Chapter 2.1 - Measurement of Length

In the previous section, we saw the basics about SI system of units. In this section, we will see details about Length.

We will write it in steps:
1. We saw that in SI system, we measure length in metre.
(One metre is the distance traveled by light in vacuum during a time interval of 1/299,792,458 of a second).
Let us see some practical situations:
• Length of a study table is usually around 120 cm.
• The table that I am using right now, has a length of 122 cm.
• We can write it in three ways:
    ♦ Length of the table = 1.22 m.
    ♦ Length of the table = 122 cm.
    ♦ Length of the table = 1220 mm.
2. For objects like tables and chairs, the third method involving mm, is not generally used
• mm is used for smaller objects. Let us see an example:
• The length of the pencil that I am using right now, has a length of 14.5 cm.
• We can write it in 3 ways:
    ♦ Length of the pencil = 0.145 m.
    ♦ Length of the pencil = 14.5 cm.
    ♦ Length of the pencil = 145 mm.
• For objects like pens and pencils, the first method involving m, is not generally used.
• m is used for larger objects.
3. Now consider a small cylinder which has a length about the same as a pen.
• This cylinder is an important part of a precision instrument. It’s diameter is very important.
4. So a technician took an ordinary ruler and measured it’s diameter at one end.
• He got it as 2.7 cm. That is., 27 mm.
5. But this cylinder is to be used in a precision instrument. We want accurate diameter.
• Upon looking closely, the technician saw that the diameter is in between 27 mm and 28 mm.
6. But unfortunately, the smallest length which an ordinary ruler can measure is 1 mm.
• 1 mm = 0.001 m = 10^-3 m.
• We say that the least count of that ruler is 10^-3 m.
7. To measure lengths which are smaller than 1 mm, we need to divide each mm on that ruler, into 10 equal parts.
• That is not very convenient because, 1 mm is a small distance.
8. In such a situation, we use an instrument called the vernier calipers.
The smallest length which a vernier calipers can measure is 'one tenth of a mm'.
One tenth of a mm = $\mathbf\small{\frac{1}{10} \times 1 \,\text{mm}}$ = 0.1 mm.
• 0.1 mm = 0.0001 m = 10^-4 m.
• So the least count of the vernier calipers is 10^-4 m.
9. The technician used it on the cylinder and found that, the actual diameter is 27.3 mm.
• We know that: 27 < 27.3 < 28
• So greater precision is achieved.
10. In situations where still more precision is required, we use screw gauge or spherometer.
• They can measure lengths as small as 10^-5 m.
11. Lengths smaller than 10^-5 m are not uncommon in our day to day life.
• For example, municipality rules do not allow us to use plastic carry bags whose thickness is less than 50 microns.
(In some municipalities, all types of plastic carry bags are banned) 
    ♦ Symbol for micron (micro meter) is 𝝻m.
    ♦ One 𝝻m = 10^-6 m.
• In nano technology, scientists deal with objects which are as small as one nano meter.
    ♦ Symbol for nano meter is nm.
    ♦ 1 nm = 10^-9 m.
12. To measure lengths smaller than 10^-5 m, we use some special indirect methods. We will see them in later sections of this chapter.

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13. What we discussed above is related to small lengths. What about larger lengths ?

Let us see some examples:
• Height of a two storeyed building used for residential purposes is around 6.4 m.
That is., 6 m plus 40 cm.
• A two storeyed building used for commercial or industrial purposes will have a greater height.
• Such heights can be measured using a measuring tape or simple survey instruments.
14. To measure the distance between two towns, we do not use m. Instead, we use km.
• Such distances can be measured using advanced survey techniques and satellite images.
15. To measure distances between planets and stars, we use astronomical unit.
    ♦ It’s symbol is au.
    ♦ 1 au is the mean distance between the center of earth and center of sun.
(Remember that, the path of earth is not circular. It is elliptical. So the distance from the sun is not a constant. That is why, we take mean distance).
    ♦ That distance is equal to 149597870700 m.
16. To measure distances between stars, we use light year.
    ♦ It’s symbol is ly.
    ♦ 1 ly is the distance traveled by light in 1 year.
(Remember that in 1 s, light travels a very large distance of 3 × 10⁸ m).
    ♦ The distance traveled in one year is equal to 9.46 × 10¹² km.
17. To measure even greater distances, we use parsec.
    ♦ It’s symbol is pc.
(We will see the definition of parsec in the next section).
     ♦ 1 pc is equal to 3.26 ly.

