Chapter 7 GravitationPosted on 22/01/2011 by amimo5095Introduction

All objects with mass attract each other with a gravitational force. The magnitude of the force between any two objects depends

on their masses and the distance between the two objects.

7.1Newton’s Law of Universal Gravitation

> Newton’s Law of universal gravitation states that the force of attraction between two objects is directly proportional to their

masses and inversely proportional to the square of their distance apart.

where G is the universal gravitational constant

G = 6.67 x 10-11 m3 kg-1 s-2

Note:

1) Gravitational forces are very weak unless it is for objects with enormous mass such as planets and stars.

2) Gravitational force is a mutual attraction. We attract the Earth with the same force that the Earth attracts us.

3) Newton’s Law of universal gravitation is an example of an inverse square law. The force decreases in proportional to the

square of the distance.

So when the distance is 2r, the force is 

when the distance is 3r, the force is 

Example 1

Why is it that the gravitational force of earth causes the object to be accelerated to earth but not otherwise ?

Solution

 

 

 

Example 2

Two stars of mass 6.0 x 1032 kg and 9.0 x 1030 kg are 1.4 x 1016 m apart. The net gravitational force is zero at a point P between

the two stars and the distance of this point from the bigger star is r. What is the distance r?

Solution

 

 

 

 

7.2 Gravitational Field Strength

A gravitational field represents a region where a body having a mass experiences gravitational force.

> Gravitational field strength is defined as the force of gravity per unit mass.

> Gravitational field strength, 

Unit: N kg-1 and it is a vector quantity.

Example 3

What is the gravitational field strength acting on an object with mass m placed on the surface of the earth? Taking the mass of

earth as M and the radius of earth as R.

Solution

 

 

 

7.3 Relationship Between g and G

> A body of mass m at a place on the earth’s surface where the acceleration of free fall is experiences a force F = mg (i.e. its

weight).

> Assuming the earth has mass M and radius R, we can also calculate the same force using, Newton’s law of gravitation.

So, 

Therefore, 

Then 

Variation of acceleration due to gravity g with height and depth on the earth

Example 4

The mass of a star is 2.0 x 1030 kg and its radius is 1.5 x 106 m.Find the gravitational field strength at:

(a) The surface

(b) 1.0 x 106 m above the surface of the star [G = 6.67 x 10-11 N m2 kg-2]

Solution

 

 

Example 5

What is the mass of the earth given that its radius is 6400 km and the gravitational field strength at the surface is 9.81 N kg-1. [G

= 6.67 x 10-11 N m2 kg-2]

Solution

 

 

 

 

7.4 Gravitational Potential

> The gravitational potential V at a point in a gravitational field is defined as the work done in bringing a unit mass from infinity

to that point. 

Unit: J kg-1 and it is a scalar quantity.

> The gravitational potential energy U of a mass m at a point in a gravitational field is defined as the work done in bringing the

mass from infinity to that point. 

Unit: J and it is a scalar quantity.

> The relationship of V and U is 

> The relationship of F and U is 

> The relationship of g (or E) and V is 

Note: Gravitational potential is defined as V= 0 at infinity. Since gravitational potential increases as height increases, V must

have negative values when it is lower than infinity.

Example 6

What is the gravitational potential V at 3.50 x 1010 m from the center of a star of mass 8.45 x 1035 kg.[G = 6.67 x 10-11 N m2 kg-2]

Solution

 

 

 

 

Example 7

What is the change in potential energy when a rocket of mass 40 000 kg moves from the surface of the earth to a height of 1.80

x 107 m above the earth? Given that the mass of the earth is 6.00 x 1024 kg and its radius is 6.40 x 106 m. [G = 6.67 x 10-11 N

m2 kg-2]

Solution

 

 

 

7.5 Satellite Motion in Circular Orbits

Satellite

A satellite is an object orbiting around a larger body by gravitational attraction.

e.g. The moon is a natural satellite of earth.

The earth is a natural satellite of sun.

There are also artificial satellites launched for the purpose of monitoring the weather, space observation and global

communications.

Energy of a satellite

• Velocity and period of satellite in a fixed orbit

To keep a satellite in circular orbit, the centripetal force needed is being supplied by the gravitational force.

So 

 since GM=gR2

If the satellite is close to earth, then r » R

Therefore v2 = gR

Once v is found, the period of satellite T is given by 

· Kinetic energy,potential energy and total energy of satellite

Kinetic energy = 

= 

Kinetic energy = 

Potential energy = 

Total energy = Potential energy + kinetic energy GMm 1 GMm

=  +

= 

Example 8

A satellite orbits the earth with period of rotation 2 hours. Given that the mass of the earth 6.0 x 1024 kg and the radius of the

earth is 6.4 x 106 m. Calculate:

(a) The height where the satellite is orbiting.

(b) The speed of the satellite. [G = 6.67x10-11 N m2 kg-2]

Solution

7.6 Escape velocity

> Escape velocity is the velocity required by an object on the surface of the planet in order to escape from the planet to infinity.

The object will not be pulled back by gravity.

> Take the earth for example, the gravitational potential energy on the surface of the earth is   and the gravitational

potential energy at infinity is 0.

So, the energy required to send an object to infinity is

0 – ( ) = 

> This energy is being supplied by the kinetic energy

So,  = 

> Therefore, the escape velocity 

Example 9

What is the escape velocity from the surface of the sun?

[Mass of the sun = 2.0 x 1030 kg, radius of the sun = 7.0 x 108 m, G = 6.67 x 10-11 N m2 kg-2]

solution

 

 

Example 10

what is the escape velocity from the surface of the earth?

g on the surface of the earth = 9.81 m s-2, radius of the earth = 6400 km]

4.0 Circular Motion

Posted on 08/02/2010 by amimo5095

Introduction

In this topic, we will study the angular motion with angular kinematics and rotation dynamics. We will investigate the relationship

between,q the angular displacement, w the angular velocity and a the angular acceleration.

 

4.1 Angular Velocity (halaju sudut)

>Defination of angular displacement,q

q =    r = radius , s = linear displacement/perimeter

unit : radian

> Definition for angular velocity

The rate of change of angular displacement for a circular motion.

Unit: Radians per second

 

> Definition of period, T

The time taken for the particle to complete a circle of the circular motion. For a complete cycle, where q  = 2p radians

Other information

            360° º 2p

 

Example 1

What angle in degree has a car travelled around a circular track if the track has a radius of 100 m and distance by the car is

300 m ?

 

Solution

Angular displacement,

 

 

                                                           

Example 2

For a point on the circumference (radius = 10 cm) which move through(from beginning to the end) the circle in 60 s. calculate:

(a) angular velocity of the circumference.

(b) linear speed in m/s.

 

 

 

4.2 Centripetal Acceleration (pecutan memusat)

> Definition

Centripetal acceleration is the acceleration where the direction of the velocity change Dv is, perpendicular to the direction of v

and is directed towards the centre of the circle.

 

> Formula

Centripetal Acceleration, a         = vw

                                                = rw2

                                                = 

 

becouse of Centripetal Acceleration, there will be centripetal force

 Fcentripetal = ma = m   = mw2r

 

Example 3

A 900 g object is whirled in a circle at the end of a string. If the string is 0.3 m long and the force in the string is 6.4 N, what is

the speed of the object?

Solution

The force acting on an object in circular motion is:          

so        

            

 

example 4

 

A rope is wound automatically by a machine at a constant angular velocity. The radius of the roller increases at a steady rate.

Sketch a graph to show:

(a) The variation of speed v of rope with radius,r.           

(b) Centripetal force F on the rope at difference distance r from the centre rotation.

solution

 

 

 

Example 5

What is the maximum speed that a car can cross a humpbacked bridge and just remain in contact with it? The radius of the

bridge is 20 m.

Solution

Push of bridge on car wheels = P

Weight of car = W

The resultant force F = W – P

This resultant force is the centripetal force.

 

 

4.3 Centripetal Force

> Definition

The force on a body moving in a circle and always directed towards the centre O of the circle.

Ø      Formula F= ma

Ø      F =   = mv w = mrw2

 

Horizontal circle

 

•Tension, T = 

v = speed of the particle

r = radius of the circle

 

•Other information

-When the speed increases gradually, the tension in the string increases until the braking tension is attained.

- Then the object continues the motion in straight line with uniform speed at the direction tangential to the circle. (First Newton’s

Law)

 

 

 

 

 

 

 

 

Vertical Circle

•Tension, T =  + mg cos q

 

Tension maximum Tension minimum

When q  = 0°,

T is maximum

 

Tension, T =  + mg

When q = p

T is minimum

 

Tension, T =   -mg

 

•To maintain the movement at vertical circle:.

T ≥ 0

≥ mg

v ≥  -The minimum velocity of the object at the highest point.

 

·        Other information

The object remains without dropping because the centripetal force on the object is greater than the weight of the object.

 ≥ W

 

Example 6

Plane flying in a uniform horizontal circle at constant speed and height,

(A) has no resultant force acting on it.

(B) experiences a resultant force acting away from the centre of the circle.  

(C) experiences a resultant force acting towards the centre of the circle.

(D) experiences an increasing force acting towards the centre of the circle.

Solution

Objects moving with uniform circular motion experience a resultant force acting towards the centre of the circle.It is called the

centripetal force.

 

Example 7

An object is whirled in uniform circular motion in a vertical circle. If the weight of the object is W the tension  in the string is T.

what is the centripetal Force. F, at the moment that the object passes through the bottom point of the circle.

Solution           

Value  of the centripetal force must remain constant at all points in the circle if the motion of the object  is uniform circular

motion.

At the bottom of the circle the weight of the object is acting down and the tension in the string I acting up. The resultant of these

two forces must be the centripetal force. Add the two forces (which are vectors, of course) and you get:

Fc= T-W

This means that the tension in the string has to be large enough to overcome the weight of the object and also provide the

centripetal force.

Note that at the top of the circle both the tension and the weight act down. Because the two forces act in the same direction, the

tension must be smaller than it is at the bottom if the resultant they produce (the centripetal force) is to be the same.

 

Example 8

Which of the following statements is correct for an object moving with uniform horizontal circular motion?  

(A) The speed of the object varies but the velocity is constant.

(B) The kinetic energy of the object varies but its momentum is constant. 1

(C) The momentum of the object varies but its speed is constant.            1

(D) The linear velocity of the object varies but its momentum is constant. I

Solution

The only answer to satisfy these conditions is C.

It an object is travelling in a circle, its direction is always changing, so no vector quantities can remain constant e.g. velocity or

momentum. Scalar quantities can remain constant, however e.g. 1 speed and kinetic energy.

Note that although the value of acceleration remains the same and it is always towards the centre of the circle, the direction of

the acceleration is always changing.

 

Example 9

An object of mass 2 kg is attached to the end of a string length 1 m and whirled in a vertical

circle at a linear speed of 4 ms-1. Find the tension in the string at the top and the bottom of the circle.

 

Solution           

 

Example 10

 A pilot of mass 60 kg flies in a vertical circle of radius 1 km at a speed of 150 m/s.

What is the push of the seat on the pilot at the top of his flight?

 

 

 

 

Conical Pendulum

 

Horizontal component Vertical component

T sin q = 

            = mrw2

            = m(l) sin q w2

            = mlw2

T cos q = mg

ml w2 cos q = mg

 

q = cos-1 

This method can be used to calculate the value of g but is not suitable because:

(i) It is difficult to measure accurately the angle q.

(ii) It is hard to fix q at a constant value when measuring the period T.

