

being whirled in a horizontal circular path Let us assume that the ball move

with constant speed Because the velocity changes its direction continuously

( 29 )

during the motion the ball experiences a centripetal acceleration directed

toward the center of motion with magnitude

ac=υt

2

r

The strings exerts a force on the ball that makes a circular path This force is

directed a long the length of the string toward the center of the circle with

magnitude of

F c=mυt

2

r

This force is called centripetal force It can be a frictional force a gravitational

force or any other force

( 30 )

Problem 1

A car travels up a hill at a constant speed of 37 kmh and returns down the hill

at a constant speed of 66 kmh Calculate the average speed for the whole trip

Solution

By definition the average speed is the ration of the total traveled distance and

the total traveled time Let us introduce the total traveled distance of the car as

L Then the time of the travel up the hill is

The time of the travel down the hill is

The total traveled time is

Then the average velocity is

( 31 )

Problem 2

A car is accelerating at Find its acceleration in

Solution

To find an acceleration in we need to use the relations

Then we can write

Problem 3

We drive a distance of 1 km at 16 kmh Then we drive an additional distance

of 1 km at 32 kmh What is our average speed

Solution

By definition the average speed is the ratio of traveled distance and traveled

time

The traveled distance is 2 km The traveled time is the sum of two contributions

time of the motion a distance 1 km with speed 16 kmh It is


Then the average speed is

( 32 )

Problem 4

A speedboat increases its speed uniformly from 20 ms to 30 ms in a distance

of 200m Find

(a) the magnitude of its acceleration and

(b) the time it takes the boat to travel the 200-m distance

Solution

(a) This is the motion with constant acceleration We can use the following

equation to find the magnitude of acceleration

Where

Then

(b) We can find the time of the travel from the following equation

( 33 )

Problem 5

A car is moving with a velocity of 72 kmh Its velocity is reduced to 36 kmh

after covering a distance of 200 m Calculate its acceleration

Solution

The first step is to check that we have correct units for all variables The

velocity should be measured in ms and the distance in meters Then initial and

final speeds are

The covered distance is

To solve this problem we need to use the following equation which relates

initial speed final speed acceleration and travelled distance

From this equation we can find acceleration

The acceleration is negative which means that the direction of acceleration is

opposite to the direction of velocity ndash the car is slowing down

( 34 )

Problem 6

A particle moves in a circle of radius 1 m Its linear speed is given by

where t is in second and v in metersecond Find the radial and tangential

acceleration at

Solution

The tangential acceleration is defined as the change of the speed (magnitude

of the velocity) of the particle Therefore the tangential acceleration is

The radial acceleration is centripetal acceleration which changes the

direction of velocity The centripetal acceleration is determined by the speed

(not the change of the speed) and the radius of circular orbit

The velocity at

is then

( 35 )

Problem 7

The force required to stretch Hookes Law spring varies from 0 N to 65 N as

we stretch the spring by moving one end 63 cm from its unstressed position

Find the force constant of the spring Answer in units of Nm

Solution

We understand from the formulation of the problem that the spring force is 65

N when the spring is stretched by

which in SI units is

At the same time from the Hookes law we know that the spring force is

Where

k is a force constant (force constant) of the spring We substitute the known

values on the above expression and obtain

Then

( 36 )

Problem 8

The work done by the heart is 1 joule per beat Calculate the power of the

heart if it beats 72 times in one minute

Solution

In this physics problem we need to use definition of the power The power is

the work done by unit time ndash per one second

We know that the heart does 1 joule of work per beat and there are 72 beats

per minute

It means that the heart does work of

per minute ndash per 60 seconds

Then we can easily find the work done per one second which is power We

just need to divide the total work by the time

( 37 )

the + sign indicates that the motion is in the positive direction The

displacement is a vector quantity

Average velocity

Which is the ratio of the displacement that occurs during a particular

time interval to that interval A common unit of velocity is the meter per second

(ms)

iquestυgtiquest ∆ x∆ t

=x2minusx1

t 2minust 1

Average speed

Is a different way of describing ldquohow fastrdquo a particle moves and involves

the total distance covered independent of direction(ms)

iquest sgtiquest totaldistance∆ t

Average acceleration

When a particlersquos velocity changes the particle is said to accelerate For

motion along an axis the average acceleration over a time interval is

iquestagtiquest ∆υ∆t

=υ2minusυ1

t2minust1

Acceleration is vector quantity The common unit of acceleration is the meter

per second per second (mssup2)

In many types of motion the acceleration is either constant or approximately

so In that case the instantaneous acceleration and average acceleration are equal

( 19 )


t 2minust 1



υ =υ0+at


x=x0+υ0 t+12a t 2



Time


υ2=υ02+2ax








approximately

( 20 )





Example

υ =υ0+at = 7 + 08x2 =86 ms

Example

d = 220 m υ = 1097 kmsec = 1097 x103 msec a


= 2a d (υ0 = 0 x0 = 0 )

a= υ2

2d=1097 x103

2 x220=27 x105asymp28 x104g ( g = 98 ms2 )

( 21 )

Exercises
















( 22 )


gravitational

electromagnetic











W=Fg=mg


( 23 )

There many forces







( 24 )

Example


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )


t 2minust 1



υ =υ0+at


x=x0+υ0 t+12a t 2



Time


υ2=υ02+2ax








approximately

( 20 )





Example

υ =υ0+at = 7 + 08x2 =86 ms

Example

d = 220 m υ = 1097 kmsec = 1097 x103 msec a


= 2a d (υ0 = 0 x0 = 0 )

a= υ2

2d=1097 x103

2 x220=27 x105asymp28 x104g ( g = 98 ms2 )

( 21 )

Exercises
















( 22 )


gravitational

electromagnetic











W=Fg=mg


( 23 )

There many forces







( 24 )

Example


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )





Example

υ =υ0+at = 7 + 08x2 =86 ms

Example

d = 220 m υ = 1097 kmsec = 1097 x103 msec a


= 2a d (υ0 = 0 x0 = 0 )

a= υ2

2d=1097 x103

2 x220=27 x105asymp28 x104g ( g = 98 ms2 )

( 21 )

Exercises
















( 22 )


gravitational

electromagnetic











W=Fg=mg


( 23 )

There many forces







( 24 )

Example


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Exercises
















( 22 )


gravitational

electromagnetic











W=Fg=mg


( 23 )

There many forces







( 24 )

Example


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )


gravitational

electromagnetic











W=Fg=mg


( 23 )

There many forces







( 24 )

Example


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )







( 24 )

Example


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )


acceleration










( 25 )




backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )



backward




Work



displacement


( 26 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Kinetic energy

( 27 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Example



and velocity



( 28 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )


=∆ p∆t



Example




( 29 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )



ac=υt

2

r



magnitude of

F c=mυt

2

r



( 30 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 1



Solution







( 31 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 2


Solution


Then we can write

Problem 3



Solution


time





( 32 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 4


of 200m Find



Solution



Where

Then


( 33 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 5



Solution



final speeds are







( 34 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 6



acceleration at

Solution






The velocity at

is then

( 35 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 7




Solution





Where



Then

( 36 )

Problem 8



Solution




per minute





( 37 )

Problem 8



Solution




per minute





( 37 )

· web viewclassical mechanics describes the relationship between the motion of an object and the...

Documents