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Measurement of large distances

• Three examples of large distances are given below:
(i) The distance between earth and the moon.
(ii) The distance between earth and another planet.
(iii) The distance between earth and another star.
• In such cases, we cannot use a measuring tape. So we use indirect methods.
• Before we learn the details of those indirect methods, we must see some fundamental properties related to ‘large distances’. We will write them in steps:

1. In fig.2.3 below, a person PQ is moving in a car along a straight line.

Fig.2.3

• The top of a tree is marked as T. it is shown in green color.
• The tree is on the side of the road.
• When the man is at position A, his eye level is marked as P_A
2. We know the science behind vision. Based on that, we can write:
The man sees the tree top T because the light rays reflected from T reaches his eyes.
• One such light ray is shown in magenta color. It is named as TP_A
• Similarly, when the man is at position B, he sees T because of the ray TP_B
• Note the angle between the two rays TP_A and TP_B. It is marked as θ₁
3. Consider a hill on the side of the road. It’s top point is marked as H.
• The hill is behind the tree and is far away from the road.
• So H is far away from the man.
    ♦ When the man is at A, he sees H because of the ray HP_A
    ♦ When the man is at B, he sees H because of the ray HP_B
• The angle between those two rays is marked as θ₂
4. We see that θ₂ less than θ₁
• This is because, H is further away from the man than T.
5. Consider a mountain on the side of the road. It’s 'snow covered top point' is marked as S.
• The mountain is behind the hill and is very far away.
• So S is very far away from the man.
    ♦ When the man is at A, he sees S because of the ray SP_A
    ♦ When the man is at B, he sees S because of the ray SP_B
• The angle between those two rays is marked as θ₃
6. We see that θ₃ is less than θ₂
• This is because, S is further away from the man than H.

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7. So we find that, as the distance increases, the angle decreases.

• The ray TP_B is very different from TP_A
• The man feels that, the tree moves past him in the opposite direction, at great speed.
8. The ray HP_B is also very different from HP_A
• But difference is not so high as in the case of the tree.
• The man feels that the hill moves past him in the opposite direction, but at a slow speed.
9. The ray SP_B is also very different from SP_A
• But difference is smaller than that in the case of the hill.
• The man feels that the mountain moves past him in the opposite direction, but at a very slow speed.

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10. So we find that, when the distance increases, the 'movement of objects' in the opposite direction becomes slower and slower.

■ What happens if the distance is very large?
Ans: Obviously in that case, the movement will be so slow that, it will appear to be stationary. We experience such a situation in the case of the moon and the stars.
11. The height of Mount Everest is 8,848 m.
    ♦ That means, the tip of Mount Everest will be at a distance of 8,848 m from a person at ground level.
• The distance of the moon from the earth is 384,400,000 m.
    ♦ That means, the Moon will be at a distance of 384,400,400 m from a person at ground level.
12. Suppose that in the above fig.2.3, we mark the moon as M.
• The rays coming from M will be MP_A and MP_B
• M will be very high above S.
• So the angle θ between the two rays will be so small that, both rays will coincide.
• That means, the two rays are practically the same.
• So the moon will appear to be stationary.
13. But in our actual experience, we feel this:
■ The trees and hills are moving past us. But the moon is moving along with us. Why is that so?
Ans: At every point of our travel, practically the same ray from the moon reaches our eyes.
• So the moon is actually stationary with respect to us.
• But our minds know that we are moving.
• So we feel that, the moon is moving along with us.
(Suppose that we are moving in a very smooth train. Looking upwards through the sky roof, if we see only the moon and no trees,clouds, or any other objects, we will truly feel that the moon is stationary).
14. So we can conclude that, far away objects appear to be stationary when we are moving.
• We will be using this information while finding distances of far away objects.

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In the next section, we will see Parallax method.

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Chapter 2 - Units and Measurement

In this chapter, we will see various details about units and measurements.