 

 

Motion of a motorcyclist round a curved track

F = 

The force F produces a moment in the clockwise direction. So, to prevent from toppling over, the person has to bend inwards

so that the weight mg and the reaction R form a couple to balance the moment produced by F.

a = distance

 

Clockwise moment = anticlockwise moment

F x h = mga

x F = mga

= 

The angle of q increases if:

(a) The speed v increases.        (b) The radius, r of the track is small.

The centripetal force is provided by the lateral frictional forces F1and F2 between the inner and outer pair of tyres.

F1 + F2 =    

There is no acceleration in vertical direction, the normal reactions at the inner and the outer pairs of tyres

R1 +R2 = mg

The clockwise moments = Total anticlockwise moments

(F1 + F2 )h + R1a = R2a

R2-R1 = 

Solving the equation simultaneously

 

R1 = 

R2 = 

For the normal reaction R1 at the inner pair of tyres is less than R2the normal reaction at the outer pair of tyres.

 

The car over turn when

R1 =   = 0

 

v is the maximum speed of the car to round the curved track without overturning. The speed can be increased when:

(a) The radius r of the circular track is bigger.

(b) The centre of gravity G of the car is lower, h is smaller.

(c) The wheel base is bigger.

 

Example 11

A bicycle and rider of total mass 120kg are rounding a bend of radius 20m. The maximum sideways frictional force available is

600N. Find the greatest linear speed of the bike before it will skid out of the bend.

 

Solution

F =ma  

Gives :  600N = 120 x  

            V = 10 ms-1

 

Example 12

Pendulum moves in a horizontal circle at an angle of 60° to the horizontal

as shown. Calculate the speed at  which it must move to maintain this angle.

 

Solution

T-tension in the rope      W – the weight of the mass

 

Resolve the tension vertically and horizontally   

T sin 60°

T cos 60°

 

 

(i ) Since the pendulum is not moving up or down the vertical forces must be equal.

T cos 60°  = mg            ……………1

 

(ii)The centripetal force is given by the force acting towards the centre of the circle. 2

            T sin 60° =            …… ………. 2

Its all rather neat divide ………2  by……..1

 

 

 

 

 

 

 

v=2.94 ms -1

Hence speed of the pendulum is 2.9 ms-1.

 

Example 13

An aircraft is flying at a constant speed of 100ms-1 in a horizontal circle of radius 25 km. A plumb line, attached to the roof of the

cabin is inclined at an angle q to the vertical while the aircraft is turning.

(a) What is the centripetal acceleration of the aircraft?  

(b) Calculate the angle q.

Chapter 6 staticPosted on 02/10/2010 by amimo50956.1 Equilibrium of Particles

> Any object is said to be in equilibrium if its acceleration is zero.

> If the velocity is also zero, the object is said to be in static equilibrium.

> Since F = ma, if a = 0, then F = 0. Thus the conditions for a point object (particle) in equilibrium is:

the vector sum of all the external forces acting on the particle must be zero, ΣF = 0

> For forces in a plane, (x-y plane) in order that a particle is in equilibrium, the sums of the resolved parts of the forces in the x

and y directions must be zero. ΣFx = 0 and ΣFy= 0. (In general, to prove that a particle is in equilibrium, we must show that the

sums of the resolved parts of the forces on the particle in any two directions are each equal zero.)

6.2 Closed Polygon

> If the forces as mention in 6.1, drawn to scale and in the appropriate directions, a close triangle or a closed polygon is

formed.

For example,(summary)

(a) Two forces in equilibrium along the x-axis ΣF = 0

F1 = F2

(b) Three forces in equilibrium in the x-y plane

The forces will formed a closed triangleΣF = 0ΣFx = 0 and ΣFy = 0

(c) Four forces in equilibrium in the x-y plane

forces will formed a closed polygon

ΣF = 0ΣFx = 0 and ΣFy = 0

Example 1

Coplanar forces F1,F2 and F3 act on a point mass A. as shown in figure above. State or show on a labeIled diagram, the

condition for the mass to be in equilibrium.

solution

Example 2

a small ball with weight W = 20 N is suspended by a light thread. When strong wind blows horizontally, exerting a instant force

F on the ball, the thread makes an angle 30° to the vertical as shown. What are the values of T and F ?

solution

Example 3

A cable car travels along a fixed support cable and is pulled along this cable by a moving draw cable. For the situation shown,

the cable car and passengers with weight 50 x 104 N is considered to be stationary with forces T, and T, acting on it. Assuming

the draw cable exerts negligible force on the cable car.

(a) sketch and show the forces acting on the cable car.

(b) find the magnitude of T1 and T2

Solution

Example 4

An object of weight W is placed on a smooth plane inclined at an angle q to the horizontal. A force F is applied in a direction

parallel to the plane so that the body is in equilibrium. Find F and R, the I normal reaction in terms of W and q. ,

Solution

Resolving forces in a direction perpendicular to the plane, R = W cos q

Resolving forces in a direction parallel to the plane, F = W sin q

Example 5

An object of weight W is placed on a smooth plane inclined at an angle 0 to the horizontal. A force F is applied horizontally so

that / the body is in equilibrium. Find F and R, the normal reaction in / terms of W and 0.

Solution

Turning effects of forces

• Torque or moment of a force about a point is the product of that force and the perpendicular distance from the line of action of

the force to the point.

(a)Torque/Moment, t = F x d

(b) Torque/Moment = F d sin q

• moment of a force = force x perpendicular distance from pivot to the line of action of force

• Unit of moment is Newton metre (N m)

· torque is a vector quantity

• A torque can turn in a clockwise direction or an anticlockwise direction.

Example 6

The crank of a bicycle pedal is 16 cm long and the downwards push of a leg is 300 N. Calculate the moment due to the force

exerted on the pedal when crack has turned through an angle of 60° below the horizontal.

Solution

Moment of the force = 300 x 0.16 cos 60° = 300 x0.16 x0.5 = 24 Nm

Couple/Gandingan

•When two equal forces are acting in opposite direction but not along the same straight line, they form a couple.

•A couple has no resultant force.

•A couple only produces turning effect (rotation).

•The moment of a couple is given by:

• The moment of a couple is the product of one of the forces and the perpendicular distance

between the two forces.

Summary

Torque/Moment* CoupleTurning Effectt = F x dF : forced : jejari/radiusMoment = F d sin qEquilibrium when moment clockwise = anticlockwise*in physic moment of force(moment) and Torque, they have the same meaning

When two equal forces are acting in opposite direction

Example 7

Calculate the moment of the Couple produced by the forces acting on the steering wheel of a car in the diagram given.

Solution

Moment of couple = 25 x 0.30 = 7.5 N m

6.3 Equilibrium of Rigid Bodies

> For a rigid body (a non-point object or extended object) in equilibrium,

(a) there must be zero resultant force. ΣF = 0

(b) there must be zero resultant torque. Στ = 0

Alternatively, we can apply the principle of moments which state that for any body in equilibrium, the sum of the clockwise

moments about any pivot must equal the sum of , the anticlockwise moments about that pivot.

> Forces that act on rigid body may be concurrent forces or non-concurrent forces.

1.concurrent forces(Daya bersetemu)

- Concurrent forces are forces whose lines of action pass through a single common point.

- Concurrent forces will only cause translational motion.

- To determine whether concurrent forces acting on a rigid body are in equilibrium, all that is required is to check whether the

resultant forces is zero.

2. Under non concurrent forces

- Non-concurrent forces are forces whose lines of action do not pass through a single common point.

- To determine whether non-concurrent forces acting on a rigid body has translational and rotational equilibrium, it is necessary

to check whether condition ΣF = 0 as well as condition Στ = 0.

Example 8

The seesaw in the diagram is balanced. Use the principle of moments to calculate the weight, W.

Solution

W(15) = 300(1.0) + 450(1.5) = 650N

Example 9

uniform rod XY of weight 20.0 N is freely hinged to a wall at X. It is held horizontal by force F acting from Y at an angle of 60° to

the vertical as shown in the diagram. What is the value of F?

solution

The diagram shows the forces acting on your forearm when you hold a weight of 60 N with your arm horizontally. Your elbow

joint acts as a fulcrum. The weight of the arm is 20 N and its

is considered to act at a distance 14.0 cm from the fulcrum. Use the principle of moments to calculate the force T exerted by

your biseps. What is the reaction force R at the fulcrum?

solution

6.4 Frictional Forces

> Friction acts whenever two surfaces move or try to move relative to one another.

> There are two kinds of frictional forces:

1. Static friction

- It is a force at the contact surfaces which prevents the surfaces from sliding over each other.

-The frictional force always acts in the opposite direction to the pulling force P. It is always self-adjusting, constantly equalising

itself to P, maintaining static equilibriumas long as the limiting friction is not exceeded.

-If the pulling force is greater than the limiting friction, the block moves and anoth frictional force known as kinetic

friction comes into effect.

Limiting friction:

(a) Depands on the nature of the surfaces.

(b) Is independent of the area of contact.

(c) Is proportional to the normal reaction,R.

Thus limiting friction has a value F = mR

2.Kinetic friction

-It is a force between two moving surfaces which opposes the sliding motion.

- When the pulling force P exceeds the limiting friction, the resultant force accelerates the block.

- Once in motion, the frictional force decreases. The frictional force involved now is the kinetic friction.

- To maintain constant velocity, the pulling force P has to be decreased to the same magnitude as the frictional force (kinetic

friction).

- The kinetic friction is independent of the relative velocity of the surfaces.

Chapter 5 Rotation Of Rigid   Body

Posted on 15/04/2010 by amimo5095

Introduction

Particles make a rigid body. The relative positions of a particles remain static when the body moves or acted upon by a force. In

this chapter, no new law is introduced in rotational motion. Newton’s laws of motion are adapted to this topic.

5.1 Rotational Kinematics

> It involves the movement of an object during rotating.

> The formulae are listed:

Linear motion Rotational motion

v = u + at

v2 = u2 + 2as

 

w = w0 + at

a  = 

a  = w0 t +   at2

w2 =w02 + 2aq

> When doing calculation for rotational kinematics, students are not allowed to write the linear motion formulae.

 

Example 1

A disc is rotated from rest at vertical axis at the centre O with the angular acceleration of 4.0 rads-1 . The total angular

displacement is 18 rad. Calculate:

(a) The angular velocity.

(b) The duration for the movement.

Solution:

 

Example 2

A ball with diameter 4 cm rolls down from rest. After 3.0 s the angular velocity of the ball is 1.5 rads-1 1. Calculate:

(a) The angular displacement.

(b) The linear displacement along the plane.

(c) The linear velocity.

Solution

 

5.2 Rotational Dynamics

> Definition

Moment of inertia is define as the resistance of a body to change to its rotational motion.

> Formulae

I = mr2

I – the moment of inertia

m – mass of a particle

r – perpendicular distance from the axis of rotation.

The moment of inertia of the rigid body is

I = ( m1r12 + m2r2

2 + m3r32………..miri

2)

= 

The I, the moment of inertia depends on:

-The mass.

-The distribution of the mass from the axis of rotation (shape of the body).

-The position of the axis of rotation.

> the centre of mass of a rigid body, 

The point where the mass of the body is assumed to be concentrated

If a force is applied through [lie Centre of mass

- the motion is a linear motion with no rotational motion.

- the body accelerates in the direction of the force.

 

Example 3

Particles with mass 0.30 kg and 0.60 kg are at 30.0 cm and 40.0 cm mark respectively on a uniform metre rule of mass 0.2 kg.

Find the position of the centre of mass of the system.

Solution

Moment of inertia of different shape of body

Shape Illustration Formulae

Uniform rod (a) Axis through the centre of the mass G and perpendicular to rod

(b)Axis through one end and perpendicular to rod

Ring Axis through centre of mass G I = MR 2

Disc Axis through the centre of mass G

Hollow sphere Axis through the centre of mass G

Solid sphere Axis through the centre of the mass G

Solid cylinder Axis of rotation about centre of cylinder

5.3 Theorems to Find Moment of Inertia, I of A Rigid Body

> Parallel axis theorem

If the moment of inertia of a rigid body IG about an axis through its centre of mass G is known, the moment of inertia of the body

about an axis which is parallel to the first axis can be obtained by using the parallel axis theorem.

I = IG +Mh2

IG refers to the moment of inertia about the axis through the centre of mass G, and h is the distance between the two axes.

> Perpendicular axis theorem

Suppose IX and IY are the moment of inertia of a body about axis OX and OY respectively, the OX is perpendicular to OY. The

the moment of inertia I about an axis OZ which is perpendicular to both OX and OY is given by

I = IX + IY =   IX = IY = 

Example 4

A chlorine molecule consists of two atoms each of mass 3.15 x 10-32 kg. The atoms are separated by a distance of 1.78 x 10-

11 m. What is the moment of inertia of a chlorine molecule about an axis through its centre of mass and perpendicular to the line

joining the two atoms`?

Solution

Moment of inertia, I = S mr2

                                   = mr2 + mr2

                                   = 2 mr2

5.4 Angular Momentum

> There are three types of momentum

-Linear momentum

p=mv = mrw

-Angular momentum

p=mvr =mrwr = mr2w

Angular momentum of a rigid body

L = (Smr2)w

    =Iw

Example 5

Calculate the angular momentum of a bowling ball of mass 5.5 kg which moves in a circle of radius 0.15 m with angular velocity

3.0 rads-1.

Solution

Angular momentum, L = (Smr2)w = Iw

 

5.5 application of Angular Momentum

Conservation of angular momentum, I1w 1 = I2w2,

> Swimmer jumps off from the diving board

-The swimmer starts to rotate because a torque is produced by his weight.

- During rotation, he pulls himself closer to his axis of rotation. This mechanism reduces his moment of inertia, I. Then,

his angular velocity, w increases and the swimmer is able to make more rotations.

-Before hitting water, the swimmer needs to straighten his body, the moment of inertia, I, increases, and the angular

velocity w decreases.

- When the swimmer falls, the path taken by his centre of mass is a parabolic path (the path of projectile).

> Determining if an egg is hard boiled or half boiled

-When a raw egg is spun, the yolk which is denser moves away from the axis of rotation. Since moment of inertia I = mr2, when

the distance r from the axis of rotation increases, I increases.

- When I increases, w decreases. – It is hard to spin a raw egg.

- When a hard-boiled egg is spun, its yolk does not move away from the axis of rotation.

-The moment of inertia, I remains constant and the egg spins easily.

> A ballet dancer

- When she starts to spin and stretches her arm, the moment of inertia I is large and the angular velocity w is small.

- When she lifts her arms above her head, the moment of inertia decreases and the angular of velocity m increases.

5.6 Rotational Kinetic Energy

> Definition – Rotational kinetic energy is the total energy of all the particle in the body when a rigid body rotates about a fixed

axis.

> Formula apply – 

> Application kinetic energy of a rolling body

Velocity v of the centre of mass where

R = radius of rolling body

 w= angular velocity

v = Rw

Total kinetic energy

= rotational kinetic energy + translational kinetic energy

Kinetic energy of a hollow sphere

Acceleration of a hollow sphere

Example 5

the moment of inertia of a ball of mass 30 kg and radius 0.50 m about its axis is 3.2 kg m2. What is its total kinetic energy when

it rolls along a horizontal surface with a speed 3.0 ms-1.

Solution

Example 6

A uniform rod, length 2.40 m is hinged freely at one terminal. The rod wised so that it is 1 horizontal and then released from rest

so that it rotates about an axis in a vertical plan. Calculate by using l(length) and g the angular velocity when the rod is vertical.

[moment inertia of the rod about its axis  ]

Solution

Example 7

A cylinder of radius 10 cm and mass 600 g rolls on a floor without slipping with a speed v. The cylinder then rolls up an inclined

plane.

(a) Calculate the speed of v if the maximum height h reached by the cylinder is 3 m.

(b) After 5 second the cylinder rolls down. Calculate the angle of inclination of the inclined plane, q and the deceleration.

(Given that the moment of inertia of a cylinder about its axis is  )

Solution

Torque

•Definition – Torque or the moment of the force is the turning effect of force.

• Formula -Torque,G

Force x Perpendicular distance of the force from the axis of rotation = F x r

Torque is measured in the unit of Nm

G = Fr = la = I  where a = 

Example 8

CD player rotates about a vertical axis with uniform angular velocity of 30 revolutions per minute. A CD is placed onto the

recorder. The friction between the CD and the recorder causes the CD to accelerate uniformly until angular velocity of 30 rev

per minute in 5.0 s. The moment (d f inertia of the CD about its axis of rotation is 2.0 x 10-2 kg m2. Calculate:

(a) The torque on the CD.

(b) Rotational kinetic energy gained by the CD.

(c) The total work done by the torque in 5.0 s.

From the findings of b and c, what inference can be concluded?

Solution

Example 9

The diagram shows a disc that is attached to an object with the mass of 1.0 kg. When the mass is released from rest to fall

vertically, the disc starts to rotate. The radius of the disc is 12.0 cm and its moment of inertia about the axis of rotation is 0.030

kg m2, calculate the acceleration of the mass.

Solution

Translational motion Rotational motion

Displacement,s

Angular displacement

a  = 

Velocity, v

Angular velocity w = 

Acceleration, a

Angular acceleration a = 

Mass, m Moment of inertia, I = Smiri2

Linear momentum, p = mv Angular momentum L = lw

Impulses; Ft = mv – mu Angular impulse Gt = Iw – Iw0

Work W = Fs Work W = Gq

Translational kinetic energy =  mv2 Rotational kinetic energy =  ,

Power P = Fv Power P = Gw

Chapter 2 KinematicPosted on 04/07/2012 by amimo5095

Introduction•Kinematics – A study of a motion (gerakan) of an object without considering the effect that produces motion. Kinematics analyses the position of an object relation to time. 

2.1 Rectilinear Motion (Gerakan Linear) 

 

> Definition

Distance (jarak) and displacement (sesaran) are both ways of measuring how far an object has moved.> Distance,l is a measure of how far an object has moved.> Displacement,s is defined as the distance moved in a particulardirection. (It is the change in position)            Ex:            To the right = +  (+5km)            To the left = – (-10 km)> Speed is defined as the distance moved per second (or the rate of change of distance)> Velocity is defined as the rate of change of displacement.Average velocity = 

Instantaneous velocity =   =

The instantaneous velocity of an object is its velocity at a particular instant or time.> Acceleration is defined as the rate of change of velocity.Average acceleration =   

Instantaneous acceleration = 

 

Motion with Constant Acceleration 

> From the definition of acceleration, we learn that

v = u + at……………………… 1  

> Displacement = average velocity x time…………………. 2

Replace (1) to ( 2)

From (1), v-u = at and (2)  v + u = 

v2 = u2 + 2as 

Summary of equation:  The idea of  ‘s u v a t’ 

Formulae Uses when absence of

Eq. No 

s, displacement 1 

 a, acceleration 2

 

v, final velocity 3 

v2 = u2 + 2as t, time 4 

 

Example 1Starting at time t = 0 seconds, an object accelerates from 12 ms-1 to 27 ms-1in 4 seconds. Find  the value of time, t (to the nearest second), when its total displacement is 110 m. 

Example 2A bus travels at straight road with the speed of 24 m/s decreases uniformly to 8 m/s by travelling 40 m. Calculate:(a) The decelerations of the bus.(b) The time taken for the deceleration.(c) Total distance travels before the bus stops. 

 

Graphical representation of motion 

 

 

 

Example 3The velocity-time (v-t) graph for a car is as shown in the graph.Graf halaju-masa  (v-t) untuk pergerakan sebuah kereta ditunjukkan di bawah. 

Which of the following is the displacement-time (s-t) graph for the car?Graf sesaran-masa (s-t) yang manakah yang betul? 

 

Example 4The following is the velocity-time graph of a car.

Based on the graph,(a)           Calculate the acceleration of the car between                 (i) JK                      (ii)  KL                   

      (iii) LM                    (iv) MP                (vi)    PN                 (vii) NO(b)           State the type of motion of the car between                (i) JK                       (ii) KL   

     (iii) LM                     (iv)  MP                (vi) PN                    (vii)   NO(c) Calculate the total displacement travelled of the car  during

(i)   the first  10 s of motion(ii)  the car moves with uniform velocity(iii)the last 10 s of motion

(d) Calculate(i) the total displacement for the whole journey.(ii)  the total distance for the whole journey.

(e) Calculate                (i)            the average velocity                (ii)           the average speed 

Acceleration due to gravity (ignoring friction(geseran)) 

 

•Objects in gravitation field experience a downwards force – their weight. If unbalanced, this produces acceleration downwards.• There is an acceleration free fall (jatuh bebas) or acceleration due to gravity when the object moves at the same acceleration.•Acceleration free fall is represented by the symbol “g” as assume constant 9.81m s-2.• The value of g depends on the displacement or location of the object. 

Study the following figure: 

 

 

Case 1 Case 2 Case 3

An object travelsupwards

An object travelsdownwards

An object travels below the level

Displacement = +sVelocity = +v

Acceleration = -g

Displacement = +s

 Velocity =-vAcceleration = -g

Displacement = -sVelocity = -v

Acceleration = +g 

 

 

 

 

 

• Useful information for calculation:

Information  AnalysisThe velocity decreases to zero when the object reaches the highest point.

 1. Calculation of maximum displacement:

   

Time taken for the object to reach the highest point is thesame as the time it takes to drop back to its initial point.

  2. Calculate total time for the objects to travel.

Using the formula:

s is H and a is g

Total time for traveling is 2 x t.2. Calculate velocity before reaching the ground.v2 = u2 + 2asv2 = u2 + 2gHu= 0v2 = 2gH

 

Example 5

Sebiji batu di lontarkan mencancang ke atas dengan laju 30 ms-1. Kirakan selang masa di antara dua ketika apabila 25 m dari Bumi. 

Example 6A ball falls under gravity from the top of a tall building. Which of the following is thegraph of the height h of the ball from the ground against time t?Sebiji bola jatuh dari atas sebuah bangunan dengan pengaruh graviti. Graf yang manakah menunjukkan hubungan di antara ketinggian h dan masa t? 

 

Example 7Satu jasad diluncurkan tegak ke atas dengan laju 20 ms-2 dari satu titik  P, 25 m di atas Bumi. Lakarkan graf.

(a)     sesaran-masa(b)     halaju-masa(c)     laju-masa

 

 

2.2 Projectile 

  

> The motion equations can be used with object projected or thrown through the air at an angle. Examples of projectiles : motion of missiles, throwing basketball and long jump> Consider the ball thrown at a velocity  at an angle  to the ground.Ignoring air resistance, the only force acting on the ball during its flight is the gravity. Analysis shown

-A downward acceleration that only affects the vertical component of the velocity.

-The horizontal velocity remains constant.-The ball follows a parabolic path through the air.-The time of flight depends on the vertical velocity.

The horizontal distance travelled depends on the horizontal velocity and the time of – flight.> The horizontal and vertical motions of a projectile are independent and can be treated separately in calculations.