To measure any quantity, we need a unit. Let us see the significance of ‘unit’ by considering an example. We will write it in steps:
1. Fig.2.1(a) below shows a block of wood brought into a town for making furniture. The carpenter wants to know it’s length ‘l’.

Fig.2.1

2. Fig.b shows a steel rod that he usually uses to measure lengths. He calls that rod: cubit.
• That means, the length of that steel rod is considered as 1 cubit.
3. To find the length of the block, he places that steel rod end to end as shown in fig.c.
• We see that, the length of the block is 3 times the length of the steel rod.
• So the carpenter notes down in his book:
Length of the block = 3 cubits.
4. Here, 'cubit' is the unit used by the carpenters in that town to measure lengths.
• All carpenters in that town use the same rod (or it’s duplicates) to measure length.
5. Similarly, a vessel having appropriate volume ('appropriate' indicates not too small or too large based on the situation) can be selected to measure volume of oil, kerosene etc.,
• One such vessel was called 'amphora' in olden days.
• So 'amphora' is the unit for measuring volume in that town.
• All people in that town use that same vessel (or it’s duplicates) to measure volume.     

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• So we see the importance of ‘units’. With out specifying a unit, we cannot measure quantities.
    ♦ That unit should be acceptable to all.
• Next we will see why it is important to standardize various units:

Here also, we will see an example. We will write it in steps:
1. An engineer and his assistant is measuring the length of a wall.
• After taking the measurement, the engineer tells his assistant: 3.6
• The assistant notes it down.
2. After some time, another engineer and his assistant reaches there.
• They also measure the length.
• After taking the measurement, the engineer tells his assistant: 12
• The assistant notes it down.
3. The onlookers are puzzled. How can the same wall have two different lengths?

Another example:
1. Some oil is kept in a vessel in a town.
• An officer and his assistant reaches there and measure the volume of oil available.
• After taking the measurement, the officer tells his assistant: 800
• The assistant notes it down.
2. After some time, another officer and his assistant reaches there.
• They also measure the volume.
• After taking the measurement, the officer tells his assistant: 211
• The assistant notes it down.
3. The onlookers are puzzled. How can the same oil have two different volumes?

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Let us analyze each incidents described above:

In the case of the wall:
1. The first engineer was using the unit ‘metre’.
• When he places '3 units plus 0.6 of his units'  end to end, he will reach the other end of the wall.
2. The second engineer was using the unit ‘feet’.
• He has to place 12 of his units end to end to reach the other end of the wall.
■ Thus the two measurements gave different values.

In the case of oil:
1. The first officer was using the unit ‘litre’.
• When he uses his vessel 800 times, he will have measured out all the oil.
2. The second officer was using the unit ‘gallon’.
• When he uses his vessel 211 times, he will have measured out all the oil
■ Thus the two measurements gave different values.

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This is a very serious issue. Suppose that, scientists and engineers from different parts of the world are collaborating on a project. Even if all of them speak English, none of them will be able to understand what the others are saying. The project will not materialize. The importance of using standardized units is obvious.

Let us first see a brief history of units:
• In earlier times, scientists of different countries were using different systems of units.
• 3 such systems were prevalent until recently. They were:
1. The CGS system.
• In this system the base units were as follows:
    ♦ Base unit for Length was centimetre.
    ♦ Base unit for mass was gram.
    ♦ Base unit for time was second.
2. The FPS system.
In this system the base units were as follows:
    ♦ Base unit for Length was foot.
    ♦ Base unit for mass was pound.
    ♦ Base unit for time was second.
3. The MKS system.
In this system the base units were as follows:
    ♦ Base unit for Length was metre.
    ♦ Base unit for mass was kilogram.
    ♦ Base unit for time was second.

• In the year 1971, scientists from different parts of the world met at the General conference on weights and measures.
• At that conference, a new system was developed and recommended to be used internationally.
• This system was called Systeme Internationale d’ unites.
• ‘Systeme Internationale d’ unites’ is the French for 'International System of Units'.
• In every day usage, the scientific community refer to this system in an abbreviated form: SI system.