The table below shows the horizontal and vertical components of projectile motion.

 

  Horizontal component (x axis) Vertical component (y axis)

Initial velocity u cos  u sin 

Acceleration 0 -g

Time takenUsing

formula:v = u + at

To reach RTotal time to R is twice the

timeto reach H.

So,tR=2tH

To reach H, v = 00 = usin  – gtH

tH=

Height H /Range Rtravelled

R = (u cos ) 2 tH

= 2 (u cos  ) 

R =

Using formulaV2 = u2 + 2 as

0 = (u sin )2 – 2gH

H =

Displacementat any

instant tx = (u cos )t

 

y = (u sin )t-  gt2

Extrainformation

Maximum R occurs when  = 45°

The range of projection is thesame for  and (90 – ).

 

Example, 0 = 15° and 75° 

 

Example 8Satu jasad diluncurkan ke atas dengan laju 30 ms-1 pada sudut dongakan 60. Kirakan :

(a)     masa yang di ambil oleh jasad untukmencapai tinggi maksimum(b)     tinggi maksimum yang dicapai(c)     Julat(Range)(d)     Masa yang diambil untuk mencapai ketinggian 30m(e)    Halaju seketika jasad di ketinggian 30 m

3.0 Work,Energy And   Power

Posted on 08/02/2010 by amimo5095

Introduction

we apply work in our daily life. In this chapter, we will study work, different types of energy ,such as kinetic energy, potential

energy, conservation of energy and power.

 

3.1 Work

> Definition

Work is something done whenever it transfer energy from one form to another. It also can be defined as force acting on a body

moves the body.

> Formulae of work in different situation

(a)        W = Fs

(b)        W = F cos q s

 

 

(c) Work can be calculated by getting the area under the force – displacement graph.

 

 

(d)        Work due to spring extension (pemanjangan)

W =   Fe; F = force; e = spring extension

 

extension , e = y-x

 

(e) The work of gas inside the container

 

DW       = FDx  

so pressure,    then  F = PA

            = P ADx

            = P V

 

3.2 Energy

 

Kinetic energy, K Potential energy, U

Kinetic energy is the energy of an object due to

motion.

K.E=  mv2

 

Kinetic energy stored in an object that is

moving equals to the amount of work done

accelerating it to that speed.

Work done=   mv2

Potential energy is the energy of an object due

to its relative position or physical state.

P.E = mgh

 

Potential energy stored in objects equals the

work done to lift the object against a

gravitational field.

 

Work done = m x g x Dh

 

- According to principle of conservation energy, energy cannot be created or destroyed, but it can be converted to other form of

energy.

            K + U = constant

            DK = -DU

 

For example when a ball falls from a building and the ball rebounds, the energy conservation is shown in the diagram below.

Forces can be calculated by differentiating the potential energy.

Force = 

Example 1

A car  is pushed 31 m along a horizontal surface. It the total frictional force is a constant 300N and the work done by the person

pushing the car is 14 kJ, how much energy is transform to the kinetic and heat energies ?

solution

the work done against friction is 300 N x 31 m = 9 300 J

this become heat energy.

If total  work done is 14 000 J, then 14 000 J – 9300 J = 4700 J must have become kinetic energy.

 

 

 

 

Example 2

Calculate the retarded force on the object that is shown in the figure below.

Solution

The force is obtained by calculating the slope/gradient (kecerunan) of the graph

F=  =250N

 

Example 3

A force of 10 kN is used to push a body weighing 30 kN up a slope. The distance moved up the

is 40 m. The total height gained is 10 m. If you assume that there is no friction between the slope and the body, how much of

the work done ends up as kinetic energy of the body?

 

Solution

The, total work done by the force pushing the body up the slope is:

Work = force x distance

=10 x 103 x 40m = 400 kJ

 

The  gain in potential energy due to it going up the slope is:

Ep = weight x height

= 30 x 103 x 10m = 300 kJ

Total work done that become kinetic energy = 400 kJ – 300 kJ = 100 kJ

 

Example 4

An object is thrown vertically upwards, what graph will describe its variation of Ep and Ek with height’?

Example 5

If an object of mass 40 kg is falling at its terminal velocity of 100 ms t through air, what is the  energy lost to heat each second

as it falls? (g = 9.81 ms-2)

 

Solution

The object is losing EP as it falls. When it is at its terminal velocity, the lost EP is all becoming heat. None of it is becoming Ek as

the object is not gaining Ek because it is not speeding up.

Each second the object loses 100 m of height. Hence, EP lost each second is:

EP = mgh

 = 40x 9.81 x 100 

= 39 240 J

= 39 kJ

 

Example 6

If an object of mass m is thrown vertically upwards with velocity v and reaches height h, what  height would a second object

with mass 2 m reach if it is thrown with velocity 4v ?

Solution

The Ek becomes EP so: mgh =   mv2 

cancel off the m and reorganizes to        h = 

That’s for the first object.            z

 

For the second object:  h’ =   =   = 16 h

 

In this case, the change in the mass of the object does not matter because the object will have twice the mass but also twice

the Ekinitially as it is going at twice the speed.

Example 7

The graph  shows a movement of object 2 kg after it is applied with a force F which changes with displacements. Calculate:

(a) work done by the force.

(b) The height the object could travel.

 

Solution

Work done by the force = Area under (F-x) graph

= 6 x 20+  (2) x 20 +  x2(-10) = 130 J

(b) mgh = work done

2 x 9.81 x h = 130 J

h = 6.625 m

 

3.3 Power

> Definition

Power is the rate of doing work and time taken

Power =   

                        

 

Unit: watt (W), joule per second (Js-1)

 

Example 8

A wheel is connected to a motor and turned clockwise at a constant speed of 10 revolutions per second. Two weights are  hung

from a string passing over the wheel as shown. With the  wheel turning, the system is in equilibrium. The length of string in

contact with the wheel is 0.5 m. What power must the motor produce to turn the wheel?

 

Solution

Use the equation Power = Force x Velocity.

Here, the resultant force due to the weights would be 20 N down on the left hand side, but the friction between the wheel and

the I string must provide enough force to balance the system. i.e. I friction must be 20 N.

 

If 0.5 m of string is in contact with the wheel and the wheel rotates at 10 revolutions per second, then the wheel rubs on

the  string for a total distance of 10 s-1 x 0.5 m = 5 ms-1 .

 

So, Power = Force x velocity  = 20N x 5ms-1 =100 W

 

Example 9

What is the maximum power that you could achieve from a wind turbine that has a blade radius of 14 m in a wind of velocity 2

ms-1? (Take the density of air as 1 kgm-3)

 

Solution

The energy for the turbine comes from the kinetic energy of the air. The maximum power achievable is if the blade removes

100% of the kinetic energy of the air each second.

 Energy in the air =   mv2

In one second, the mass of air passing the blade (which is the air that can give its energy to the  blade) is found by calculating

the volume of air that passes the blade and multiplying it by the density of the air.

Mass of air = volume of air x density of air     

= (pr2 x velocity) x density

where r is the radius of the blades.

Substitute in the numbers and get:

 

Mass of air = 1232 kg per second.

Hence, Ek available per second =  (1232 kgs-1) x 22

= 2464 W

And finally the power is the energy per second which is 2464 W.

 

Example 10

Calculate the work done within 20 seconds.

 

Solution

To find the area under the graph you need to multiply Power x Time (that’s the quantity on one axis times the quantity on the

other axis).

 

you need to recognize that

Work = Power x Time

 

Hence the units of the quantity that you calculate must be the units of work, i.e. joules.

Work=  (5+10)x20

= 150 J

 

 

3.4  Efficiency (kecekapan)

Definition

The efficiency of a machine is the ratio of the useful work done to the energy input.

 

> Formulae

Work done        x 100%

Energy input

 

Work done per second x 100%

Energy input per second

 

Output power x 100%

Input power

 

Useful energy out x 100%

Total energy in

 

When work done = energy input     \ 100% efficiency.

However, most machines do not achieve 100% efficiency because heat is produced and energy is needed to overcome friction.

 

Example 11

A motor of efficiency 80% is used to haul a lift up its shaft in 10 seconds. It the total length of the shaft is 50 m and the mass of

the lift is 1000 kg, what will the electrical power supplied  the motor be?

 

Solution

 

The weight of the lift is ,W = mg = 1000 kg x 10 N kg-1 = 10 kN 

 

The power needed to raise the lift is

= 50 kW

But that is the output power of the motor. We need to put in more than that as it is only 80% efficient. So the electrical input

power is:

 

Example 12

A machine is used to transport goods at Port Klang. The machine is equipped with 15 A and 30 V. The machine uses 10 s to

move goods of 10 kg from ground to the stage which is 40 m above the ground.

Calculate:

(a) Work done by machine.

(b) Potential energy at goods.

(c) Efficiency of machine.

If the goods are transported to cargo at ground with a slop, calculate the velocity before reaching the ground. [g = 10 ms-2]

Solution

(a) P     = VI

            = 30V x 15A     

            = 450 W

     W = P x t

     W = 450W x 10s

            = 4500 J = 4.5 kJ

 

(b) Potential energy       = mgh

                                    =10kg x10ms-2 x 40m = 4000 J

 

(c) effiency =  = = 89%

 

Example 13

The graph shows the work done by a machine and the input when lifting a cargo in Port Klang

Calculate:

(a) the efficiency of the machine.

(b) Discuss the efficiency of the machine.

 

solution

Chapter 16 Thermal   Conduction

Posted on 28/03/2011 by amimo5095

16.1 Mechanism of Thermal Conduction through Solids

> For non-metals heat is conducted by vibration of atoms or molecules. When one end of a solid is heated, the molecules at the

hot end receive thermal energy and vibrate more vigorously. These, in turn, cause the neighboring molecules to vibrate more.

Eventually the molecules at the other end of the solid vibrate more vigorously and the increased vibration represents transfer of

energy and an increase in temperature.

> For metals, besides the above mechanism, thermal conduction also occurs through the collisions of free electrons with the

lattice atoms. Metals contain many free electrons. When one end of metal is heated, the electrons at the hot end receive

thermal energy and diffuse to the other parts of the metal. Energy is transferred through the collisions of the free electrons with

the lattice atoms.

Note: A good electrical conductor is also a good thermal conductor because a good electrical conductor has a large number of

free electrons in it.

16.2 Thermal Conductivity

> Thermal conductivity (k) of a material is the rate of heat flow per unit cross-sectional area per unit temperature gradient, in

the steady state.

(Steady state is a condition when all temperatures have become constant)

 = -kA   (negative sign shows that the temperature q decreases as distance x increases)

The unit for k is given by W m-1 K-1.

16.3Temperature Distribution Along Uniform Rod

Rod insulated

• Figure below shows a rod of length l and cross-sectional area A is perfectly insulated. Heat flows along the rod and none

escapes from the sides.

When steady state is attained,

(i)the rate of heat flow dQ is the same at all points along the rod.

(ii)the temperature decreases uniformly with distance x along the rod.

(iii) the temperature gradient   is uniform along the rod.

Temperature gradient   = – 

Rate of heat flow   = -kA 

Example 1

Window glass has a thermal conductivity of 0.80 Wm-1 K-1. Calculate the rate at which heat is conducted through a window of

area 2.0 m2 and thickness 3.0 mm if the temperature in at air-conditioned room is 20°C and the outdoors’ temperature is 32°C.

Solution

Example 2

An Eskimo igloo is made of blocks of compressed snow, 0.50 m thick. What thickness of expanded polystyrene would be

required to give the same thermal insulation (i.e. the same rate I of loss of heat for the same temperature difference across the

walls)?

Thermal conductivity of compressed snow = 0.20 Wm-1 K-1

Thermal conductivity of expanded polystyrene = 0.012 Wm-1 K-1

Solution

Example 3

A saucepan has a flat steel base with an area of 0.032 m2 and thickness 0.0040 m. It is in direct thermal contact with a hotplate.

The two surfaces in contact are at the same temperature. The rate of heat flow through the base of the saucepan is 1500 W

when water is boiling steadily at 1100°C in the saucepan. Calculate the surface temperature of the hotplate.

[Thermal conductivity of steel is 46 Wm-1 K-1]

Solution

Rod not insulated

•Figure below shows a rod of length l and cross-sectional area A is not insulated. As the heat flows along the rod, some heat

escapes from the sides.

When steady state is attained,

(i) the rate of heat flow dQ decreases with distance x along the rod.

(ii)the temperature decreases with distance x along the rod.

(iii) the temperature gradient d0 decreases with distance x along the rod.

16.5 Heat Conduction through Composite Rods of Different Materials

Introduction

Example 1

A container with total surface area 40 m2 is used for keeping objects cool. The walls of the container consist of wood and a poor

conducting material. The thickness of wood and the poor conducting material is 5.0 mm and 60.0 mm respectively. If the

outside and inside surfaces are at temperatures 35°C and 15°C respectively, what is the rate of heat flow through the walls of

the container with and without the poor conducting material.

Thermal conductivity of wood = 0.24 Wm-1 K-1

Thermal conductivity of poor conducting material = 0.04 Wm-1 K-1

Solution

16.4 thermal resistance

> The above example can be solved by using the concept of thermal resistance. From the equation of rate of heat flow   =

kA 

= 

 is known as the thermal resistance

> For composite rods, the total thermal resistance   can be determined and

= 

Example 2

Three different materials X, Y and Z with length l, 3l and 2l are joined in series to formed a rod.The thermal conductivity of the

materials are 2k, k and 4k respectively. The free ends of the rod is kept at 100°C and 0°C as shown in the diagram. Find the

equilibrium temperature at the X-Y and Y-Z interface.

Solution

16.6 Determination of thermal conductivity

Determination of thermal conductivity of good conductors – Searle’s method

•The sample is in the form of a thick, long insulated rod.

•Rod is insulated so that very little heat is lost from the sides of the rod. This will ensure that the rate of heat flow and the

temperature gradient are constant along the rod.

• Rod is of large cross sectional area so that the rate of heat lost from the sides is negligible compared to the rate of heat flow

along the rod.

• Rod is long so that the temperature difference across the long length of the rod is measurable. A bigger temperature

differences reduces the percentage error in the measurement of the temperature difference.

• One end of the rod is heated by a steam chest X to maintain the temperature at about 100°C. • At the other end of the rod, the

heat is carried away by a slow, steady stream of cold water flowing through the pipe Y maintained by a constant pressure

apparatus.

• When the steady state (when all temperatures are steady) is attained, the temperature q1, q2, q3 and q4 are recorded.

• In the steady state   = kA  where A = cross sectional area of the rod

=   (d is the diameter of the rod)

•The rate of heat flow is equal to 

where m is the mass of cold water passing through the pipe in t sec c is the specific heat capacity of water

q3 and q4 are the steady temperatures of the water entering and leaving the constant pressure apparatus

•   = kA   = 

k     = 

k =

Determination of thermal conductivity of poor conductors – Lee’s method

•The sample is in the form of a thin disc.

•Sample is thin so that steady state can be achieved in a short time.

•Sample is of large cross sectional area so that the rate of heat flow is adequate.

•Sample is not lagged because the heat loss from the sides of the disc is negligible.

•The sample is sandwiched between two thick conducting discs P and Q (e.g. brass). Q is part of a steam chest R.

• The apparatus is suspended in mid air and Q is heated by passing steam through the steam chest.

• When the steady state is attained, the temperature q1 and q2 are recorded.

• In the steady state   = kA   where A = cross sectional area of the disc

=   (D is the diameter of the disc)

l = thickness of the disc and the thickness is measured by using micrometer screw gauge. •

· The rate of heat flow   is also the rate of loss of heat from the exposed surface of P.

• To measure 

• The steam chest is removed, leaving only the sample resting on the disc P, suspended as before.

• P is gently heated by a Bunsen until its temperature is several degree above its previous temperature q1.

• The disc P is then allowed to cool under the previous external conditions and a cooling curve q against t is plotted.

• The rate of fall of temperature   at temperature q1 is obtained from the graph. The rate of heat flow is equal to mc

 where m is the mass of disc P and c is the specific heat capacity of disc P

•   = kA   = mc

k     = mc

k =

Analogy between thermal and electrical conduction

Thermal conduction Electrical conduction

 = -kA 

or  =

=

= 

Heat Q Electric charge Q

Temperature q Electric potential V

Temperature gradient Potential gradient 

Thermal conductivity k

Electrical conductivity 

Chapter 15 Thermodynamics of gasesPosted on 27/03/2011 by amimo5095Introduction

Thermodynamic is the study of the laws that govern the conversion of energy from one form to another, the direction in which

heat will flow and the availability of internal energy to do work.

15.1 Heat Capacity

> Heat capacity, C of an object is the heat required to raise the temperature of the object by 1 K (or 1°C).

Unit for C is J K-1 or J °C-1

The heat, Q required to raise the temperature of a body by DT is given by: Q =CDT

> Specific heat capacity, c of a material is the heat required to raise the temperature of 1 kilogram of the material by 1 K (or

1°C).

 where m = mass of body

Unit for c is J kg-1 K-1 or J kg-1 °C-1

Note that 

The heat, Q required to raise the temperature of m kilogram of a material by DT is given by:Q = mcDT

>Molar heat capacity, Cm of a material is the heat required to raise the temperature of 1 mole of the material by 1 K (or 1°C).

 where n = number of mole

Unit for Cm is J mol-1 K-1 or J mol-1 °C-1

Note that 

The heat, Q required to raise the temperature of n mole of a material by DT is given by, Q = nCm DT

For gases

• Gas has two heat capacities, one at constant volume and another at constant pressure.

• Molar heat capacity at constant volume, Cv,m of a gas is the heat required to raise the temperature of 1 mole of the gas by 1

K (or 1 °C) at constant volume.

• The heat, Q required to raise the temperature of n mole of a gas by DT at constant volume is given by

Q = n CV,m DT.

• Molar heat capacity at constant pressure, Cp,m of a gas is the heat required to raise the temperature of 1 mole of the gas by

1 K (or 1 °C) at constant pressure.

• The heat, Q required to raise the temperature of n mole of a gas by AT at constant pressure is given by Q = n CP.m DT.

• Cv,m and Cp,m depend on the degree of freedom.

•CV,m =   and Cp,m =   where f is the degree of freedom. (To be derived later)

Example 1

A kettle made of aluminium has a mass of 2.0 kg and the heat capacity of the kettle 1.8 x103JK-1.

(a) What is the heat required to increase the temperature of the kettle by 40°C?

(b) What is the specific heat capacity of aluminium?

Solution

Example 2

How much heat is released when 2.5 kg of water is cooled from 90°C to 30°C?

[specific heat capacity of water is 4.2 x 103 J kg-1 K-1]

Solution

Example 3

The specific heat capacity of copper is 385 J kg-1 K-1. The molar mass of copper is 63.5 g. What is the molar heat capacity of

copper?

Solution

15.2 Work

> When a gas expands, work is done by the gas.

When a gas is compressed, work is done on the gas.

> When the gas at constant pressure expends by pushing a light frictionless piston as shown in Figure above the work done by

the gas is:

dW = F dx = pA dx = p dV

> Hence, the work done by a gas when its volume increase from V1 to V2 is given by

 = = area under the p-V graph

> The work done by gas for various types of changes will be discussed in the subsequent section. The changes involved are:

- Isobaric change (constant pressure)

- Isometric change (constant volume)

- Isothermal change (constant temperature)

- Adiabatic change (no heat flow in or flow out from the system)

Example 4

A gas undergoes the cycle of pressure and volume changes W ® X ® Y® Z ® W as shown in the diagram. What is the net work

done by the gas?

Solution

15.3 First Law of Thermodynamics

> The first law of thermodynamics is a law of conservation of energy.

> The first law of thermodynamics states that the heat supplied to a system equals to the sum of increase in internal energy of

the system and the work done by the system. Q= DU+W

> The sign convention for Q, DU and W used in the formula above is shown below:

Q is positive when heat is supplied to the system.

Q is negative when heat is released from the system.

DU is positive when the internal energy (temperature) increases.

DU is negative when the internal energy (temperature) decreases.

W is positive when work is done by the system.

W is negative when work is done on the system.

15.4 Internal Energy

> The internal energy, U of a system is the total potential and kinetic energy of the molecules.

> For an ideal gas, because the force between the molecules is negligible, the potential energy equal zero. The internal energy

is the total kinetic energy of the molecules only.

> From chapter 14, for gas with f degree of freedom,

kinetic energy (internal energy) for one mole of gas =   kinetic energy (internal energy) for n mole of gas = (n) 

> The internal energy of the ideal gas depends on:

(a)Amount of gas (number of moles)

(b)Absolute temperature

(c) Degree of freedom

> When temperature of gas increases, internal energy increases and DU is positive. When temperature of gas decreases,

internal energy decreases and DU is negative.

Example 5

A fixed mass of gas undergoes changes in pressure and volume as shown in Figure above. When the gas is taken from state P

to state R by the stages PQ and QR, 16 J of heat are absorbed by it and 6 J of work are done by it. When the same resultant

change is achieved by stages PS and SR, 2 J of work is done by the gas.

(a) What is the change in internal energy from state P to state R?

(b) What is the heat absorbed by the gas when the gas undergoes the changes by stages PS and I SR?

Solution

15.5 Isometric Change

> An isometric process (short term for isovolumetric) is a constant-volume process.

> Since volume is constant,

DV = 0 and W=p DV=0

> Q =DU+W = DU+0 thus Q = DU

> To raise the temperature of n mole of gas by DT at constant volume, the heat required Q = nCv,m DT = DU

15.6 Isobaric Change

> An isobaric process is a constant-pressure process.

> Since there is a change in volume and pressure is constant at p,

W=pDV=p(V2-V1)

> Q = DU+W = DU+p (V2-V1)

>To raise the temperature of n mole of gas by DT at constant pressure, the heat required

Q = nCP,M DT= DU + p(V2-V1)

15.7 relation between Cv,m and Cp,m

> n Cp,m DT=DU+p(V2-VI)

n Cv,m DT = DU

n Cp,m DT – n Cv,m DT = p(V2 – V1 )

> From the ideal gas equation pV = nRT

pV2 = nRT2

pV1 = nRT1

pV2-pV1 = nRT2 – nRT1

p(V2 – V1) = nR(T2 –T1) = nR DT

n Cp,m DT – n Cv,m DT = p(V2 – V1 ) and CP,m – Cvm = R

= nR DT

Example 6

The molar heat capacity at constant volume of an ideal gas is 20.8 J mol-1 K-1.

(a) How much heat is required to raise the temperature of 3.0 mole of the gas from 300 K to 450 K at constant volume?

(b) How much heat is required to raise the temperature of 3.0 mole of the gas from 300 K to 450 K at constant pressure?

(c) What is the internal energy of the gas at 300 K?

(d) What is the increase in internal energy when the temperature changes from 300K to 450K?