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Now we will see the details of the SI system. We will write them in steps:

1. In the SI system, there are seven base quantities:
Length, mass, time, electric current, temperature, amount of substance and luminous intensity.
• The units of the seven base quantities are taken as the base units.
So the seven base units are:
1. Unit of length. It is called the metre. It's symbol is m.
2. Unit of mass. It is called the kilogram. It's symbol is kg.
3. Unit of Time. It is called the second. It's symbol is s.
4. Unit of electric current. It is called the ampere. It's symbol is A.
5. Unit of temperature. It is called the kelvin. It's symbol is K.
6. Unit of amount of substance. It is called the mole. It's symbol is mol.
7. Unit of luminous intensity. It is called the candela. It's symbol is cd.

Note: When mole is used, we have to specify the ‘type of entity' as well.
• Some examples:
    ♦ 2 moles of atoms.
    ♦ 5 moles of molecules.
    ♦ 3 mols of ions.

• Besides the seven base units that we saw above, there are two more:
(i) plane angle dθ (ii) solid angle d𝛀.
We will write about them in steps:
1. Consider fig.2.2(a) below:

Fig.2.2

• ds is the length of the arc AB.
• arc AB is a portion of a circle.
• O is the center of that circle.
• Arc AB subtends an angle dθ at O.
2. OA and OB are radii.
• Length of OA = Length of OB = r
• From our math classes, we know that $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$.
• So we get $\mathbf\small{d\theta=\frac{ds}{r}}$
■ Thus we get the definition for plane angle:
Plane angle dθ is the ratio of the arc length ds to the radius r
3. Consider fig.2.2(b).
• A 'portion of the surface of a sphere' is shown in red color. It has an area of dA.
• O is the center of that sphere. So the yellow lines are radii.
• The lengths of all the yellow lines are equal to r.
• The area dA subtends a solid angle d𝛀 at O.
4. The solid angle d𝛀 is given by $\mathbf\small{d\Omega=\frac{dA}{r^2}}$.
■ This gives the definition for solid angle.
Solid angle d𝛀 is the ratio of the area dA to the square of the radius r.

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• So now we have a basic idea about base units. Next we will see derived units.

• For that, first, we have to see derived quantities.
• Let us see an example. We will write it in steps:
1. Consider the quantity: volume.
• Usually, we obtain volume by multiplying three items:
length, width and height.
2. width and height are also ‘lengths’.
• The quantity ‘volume’, is derived by multiplying the base quantity length, 3 times.
• So ‘volume’ is a derived quantity.

Another example:
1. Velocity is derived by dividing distance by time.
2. Both distance (length) and time are base quantities.
• So velocity is a derived quantity.

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The units of derived quantities is called derived units.
An example:
• We get volume by multiplying length 3 times.
• So the unit of volume is obtained by multiplying the 'unit of length' 3 times.
• The SI unit of length is metre.
• Thus we get unit of volume as: m × m × m = m³

Another example:
• We get velocity by dividing distance by time.
• So the unit of velocity is obtained by dividing:
'unit of distance'   by   'unit of time'. 
• The SI unit of distance is metre.
• The SI unit of time is second.
• Thus we get unit of velocity as: m/s.

Now we can appreciate the difference between base units and derived units.
• We see that base quantities are independent quantities. Each of the seven base quantities can exist independently, with out any help from other quantities. Also none of the base quantities can be expressed in terms of other quantities. For example, mass cannot be expressed in terms of length or time or any other quantity.
• But derived quantities are dependent on base quantities. For example, we can write the volume of a wooden block only if we know the lengths (length, width and height) of that block.

Let us write the definitions:
■ The quantities which are not related to one another and cannot be expressed using other quantities are called base quantities. The units of base quantities are called base units.
■ Units which are expressed in terms of base units or those units which are dependent on base units are called derived units.

• There are only seven base units. But there are numerous derived units. We will see them in later chapters.

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At the international conference held in 1971, not only units, but also 'symbols' and ‘abbreviations’ were standardized.

Let us see an example:
• In many situations, scientists have to measure 'energy'.
• At the general conference, the following decisions were made:
    ♦ The unit in which energy is to be measured is: joules.
    ♦ The symbol to be used for joules is J

Another example:
• In many situations, scientists have to measure 'electrical resistance'.
• At the general conference, the following decisions were made:
    ♦ The unit in which 'electrical resistance' is to be measured is: ohm
    ♦ The symbol to be used for ohm is 𝛀 (Greek capital letter Omega)

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Since the SI units use the decimal system, conversions within the system can be done easily.