[R = 8.31 J K-1 mol-1]

Solution

The ratio of 

•Internal energy for n mole of gas at temperature T = (n) 

•Change in internal energy for n mole of gas when there is a temperature change of

DT = (n)  R DT

so DU = nCv,m DT

= (n)  R DT

and CV,m =  R

Cp,m = CV,m + R

= R+ R

=  R

•The ratio of Cp,m to CV,m is denoted by g

g = 

=

= 

Note: CV,m can also be obtained from the definition

CV,m = 

= 

= 

=  R

Values of Cp,m , CV,m and g at a glance

  Monoatomic

Diatomic Polyatomic

Degree of freedom at room 3 5 6

temperature 

CV,m =  R R R R

Cp,m = CV,m + R =  R R R4R

g =  =   =1.67  =1.40  =1.3315.8 Isothermal Change

> An isothermal process is a constant-temperature process.

>Since temperature is constant, DT = 0 and DU = 0

> Q= DU + W =0 + W thus Q=W

> An isothermal change obeys Boyle’s law.

pV – constant

> For an isothermal change, the gas must be held in a thin-walled, highly conducting vessel, surrounded by a constant

temperature bath. The change must take place slowly so that heat can flow into the gas or flow out from the gas to maintain its

temperature at every instant during the change.

Work done in an isothermal change

•Work done by gas in an isothermal change is given by W = 

•Using the ideal gas equation pV = nRT

p = 

so W = 

= 

= 

= 

= 

Example 7

One mole of an ideal gas undergoes an isothermal change at a temperature T so that its volume V is doubled. What is the work

done by the gas during this change?

Solution

Example 8

A given mass of air contracts isothermally from a volume of 2.0 x 10-4 m3 and pressure of 1.0 x 105 Pa to a volume of 1.5 x 10-

4 m3 at a temperature 300K.

(a) How many moles of gas are there?

(b) What is the new pressure?

(c) What is the work done on the gas?

[R = 8.31 J K-1 mol-1]

Solution

l5.9 Adiabatic Change

> An adiabatic process is a process where no heat is transferred into or out of the system. i.e. Q = 0

>Q = DU+W thus 0 = DU + W and W = -DU

> During an adiabatic expansion, W is positive and DU is negative. i.e. during adiabatic expansion, internal energy decreases

and temperature decreases.

>During an adiabatic compression, W is negative and DU is positive. i.e. during adiabatic compression, internal energy

increases and temperature increases.

> An adiabatic change observes the following equations:

pVg = constant TVg-1 = constant

> For an adiabatic change, the gas must be held in a thick-walled, badly conducting vessel. The change must take place rapidly

so that there is little time for the heat to escape.

> The magnitude of gradient of the adiabatic curve is greater than that of the isothermal.

Adiabatic equation

• Ideal gas equation pV = nRT

•First law of thermodynamics Q = DU + W

•Adiabatic change Q= 0 and W = -DU

•Change in internal energy DU = nCv,m DT

•Work done W = pDV

•Relation between Cp,m = CV,m + R

• From W = -DU pDV = – nCv,m DT

DV = – nCv,m DT

DV = – Cv,m 

(Cp,m – CV,m)  = – Cv,m 

= -

= -

(g-1) = -

(g-1) = -

(g-1) ln V = -ln T + constant

ln T + (g-1) ln V = constant

ln TVg-1 = constant

TVg-1 = constant

· From pV = nRT, substitute   into the above equation

TVg-1 = contant

Vg-1 = contant

pVg = constant

Work done in an adiabatic change

• Work done by gas in an adiabatic change is given by: W = 

• Using the adiabatic equation pVg = constant= k

p =

so W = 

= 

= 

= 

=   where 

=  [p2V2 –p1V1]

=  [T2 – T1] where p1V1 = nRT1 and p2V2 = nRT2

Note: Work done for adiabatic change may also be obtained directly from change in internal energy if the temperatures are

known.

W = -DU

= -n CVm DT

= -(n)  R DT

= -(n)  R [T2 – T1]

Example 9

Some gas, assumed to behave ideally, is contained within a cylinder which is surrounded by insulation to prevent loss of heat.

Initially the volume of gas is 2.9 x 10-4 m3, its pressure is 1.04 x 105 Pa and its temperature is 314 K. The gas undergoes

adiabatic compression until the volume becomes 2.9 x 10-5 m3 and the temperature becomes 790 K.

(a) Calculate g, the ratio of 

(b) Calculate the pressure of the gas after the compression.

(c) What is the work done to compress the gas?

Solution

Example 10

The gas in a heat pump can be considered to undergo a cycle of changes of pressure, volume and temperature. One such

cycle, for an ideal gas, is shown on the graph.

The table below shows the changes in internal energy in section A to B, B to C and C to D. 1 also shows that in sections A to B

and C to D, no heat is supplied to the gas.

Section ofcycle

Increase in internalenergy of gas/J

Heat suppliedto gas/J

Work doneby gas/J

A to B 1200 0  B to C -1350    C to D -600 0  D to A      Using the first law of thermodynamics and the data from the graph, determine:

(i)Work done by gas for A to B.

(ii)Work done by gas for C to D.

(iii) Work done by gas for B to C.

(iv) Heat supplied to gas for B to C.

(v)Work done by gas for D to A.

(vi) Increase in internal energy for D to A.

(vii) Heat supplied to gas for D to A.

Solution

Example 11

The gas in the cylinder of a diesel engine can be considered to undergo a cycle of changes of pressure, volume and

temperature. One such cycle, for an ideal gas, is shown on the graph.

(a) The temperature of the gas at P and Q are 300 K and 660 K respectively. Use the data from the graph to find the

temperatures at R and S.

(b) During each of the four sections of the cycle, changes are being made to the internal energy of the gas. Some of the factors

affecting these changes are given in the table below.

Section ofcycle

Heat suppliedto gas/J

Work doneby gas/J

Increase in internalenergy of gas/J

P to Q 0 -300  Q to R 2580 740  R to S 0 440  S to P -1700    Deduce the increase in internal energy of gas for section:

(i) From P to Q.

(ii) From Q to R.

(iii) From R to S.

(iv) From S to P.

(c) What is the net work output during a complete cycle?

Solution

Chapter 14 kinetic theory of gasesPosted on 25/03/2011 by amimo5095Introduction

• Ideal gas will obey all the gas laws (Boyle’s Law, Pressure Law, Charles’ Law and Ideal gas equation) exactly. Real gas at

high pressure and low temperature will behave differently.

14.1 Ideal Gas Equation

Boyle’s law

For a fixed mass of gas, pressure is inversely proportional to volume if the temperature is constant. Boyle’s law can be written

as:

p a V or pV = constant or p1 V1 = p2 V2

Pressure law

For a fixed mass of gas, pressure is proportional to the absolute temperature if the volume is constant. Pressure law can be

written as:

p a T or   = constant or   

Charles’s law

For a fixed mass of gas, volume is proportional to the absolute temperature if the pressure is constant. Charles’ law can be

written as:

V a T or   = constant or 

Ideal gas equation

• For a fixed mass of gas, when all three variables present (pressure, volume and temperature), we need to use ideal gas

equation

PV = nRT or   = constant or 

where R = molar gas constant = 8.31 J K-1 mol-1

n = number of moles of the gas

•The number of moles of the gas can also be expressed as:

 or 

where

m = mass of gas

M = molar mass of gas

N = number of gas molecules

NA = Avogadro number (number of molecules in 1 mole of gas) = 6.02 x 1023 mol-1

•At s.t.p. (standard temperature and pressure) i.e at 273 K and 101 kPa, the volume of one mole of any gas = 22.4 x 10-3m3.

Relationship between Boltzmann constant and molar gas constant

Boltzmann constant k = gas constant for 1 molecule of gas

Molar gas constant R = gas constant for 1 mole of gas or

= gas constant for NA molecules of gas

Example 1

An ideal gas has been placed in a tank at 30°C. The gas pressure is initially 450 kPa. What will I 1 be the pressure if the

temperature is 60°C?

Solution 1

Example 1

A faulty barometer has some air at the top above the mercury. When the length of the air column is 250 mm, the reading of the

mercury above the outside level is 740 mm. When the length of the air column is decreased to 200 mm, by depressing the

barometer tube further into the mercury, the reading of the mercury above the outside level is 735 mm. Calculate the

atmospheric pressure.

Solution

Example 3

What is the pressure of 8.0 mole of a gas at temperature 77°C occupying a volume of 0.20 m3?

[R = 8.31 J mol-1 ]

Solution

Example 3

A vessel with volume 2.0 litre contains an ideal gaq at temperature 300 K and the pressure of the gas is 110 kPa.

(a) How many moles of gas are there in the vessel?

(b) How many molecules of gas are there in the vessel? Solution

Example 4

Two gases occupy two containers, A and B. The gas in A, of volume 0.12 m3, exerts a pressure of 450 kPa. The gas in B, of

volume 0.16 m3, exerts a pressure of 150 kPa. The two containers are united by a tube of negligible volume and the gases are

allowed to intermingle. What is the final pressure in the container if the temperature remains constant?

Solution

14.2 kinetic Theory of Gases

Kinetic theory of matter

•all matter is made up of small particles.

•these particles are in a state of continuous motion.

Brownian movement and diffusion provide the most direct evidence for the kinetic theory.

Kinetic theory of gases

•Particles in gases are relatively far apart and their interactions are consequently simplest. However certain assumptions have

to be made to simplify the mathematical theory.

•Assumptions made

•The gas molecules are in continuous, free and random motion.

• The gas molecules collide with one another and the walls of the containing vessel elastically.

•Forces between molecules are negligible except during collision.

•The volume of the gas molecules is negligible compared with the volume of the gas.

•The time of collision between molecules is negligible compared with the time between collision.

14.3 Pressure of A Gas

> Consider a gas molecule of mass m moving with velocity c. The components of the velocity c along x, y and z directions are

u, v and w respectively.

> Consider the component of the velocity u in the x-direction.

Change of momentum on striking surface X, DP = mu – (-mu)

= 2mu

> Assuming there is no collision between molecules as the molecule travels between two opposite surfaces, the time taken for

the molecule to collide with the surface X again is

t = 

> Hence the rate of change of momentum due to one molecule  =  = 

> Therefore the force exerted by the molecule on surface X, Fx = 

> The force exerted by N molecules on surface X,   =   (u12 + u2

2 + u32 + …… + uN

2)

> The pressure exerted by N molecules on surface X, p =   =  (u12 + u2

2 + u32 + …… + uN

2)

But <u2> = 

Therefore p = 

Since N is large and the molecules have random m

then <u2> = <v2> = <w2>

and <c2> = <u2> + <v2> + <w2> = 3<u2>

Therefore <u2> =   <c2>

Hence p =     <c2>

p =   nm<c2> n =   = number of molecules per unit volume

mn = mass of molecules per unit volume = r

p =   <c2>

Example 6

Calculate the rms speed of the atoms in a sample of argon which has a density of 1.6 kg m-3

and a pressure of 100 kPa.

Solution

Root mean square (rms) speed and mean speed

• Root mean square seed,

• Mean speed, <c> = 

Example 7

The speed of 10 particles are distributed as follows:

Speed / ms-1 1.0 2.0 3.0 4.0 5.0 6.0No. of particles 1 4 2 1 1 1Using the data in the table, find:

(a) The root mean square speed.