Let us see an example:
Volume of a container is 5 liters. Inside the container, some sand is present. Volume of that sand is 20 cm³. What percentage of the volume of the container is occupied by sand?
Solution:
1. To find percentage, both measurements must be in the same units.
• So we have to convert liters into cm³
2. One liter is the volume occupied by a cube of side 10 cm.
• That means 1 liter = 1000 cm³
• So 5 liters = 5000 cm³
• Thus we successfully converted the 'volume in liters' into 'volume in cm³'
3. Now we can find the percentage.
• The percentage volume occupied by sand = $\mathbf\small{\frac{20}{5000}\times 100}$ = 0.4 %
4. The conversion could be easily done in step (2) because, both liters and cubic centimeter are in the decimal system.
• If the conversion is not done, we would get: $\mathbf\small{\frac{20}{5}\times 100}$ = 400 %, which is absolutely wrong.

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Next we will see some basic rules to be followed while writing units.

(1) The symbols of units are normally written using small letters in the English alphabet.
Some examples:
m (metre), s (second), kg (kilogram).
(2) But there are certain occasions in which capital letters of the English alphabet are used as symbols. The units named after persons are written like this.
Let us see some examples:
(i) The unit for measuring potential difference is volt.
• This unit is named after the scientist Alessandro Volta.
• So the symbol for the unit volt is V.
(ii) The unit for measuring pressure is pascal.
• This unit is named after the scientist Blaise Pascal.
• So the symbol for the unit pascal is P.
(iii) The unit for measuring force is newton.
• This unit is named after the scientist Sir Isaac Newton.
• So the symbol for the unit newton is N.
(3) While writing the 'names of units' never use capital letters.
Let us see some examples:
(i) kelvin is the SI unit for measuring temperature.
• When writing it as the name of the scientist Lord kelvin, write it as: Kelvin.

• When writing it as the unit, write it as: kelvin.
(ii) newton is the SI unit for measuring force.
• When writing it as the name of the scientist Sir Isaac Newton, write it as: Newton.
• When writing it as the unit, write it as: newton.
(4) Never use the plural form for symbols.
Some examples:
• 10 kg (correct) 10 kgs (wrong).
• 75 cm (correct) 75 cms (wrong).
(5) Never use full stop or comma after a symbol except at the end of a sentence.
An example:
• 75 cm is the length of a table. (correct).
• 75 cm. is the length of a table. (wrong).
(6) While writing derived units a slash (/) is used to denote division. But never use more than one slash in one derived unit.
An example:
• m/s² (correct) m/s/s (wrong).
(7) When a derived unit is expressed as the product of other units use a dot or a space between them.
An example: N.m or N m
(8) Do not mix the name of a unit with the symbol.
Some examples:
• kg/m³  (correct).
• kilogram per cubic metre (correct).
• kg/cubic metre (wrong).
• kilogram per m³  (wrong).
• kg per m³  (wrong).
• kilogram/m³  (wrong).
• kilogram/cubic metre (wrong).
(9) While writing units along with a value, there must be single space between them.
Some examples:
• 273 K (correct) 273K (wrong).
• 100 m (correct) 100m (wrong).
(10) Never use more than one unit to express a physical quantity.
An example:
10.25 m ( correct) 10 m 25 cm (wrong).
■ The above given are basic rules. For a detailed information and for more such rules, the reader is advised to refer standard textbooks.

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In the next section, we will see details about Length.

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Chapter 7.30 - Power derived from Torque

In the previous section, we saw work done by a torque. In this section, we will see power

1. We have: $\mathbf\small{dW=|\vec{\tau}|\;d\theta}$
• This is the work done in the time interval during which, the body turns through dθ
• Let this time interval be dt
• So dW joules of work is done in a time interval of dt seconds
2. Thus we get:
• Power (P) produced by the torque 
= Work done by the torque in 1 second 
= $\mathbf\small{\frac{dW}{dt}=\frac{|\vec{\tau}|\;d\theta}{dt}=|\vec{\tau}|\left(\frac{d\theta}{dt}\right)}$
3. Consider the term $\mathbf\small{\left(\frac{d\theta}{dt}\right)}$
• Angular displacement is being divided by time. It will give angular velocity ω
• So the result in (2) becomes:
Eq.7.28: $\mathbf\small{P=|\vec{\tau}|\omega}$

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Now we will derive an interesting result related to rotational motion. But first, we will derive it for linear motion
1. A rigid body (of mass m) in linear motion, has an initial velocity of v₁ ms^-1
• A force F acts on it for a time interval of Δt seconds
    ♦ Direction of F is same as the direction of motion
• As a result, the velocity of the body increases to v₂
• During the Δt seconds, the body undergoes a displacement of Δs  
2. Work done by the force = Force × displacement = $\mathbf\small{F\times \Delta s}$
3. Applying work-energy theorem, we can write:
• If no work is lost against friction or air resistance, all the work done by the force will be utilized to increase the kinetic energy of the body (Details here)
• Also note that, in this case, the body is rigid. So there is no internal motion of particles. All the external work will indeed be utilized for increasing the kinetic energy of the body
4. Now, increase in kinetic energy = $\mathbf\small{0.5m\,v_2^2-0.5m\,v_1^2=0.5m(v_2^2-v_1^2)}$ 
• Equating this to the result in (2), we get: $\mathbf\small{0.5m(v_2^2-v_1^2)=F\times \Delta s}$
5. Dividing both sides by Δt, we get: $\mathbf\small{\frac{0.5m(v_2^2-v_1^2)}{\Delta t}=\frac{F\times \Delta s}{\Delta t}=F\times\frac{\Delta s}{\Delta t}}$
6. $\mathbf\small{\frac{\Delta s}{\Delta t}}$ normally gives velocity
• But in our present case, the velocity is not uniform
    ♦ This is because of the action of the force F
    ♦ Because of the F, the body will be moving with an acceleration
• So $\mathbf\small{\frac{\Delta s}{\Delta t}}$ will give us the average velocity
• That means: $\mathbf\small{\frac{\Delta s}{\Delta t}=\frac{v_1+v_2}{2}}$
7. So the result in (5) becomes: $\mathbf\small{\frac{0.5m(v_2^2-v_1^2)}{\Delta t}=F\frac{(v_1+v_2)}{2}}$
$\mathbf\small{\Rightarrow \frac{0.5m(v_2+v_1)(v_2-v_1)}{\Delta t}=F\frac{(v_1+v_2)}{2}=0.5F(v_1+v_2)}$
$\mathbf\small{\Rightarrow \frac{0.5m(v_2-v_1)}{\Delta t}=0.5F}$
$\mathbf\small{\Rightarrow \frac{m(v_2-v_1)}{\Delta t}=F}$
8. But $\mathbf\small{\frac{(v_2-v_1)}{\Delta t}}$ is the acceleration a
• So the result in (7) becomes: $\mathbf\small{m\,a=F}$
• This is Newton's second law of motion

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Thus, starting with the 'work done', we reached Newton's second law. We did it in the case of linear motion. Let us see if it is possible for rotational motion also:

1. A rigid body in rotational motion, has an initial angular velocity of ω₁ rad s^-1
• The moment of inertia of the body about the axis of rotation is I
• A torque 𝝉 acts on it for a time duration of Δt seconds
    ♦ As a result, the angular velocity of the body increases to ω₂
• During the Δt seconds, the body undergoes an angular displacement of Δθ  
2. Work done by the torque 
= Torque × angular displacement 
= $\mathbf\small{\tau \times \Delta \theta}$
3. The body is rigid. So there is no internal motion of particles. All the external work will be utilized for increasing the kinetic energy of the body
• We have:
Kinetic energy of a rotating body = $\mathbf\small{\frac{1}{2}I \omega^2=0.5I\,\omega^2}$ (see Eq.7.26) 
4. Now, increase in kinetic energy = $\mathbf\small{0.5I\,\omega_2^2-0.5I\,\omega_1^2=0.5I(\omega_2^2-\omega_1^2)}$ 
• Equating this to the result in (2), we get: $\mathbf\small{0.5I(\omega_2^2-\omega_1^2)=\tau \times \Delta \theta}$
5. Dividing both sides by Δt, we get: $\mathbf\small{\frac{0.5I(\omega_2^2-\omega_1^2)}{\Delta t}=\frac{\tau \times \Delta \theta}{\Delta t}=\tau \times\frac{\Delta \theta}{\Delta t}}$
6. $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ normally gives angular velocity
• But in our present case, the angular velocity is not uniform
    ♦ This is because of the action of the torque 𝝉 
    ♦ Because of the 𝝉, the body will be rotating with an acceleration
• So $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ will give us the average angular velocity
• That means: $\mathbf\small{\frac{\Delta \theta}{\Delta t}=\frac{\omega_1+\omega_2}{2}}$
7. So the result in (5) becomes: $\mathbf\small{\frac{0.5I(\omega_2^2-\omega_1^2)}{\Delta t}=\tau\frac{(\omega_1+\omega_2)}{2}}$
$\mathbf\small{\Rightarrow \frac{0.5I(\omega_2+\omega_1)(\omega_2-\omega_1)}{\Delta t}=\tau\frac{(\omega_1+\omega_2)}{2}=0.5\tau(\omega_1+\omega_2)}$
$\mathbf\small{\Rightarrow \frac{0.5I(\omega_2-\omega_1)}{\Delta t}=0.5\tau}$
$\mathbf\small{\Rightarrow \frac{I(\omega_2-\omega_1)}{\Delta t}=\tau}$
8. But $\mathbf\small{\frac{(\omega_2-\omega_1)}{\Delta t}}$ is the angular acceleration 𝜶
• So the result in (7) becomes:
Eq.7.29: $\mathbf\small{I\,\alpha=\tau}$
9. '$\mathbf\small{I\,\alpha=\tau}$' is analogous to '$\mathbf\small{m\,a=F}$' of linear motion
• '$\mathbf\small{I\,\alpha=\tau}$' is called the Newton's second law for rotation about a fixed axis
• Thus, in the case of rotation (about a fixed axis) also, starting with the 'work done', we reached Newton's second law

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Now we will see some solved examples

Solved example 7.33
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in the fig.7.135 below. The fly wheel is mounted on a horizontal axle with frictionless bearings
(a) Compute the angular acceleration of the wheel
(b) Find the work done by the pull when 2 m of the chord is unwound
(c) Find also the kinetic energy of the wheel at this point
(d) compare answers of parts (b) and (c)

Fig.7.135

Solution:
1. Moment of inertia (I) of the fly wheel = $\mathbf\small{\frac{MR^2}{2}=\frac{(20)(0.2)^2}{2}=0.4\,\text{kg m}^2}$
(I of a circular disc about a perpendicular axis at center)    
2. Torque $\mathbf\small{\tau}$ acting on the wheel:
• The tension in the cord will pull a particle at the periphery of the wheel. So force on that particle will be 25 N
• This force will be tangential to the fly wheel
• We know this:
Perpendicular distance between a tangent from the center of circle (axis of rotation) = radius of the circle
• So we get:   
$\mathbf\small{\tau}$ = Force × perpendicular distance from center 
= Force × radius = 25 × 0.2 = 5 Nm
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting known values, we get: $\mathbf\small{5=0.4\,\alpha}$
$\mathbf\small{\Rightarrow \alpha=\frac{5}{0.4}=12.5\,\text{rad s}^{-2}}$
• This is the answer for part (a)
4. Perimeter of the fly wheel = 2𝞹R = 2𝞹(0.2) = 0.4𝞹 m
• So 0.4𝞹 m of the circumference will cover 2𝞹 radians
• Then 1 m of the circumference will cover: $\mathbf\small{\frac{2 \pi}{0.4 \pi}=5}$ radians
• So 2 m of the cord will cover: 10 radians
■ Thus, when 2 m of the cord is unwound, the fly wheel will turn through 10 rad
5. We have: Work done = 𝞽 × dθ = 5 × 10 = 50 joules
• This is the answer for part (b)
6. Change in kinetic energy = $\mathbf\small{0.5I(\omega_2^2-\omega_1^2)}$
• Given that, the fly wheel starts from rest. So $\mathbf\small{\omega_1}$ = 0    
• Thus, change in kinetic energy = $\mathbf\small{0.5I \omega_2^2}$ 
7. We have: $\mathbf\small{\omega_2^2=\omega_1^2+2 \alpha \theta}$
• Substituting the values, we get: $\mathbf\small{\omega_2^2=0^2+2(12.5)(10)=250}$
8. Substituting the values in (6), we get:
• Change in kinetic energy = $\mathbf\small{0.5(0.4)(250)}$ = 50 joules
• This is the answer for part (c)
9. Comparing (b) and (c), we find that, the answers are the same
• That means:
Work done by the external torque = Increase in kinetic energy of the body
• We obtained this equality because, no work is lost against friction
• This is the answer for part (d)