(b) The mean speed of the particles. Solution

14.4 Molecular Kinetic Energy

> From p =   r<c2> and r =   = 

where

M = mass of 1 mole of gas

Vm = volume of 1 mole of gas

NA = Avogadro number

m = mass of 1 molecule

p =   <c2>

From ideal gas equation: p Vm = RT

Hence  <c2> = RT

( <c2>) = RT

<c2> =   = 

where k = Boltzmann constant

Mean translational kinetic energy of a molecule =  <c2> = 

Mean molecular kinetic energy a T where T is the thermodynamic temperature of the gas.

At T = 0 K,  <c2> = 0 for an ideal gas.

14.5 Rms (root mean square) speed of molecules

From  <c2> = 

Rms speed of gas molecules = 

In terms of the molar mass of the gas, the rms speed of the molecules is given by: 

Example 8

In a mixture of two monoatomic gases P and Q in thermal equilibrium, the molecules of Q I have twice the mass of those of P.

The mean translational kinetic energy of the molecules of Q is 6.2 x 10-21 J.

(a) What is the equilibrium temperature? I

(b) What is the mean translational kinetic energy of the molecules of P? I

(c) What is the ratio of rms speed of P to rms speed of Q?

[k=1.38x10-23J K-1]

Solution I

Example 9

Free neutrons in the core of a fission reactor are sometimes referred to as a `neutron gas’. These free neutrons may be

assumed to behave as molecules of an ideal gas at a temperature of 37°C.

(a) For a free neutron of mass 1.67 x 10-27 kg, calculate:

(i) Its mean kinetic energy.

(ii) Its rms speed.

(b) Determine the temperature of helium gas, assumed to be an ideal gas, whose molecules have the same rms speed as the

free neutrons. The mass of He atom is 4 times the mass of neutron.

Solution

Example 10

At what temperature is the rms speed of gaseous hydrogen molecules (molecular weight = 2) 1 1 equals to that of oxygen

molecules (molecular weight = 32) at 27°C? 1

Solution

14.6 A degree of Freedom

> The degree of freedom of a gas molecule is the number of independent ways by which energy is absorbed by the gas

molecule.

translational energy

• A gas molecule of mass m has mean translational kinetic energy. The 3 independent components of its translational motion

gives it three degrees of freedom.

<c2> =  <u2> +  <v2> + <w2>

Rotational energy

•In rotational motion, there are maximum three degrees of freedom.

I <w2> =  Ix <wx2> +  Iy <wy

2> +  Iz <wz2>

• Monoatomic

Molecules of a monoatomic gas is regarded as a point. Its moment of inertia is very small. Therefore rotational kinetic energy is

negligible.

• Diatomic

Molecules of a diatomic gas is regarded as a ‘dumb-bell’. There are 2 degrees of freedom of rotational motion.

• Polyatomic

Molecules of polyatomic gas have 3 degrees of freedom of rotational motion.

Vibrational energy

• The energy of vibration = kinetic energy + potential energy

Vibrational motion has 2 degrees of freedom.

Number of degrees of freedom, f and atomicity of gases

Atomicity of

Number of degrees

Number of degrees

Number of degrees

gases

of freedom at verylow temperature,about 50 K

of freedom at roomtemperature,about 300 K

freedom at very hightemperature,about 1000K

Monoatomicgas

3 translational 3 translational 3 translational

Diatomic gas 3 translational

3 translational +2 rotational =5 degrees of freedom

3 translational2 rotational +2 vibrational =7 degrees of freedom

Polyatomicgas

3 translational

3 translational +3 rotational =6 degrees of freedom

3 translational +3 rotational +(n x 2) vibrationalwhere n depends on the type of molecules

14.7 Maxwell’s Law of Equipartition of Energy

> From kinetic theory for gases,

Mean translational kinetic energy of a molecule =  <c2> = 

Or  <c2> =  <u2> +  <v2> + <w2> = 

i.e energy associated with 3 degrees of freedom = 

Therefore, energy associated with 1 degree of freedom = 

> The law of equipartition of energy states that the energy supplied to a system is distributed equally among all the effective

degrees of freedom and the energy associated with each degree of freedom is   .

> In general, for a gas with f degrees of freedom:

The mean energy per molecule of the gas = f ( ) = 

The mean energy for one mole of the gas = (NA)(   ) = 

The mean energy for n mole of the gas = n 

14.8 Internal Energy of An Ideal Gas

> For real gas, the internal energy of the gas is the total kinetic energy and potential energy of the molecules of the gas.

> For an ideal gas, there is no intermolecular forces between the molecules of the gas, the potential energy of the molecules =

0.

> So, the internal energy of an ideal gas = kinetic energy of the molecules of the gas. > The mean internal energy of an ideal

gas,

U = total kinetic energies of the molecules in the gas

= (No. of molecules) x (No. of degrees of freedom per molecule) x 

= N f 

> For 1 mole of gas, U = 

> For n moles of gas, U-= n 

Example 11

Assuming oxygen behaves as an ideal gas,

calculate:

(a) The root-mean-square speed of its molecules.

(b) The average kinetic energy of a molecule at 273 K.

[1 mole of oxygen has a mass of 32 g]

Solution

Example 12

The total translational kinetic energy for a monoatomic gas at a certain temperature is KT. What is the total translational kinetic

energy for a diatomic and polyatomic gas at the same temperature? 1

Solution:

14.9 distribution of Molecular Speeds

Maxwellian distribution

Figure above

•The graph is not symmetrical.

•The lowest possible speed is zero and the highest is infinity.

•The speed for which the number of molecules is a maximum is called the most probable speed, vp,

•The mean speed is   .

• Vrms >   > vp

•The area under the graph = total number of molecules in the sample.

Distribution of molecular speeds at different temperatures

• When the temperature of the gas increases, the speeds of its molecules increases. The speed distribution graph will thus be

shifted towards higher speeds.

• Since the total number of molecules in the gas remains constant, the area under the graph remains the same.

Chapter 13 Deformation of solidPosted on 24/03/2011 by amimo5095Introduction

• An elastic material returns to its original shape when deforming forces on it are removed. However if the material is stretched

past its elastic limit, it will be permanently deformed. Materials that permanently deform are plastic material.

• Materials that undergo plastic deformation before breaking are ductile(mulur). Materials, like glass, which do not undergo

plastic deformation before breaking arebrittle(rapuh).

13.1 Stress and Strain

> Stress =   F, force and A, cross sectional area

Cross sectional area, A Unit of stress is Pa or N m-2.

> Strain =   e, extension and l, original length

Strain = Strain has no dimension nor unit.

> Young’s modulus, E =   =   = 

Unit of Young modulus is Pa or Nm-2.

13.2 Force-extension Graphs and Stress-strain Graphs

> Force-extension graph

- When weight is added to a spring, the spring extended and the force-extension graph is shown in Figure below

- The graph force is a straight line until point X called the proportional limit

-Between OX, the force is directly proportional to the extension and is said to obey Hooke’s Law

- Hooke’s law states that the force is directly proportional to the extension in a spring or wire if the limit of proportionality is not

exceeded.

F a e

F=ke

where k = force constant (force needed to produce unit extension in the wire)

Note: Elastic limit is slightly above proportional limit. If Hooke’s law is stated as elastic limit not exceeded, then the proportional

limit and the elastic limit is the same point

- From Young modulus, E = 

F = 

Comparing with F = ke, the force constant, k = 

- Area under force-extension graph = energy stored in the wire = work done to extend the wire.

- For the straight line portion, area under force-extension graph =   Fe.

To extend a wire from e1 to e2, the work done is given by the shaded area shown in Figure above.

Work done =   (F1 + F2)( e2 – e1)

> Stress-strain graph

When the elastic limit is not exceeded,

the area under stress-strain graph =   x stress x strain 2\All

= 

= 

= 

= energy per unit volume

Example 1

A spring is 0.18 m long. When it is pulled by a force of 2.0 N, it stretches to 0.22 m. What is the force constant of the spring?

How much energy is stored in the stretched spring? Assume that the spring behaves elastically.

Solution:

Example 2

Force constant for spring P is 12.0 N m-1 and force constant for spring Q is 6.0 N m-1. Springs P and Q are connected in series

and pulled by a force of 1.2 N.

(a) What is the extension for each spring?

(b) What is the force constant for the combination of springs?

Solution

Example 3

When a load of 100 N is hung at the end of a long thin wire of diameter 0.8 mm and of length 50.0 cm, the extension produced

is 2.0 cm.

(a) What is the stress on the wire?

(b) What is the strain in the wire?

(c) What is the Young modulus?

Solution

Example 4

A support cable on a bridge has an area of cross-section of 0.0085 m2 and a length 40 m. It is made of high tensile steel whose

Young modulus is 2.8 x 1011 Pa. The tension in the cable is 750 kN.

(a) Calculate the extension of the cable.

(b) Calculate the strain energy stored in the cable.

Solution

13.3 Young Modulus

> Experiment to determine the Young modulus (Searle’s Method)

(a) Two long thin wires of the same materials are suspended beside each other from a rigid support.

- Long thin wire is used so that the extension produced would be longer and can be measured more accurately.

-Two wires are used to eliminate errors due to (a) changes of temperature and (b) yielding of support under heavy load.

(b) The reference wire is kept straight by a weight attached to its end and carries a scale S.

(c) The test wire carries a vernier scale V which is alongside the scale M.

(d) With loads added, the extension produced is noted. To check whether the elastic limit has been exceeded, the extension is

also measured when the loads are removed.

(e) A graph of extension e against mass of the load m is plotted.

(f) Using Young’s modulus E = 

= 

= 

= 

= 

(g) The length l of the wire is measured using a metre rule and the diameter of the wire is measured using a micrometer screw

gauge.

13.4 Strain Energy

> Behaviour of a wire

-Figure above shows the force-extension graph of a wire until it breaks.

-From O to A, extension is directly proportional to force. Hooke’s Law is obeyed.

-Until point L, the wire is elastic. Wire able to return to its original length when load is removed.

-From L to B, wire does not recover to its original length when load is removed. Plastic deformation occurred.

-The point B is known as the yield point. Beyond point B, plastic deformation due to atomic planes sliding over each other

occurred.

-At C, the force is maximum. The wire thins non uniformly and breaks at D. Efforts to save the wire from breaking by reducing

the force will not work.

> Graphs for ductile, brittle and polymeric materials

- Materials like copper and steel show a large amount of plastic deformation before breaking. They are called ductile materials.

- Materials such as glass and ceramics breaks after the elastic limit without any plastic deformation. They are called brittle

materials.

- Material like rubber produces a large extension with a small force. It is a polymeric material. When a stretching force is

applied, the coiled rubber molecules straighten and giving rise to a large extension (800%). After all the chains of molecules

have straighten out, very large force is required to increase the separation of the atoms.

Example 8

A specimen fibre of glass and a specimen of copper wire has, the same dimensions. The length of each specimen is 1.60 m

and the radius of each is 0.18 mm. Force-extension graphs for both specimens are shown in figure below.

(a) (i) Which of the two materials is brittle?

(ii) Explain using the feature of the graph that leads you to the answer in (i).

(b) Using the graphs and the data given, determine: I

(i) The area of the cross-section of the specimens.

(ii)The Young modulus of the glass. I

(iii) The ultimate tensile stress for copper. I

(iv) The approximate value for the work done to stretch the copper wire till its breaking point.

Solution

Chapter 12 State of MatterPosted on 23/03/2011 by amimo509512.1 Solid, Liquid and Gas

> Solid

- has definite volume and shape

- atoms are arranged closely in a fixed pattern

- atoms vibrate about their mean position

- strong attractive forces between atoms

- the atoms possess potential energy and kinetic energy

> Liquid

- has definite volume but assumes the shape of the vessel into which it is poured – molecules are arranged closely but not in

fixed pattern

- molecules vibrate as well as moving freely in the liquid

- attractive forces between molecules but the forces are weaker than that in solid – the molecules possess potential energy and

kinetic energy

> Gas

- has no definite volume and shape, it expands to fill any container into which it is placed

- molecules are separated far apart (size of molecule is negligible compared to the distance between molecules)

- molecules move freely in random motion

- forces between molecules are negligible

- the molecules possess kinetic energy only

12.2 Crystalline Solids

> The essence of the structure of a crystalline solid is that the arrangement of atoms, ions or molecules repeats itself regularly

over a long distance. Examples are sodium chloride crystals, copper, aluminium, diamond and silicon. In every crystal structure

there is a typical cell called the unit cell, which is repeated throughout the crystal.

> Solids where the particles are assembled in a disordered way but only show order over short distances is said to have the

structure of an amorphous solid. Glass is an example of amorphous (meaning shapeless) solid.

12.3 intermolecular force curve

> Force-separation graph (F-r graph)

- Force is required to extend a wire. This suggests that there are strong attractive force between atoms in solids.

- It is very difficult to compress solid or liquid. This suggests that there are strong repulsive forces between atoms.

- The attractive and repulsive forces between atoms are electrostatic in nature.

- In general the resultant force is given by equation

F = Frepulsive + Fattractive

- The resultant force between two atoms is shown in Figure above. Notice that the resultant force is repulsive (positive force)

when r < r0 and the resultant force is attractive (negative force) when r > ro.

- When r = ro, repulsive force = attractive force or the resultant force = 0. to is known as the equilibrium separation of the atoms.

- At r = ro± Dr

The graph F-r is a straight line.

The straight line equation can be written as F =-k(r – ro)

= -kDr

This shows that force is directly proportional to extension which is Hooke’s Law.

- The breaking force is the force that is able to overcome the maximum attraction. The separation that corresponds to the

breaking force is r1.

12.4 Potential Energy Curve

> Potential energy graph (U-r graph)

- The force F between molecules is given by 

- When the potential energy U is minimum, it corresponds to the equilibrium spacing between the molecules ro. The

temperature is at absolute zero and the matter is in solid state.

- When the solid is heated, the molecules have some energy, E, above the minimum value. The molecules oscillate between

points X and Y. Since the U-r curve is not symmetrical, the mean position G of the oscillation is greater than ro. Thus, the solid,

expands when its thermal energy is increased.

- When the energy equals E (Latent heat), the molecules are completely free and form gas.

Example 1

Figure above shows how the potential energy U changes with separation between atoms r for two types of solid A and B.

Discuss what is the difference between solid A with solid B in terms of:

(a) Arrangement of atoms.

(b) Elasticity.

(c) Molar latent heat of fusion.

Solution

(a) The minimum U for solid A occurs at a bigger value of r, this means that atoms in solid A is

further apart.

(b) The gradient of graph at ro for solid B is steeper, this means that solid B is more elastic (bigger spring constant k).

(c) Graph B has a greater minimum value, this means that the molar latent heat of fusion is greater.

Example 2

The potential energy U between two molecules at the distance r apart is given by the equation

 where a and b are constants.

(a) Write down the expression for the force F between the two molecules.

(b) Determine the equilibrium separation ro between the molecules in terms of a and b.

Solution

Example 3

The force F between two molecules at a distance r apart is given by the equation   where a and b are

constants.

(a) Determine the equilibrium separation ro between the molecules in terms of a and b.

(b) Write down the expression for the potential energy U between the two molecules.

(c) What is the minimum potential energy Umin in terms of a and b? State the physical significance of Umin.

Solution

4.3 Latent Heat

Posted on 26/02/2010 by amimo5095

What does the word “latent     heat “ mean?

Latent heat means hidden heat. This heat energy changes the state of a substance (phase change). The heat cannot be ‘seen’

because there is no rise in temperature of the substance.

 

Phase Change

When a phase change has occurred , latent  heat is absorbed or  released.

 

 

Latent heat and kinetic theory

 

In a solid,the molecules are linked to the neighbours by forces of attraction. As the solid is heated , the molecules vibrate more

strongly. When the solid reaches its melting point, the vibrations have become so strong that the links begin to give way. Extra

energy is needed to overcome these forces and separate the molecules. This is called the latent heat of fusion. No

temperature rise occurs during this process, because the latent heat of fusion is used to overcome the intermolecular binding

forces. The average translational kinetic energy does not change, so the temperature remains constant.

In a liquid, the molecules are free enough to slide  around and change neighbours, but  they are still almost as close to each

other as in a solid. The links are weaker but still effective. As the liquid is heated further, the kinetic energy of the molecules

increases more. At the boiling point, the molecules break free from each other and become a gas. Energy is needed to

overcome the remaining links. This is  called the latent heat of vaporisation.

No temperature rise occurs during this process, because the latent heat of vaporisation is used to overcome the intermolecular

binding forces. The average translational kinetic energy does not change, so the temperature remains constant.

 

The Heating and Cooling Curve (Naphthalene)

 

Heating curve

Melting point = 80o C          Boiling point = 218o C

AB= Solid ,   BC = solid  +  liquid   CD = liquid

DE = liquid  + gas     EF = gas

At     AB,CD and EF  : 

The heat supplied increases the kinetic energy of naphthalene.So the temperature rises because the temperature is a measure

of the average kinetic energy of molecules in a substance

At     BC,DE :

At  t1  and   t2 phase change has occurred.

The latent heat is absorbed to provide the energy to overcome the binding forces between the molecules. The energy absorbed

does not increase the kinetic energy of the molecules, so the temperature remains constant.

Cooling curve

Freezing point = 80o C     

Condensation point = 218o C

AB= Gas   BC = Gas + Liquid   CD = Liquid

DE = Liquid +  Solid     EF =  Solid

At AB,CD and EF :

Heat is released to the surroundings and the kinetic energy of the molecules decreases, resulting in a fall in the temperature of

the naphthalene because the temperature is a measure of the average kinetic energy of molecules in a substance

 

At BC,DE :

At  t1  and   t2 phase change has occurred.

The latent heat is released to the surroundings as the molecules become more closely pack.. The energy released does

not  decrease the kinetic energy of the molecules, so the temperature remains constant

 

Definition and the S.I unit of Specific Latent Heat     ,l

The specific latent heat of fusion , lf    :

Is the quantity heat of energy required to change 1 kg of a substance from the solid state to the liquid state , without a change

in temperature.

The specific latent heat of Vaporisation , lv    :

Is the quantity heat of energy required to change 1 kg of a substance from the liquid state to the gaseous  state , without a

change in temperature.

The S.I. unit of lf  and lv  is  J  kg-1

The relationship between     m,l     and Q

Where,

Q  =  the heat energy transferred to the substance

m =   the  mass of the substance

l   =   the  specific latent heat  of the substance  

 

Example     1

 

What is the quantity of heat required to melt  2.0 kg ice at 0 o C.

(The specific latent heat of fusion of ice = 3.34 x 105 J kg-1)

Solution

 

 

Example 2

How much energy has to be removed from 4.0 kg of water at 20o C  to produce a block of ice at   0 o C. (The specific heat

capacity of water =4.2 x 103  J kg-1 o C-1 .  specific latent heat of fusion of ice = 3.34 x 105 J kg-1)

 

Solution

 

 

Example 3

 

Calculate the heat required to convert 4 kg of ice at  - 15o C into steam at 100o C.   

( Specific heat capacity of ice  = 2.1 x 103J kg-1oC-1,

  Specific heat capacity of ice  = 4.2 x 103J kg-1oC-1,

  Latent heat of fusion of ice = 3.34 x 105 J kg-1 and

  Latent heat of vaporisation of water =    2.26 x 106 J kg-1 )

 

Solution

 

 

 

 

To determine the latent heat of fusion of ice

 

The electrical power of the heater is recorded = P

The mass of each the two empty beakers is determined using the weighing balance.

Mass of empty beaker A = m1  

Mass of empty beaker B =  m2

When water starts to drip from the filter funnels at a steady rate, the heater in  Set A is switched on.

The stopwatch is started  and the empty beakers  A and  B  are placed beneath the filter funnels.

After a period of t , the heater in Set A is switched off.

The masses of both beakers of water , A and B are determined using the weighing balance.

Mass of beaker A +  water = m3  

Mass of beaker B + water = m4

Calculate mass of ice melted by the electric immersion heater,  m  = (m3 - m1) -  (m4 - m2)

Calculate the heat supplied  by the heater = Pt

Calculate the heat absorbed by the ice during melting  = mlf

On the assumption that there is no heat loss to surroundings;,

 Pt = mlf

 lf =  Pt

                m

Precautions

 

(1)     The immersion heater must be fully immersed in the ice cubes to avoid or reduce heat loss

 

Discussions

The value of the specific latent heat  of fusion of ice ,lf determined in the experiment is larger than the standard value of lf.

This is because the experimental value of the mass of ice melted ,m less than the expected m due to some heat loss to the

surroundings.

The smaller the mass m, the greater the

 specific latent heat of fusion of ice,lf,

    lf  =     Pt

             m¯

 

 

To determine the latent heat of vaporisation of water

 

The electrical power of the heater is recorded = P

The electric heater is switched on the heat the water to its boiling point.

When the water starts to boil at a steady rate , the stopwatch is started and the reading on the balance is recorded = m1

After  a time ,t the reading on the electronic balance is recorded again = m2

Calculate the mass of water evaporated,  m = m1 - m2

Calculate the heat supplied  by the heater = Pt

Calculate the heat absorbed by the water  during vaporisation  = mlv

On the assumption that there is no heat loss to surroundings;,

 Pt = mlv

 lv =  Pt

                m

Precautions

 

(1)     The immersion heater must be fully immersed in the water to avoid or reduce  direct heat loss to the surroundings.

 

Discussions

The value of the specific latent heat  of vaporization of water  ,lvdetermined in the experiment is larger than the standard value

of lv.

This is because the experimental value of the mass of  water evaporated  ,m less than the expected m due to some heat loss to

the surroundings.

The smaller the mass m, the greater the

 specific latent heat of fusion of ice,lv,

    lv  =     Pt

              m¯

 

 

Example     4

A 800 W electric heater is used to boil water. What is the time required to reduce the mass of water by 4 kg after the water has

reached its boiling point?

[ Specific latent heat of vaporization of water = 2.26 x 106 J kg -1 ]

 

Solution

 

 

 

 

Example     5

0.5 kg of a solid is heated by a 100 W heater. The graph shows how the temperature substance varies with time.

Calculate the specific latent heat of fusion of the solid.

 

Solution

 

 

Applications of Specific Latent Heat in Everyday Life

(1)     When we are engaged in strenuous activities ,  sweating cools our bodies. The sweat evaporates and the bodies heat is

removed as the latent heat of vaporisation.thus our bodies temperature is decreased.

 

(2)   Drinks can be cooled by adding in several cubes of ice. When the ice is melting , the latent heat of fusion is absorbed from

the drinks. The temperature of the drinks is lowered.

 

(3)   Food can be cooked by using steam. Food such as cakes, eggs, fish, buns and others receive a large amount of energy

when the latent heat of vaporization of steam released from condensing steam.

Forces in PhysicsGravity Forces

Weight of a mass m near an object of gravity g

The value of g at a distance r from the centre of a planet of mass M (the minus is the magnitude)

The force of G between mass m and mass M

Gravitational PE

The potential energy of a mass m lifted a height h above the earth

The value of gravitational potential energy at a distance r from the centre of mass M

The PE of mass m and mass M at a distance r apart

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