Solved example 7.34
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution:
1. Standard axis of symmetry of a cylinder is it's axis
• I of a hollow cylinder about it's axis = $\mathbf\small{MR^2}$
• I of a solid sphere about an axis passing through it's center = $\mathbf\small{\frac{2MR^2}{5}}$ 
(Significance of 'passing through center' can be seen here )
2. Given that, the torques acting are equal. Let us denote it as $\mathbf\small{\tau}$ 
• Given that the masses are same. Let us denote it as M 
• Given that the radii are same. Let us denote it as R 
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting the values for the hollow cylinder, we get: $\mathbf\small{\tau=MR^2\,\alpha_C}$
    ♦ Where $\mathbf\small{\alpha_C}$ is the angular acceleration of the cylinder
• Substituting the values for the solid sphere, we get: $\mathbf\small{\tau=\frac{2MR^2}{5}\,\alpha_S}$
    ♦ Where $\mathbf\small{\alpha_S}$ is the angular acceleration of the sphere
4. Since the torques are equal, we can equate them:
• $\mathbf\small{MR^2\,\alpha_C=\frac{2MR^2}{5}\,\alpha_S}$
$\mathbf\small{\Rightarrow \alpha_C=\frac{2}{5}\,\alpha_S}$
5. We see that, $\mathbf\small{\alpha_C}$ is only a fraction of $\mathbf\small{\alpha_S}$.
• That means $\mathbf\small{\alpha_C}$ is less than $\mathbf\small{\alpha_S}$ 
• So after a given time, the sphere will acquire a greater angular speed

Solved example 7.35
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
1. I of a hollow cylinder about it's axis = $\mathbf\small{MR^2}$
2. Torque $\mathbf\small{\tau}$ acting on the wheel:
• The tension in the rope will pull a particle at the periphery of the cylinder
• So force on that particle will be 30 N
• This force will be tangential to the cylinder
• Perpendicular distance between a tangent from the center of circle (axis of rotation) = radius of the circle
• So we get: $\mathbf\small{\tau}$ = Force × perpendicular distance from center 
= Force × radius = 30 × 0.4 = 12 Nm
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting the values, we get: $\mathbf\small{12=(3)(0.4)^2\,\alpha}$
$\mathbf\small{\Rightarrow \alpha=}$ 25 rad s^-2
• This is the answer for part (i)
4. The cylinder starts from rest
• So initial angular velocity $\mathbf\small{\omega_0=0}$
5. Let us find the angular velocity after any convenient interval of time, say 2 s
• We have: $\mathbf\small{\omega=\omega_0+\alpha \, t}$
• Substituting the values, we get: $\mathbf\small{\omega=0+(25) \, (2)}$ = 50 rad s^-1
6. So at the instant when the stop watch shows 2 seconds, the cylinder will be rotating with an angular speed of 50 rad s^-1 
• Every particle in the cylinder will be rotating with the angular speed of 50 rad s^-1 at that instant
• Any particle at the periphery will also be rotating with the angular speed of 50 rad s^-1 at that instant
7. Now, we use the relation between linear velocity and angular velocity
• We have: $\mathbf\small{v=r\;\omega}$
• We apply it to a particle at the periphery: v = 0.4 × 50 = 20 ms^-1.
• So, when the stop watch shows 2 seconds, any particle at the periphery will be moving with a linear speed of 20 ms^-1.
8. Given that, there is no slip between the cylinder and the rope
• So the rope will also be moving with a linear speed of 20 ms^-1
• We will use the relation $\mathbf\small{v=v_0+at}$
• The rope also started from rest. So $\mathbf\small{v_0=0}$
• Substituting the values, we get: 20= 0 + a × 2
⇒ a = 10 ms^-2.

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In the next section, we will see angular momentum

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