Chapter 12.8 - Efficiency of a Carnot Engine

In the previous section, we saw the basic processes in a Carnot engine. In this section, we will see the work done during each of those processes. For easy reference, the final PV diagram is shown below:

Fig.12.23

First segment from A to B
• In this segment, the gas is subjected to an isothermal expansion
• We know that, the work done by the ideal gas during the isothermal expansion will be equal to: $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$ (see Eq.12.3 at end of section 12.2)

Second segment from B to C
• In this segment, the gas is subjected to an adiabatic expansion
• We know that, the work done by the ideal gas during the adiabatic expansion will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$ (see Eq.12.6 at end of section 12.3)

Third segment from C to D
• In this segment, the gas is subjected to an isothermal compression
• We know that, the work done on the ideal gas during the isothermal compression will be equal to: $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$

Fourth segment from D to A
• In this segment, the gas is subjected to an adiabatic compression
• We know that, the work done by the ideal gas during the adiabatic compression will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$

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Now we can find the efficiency of the Carnot engine. It can be written in 9 steps:
1. Net work, W = Work done by the gas - Work done on the gas
   ♦ Work done by the gas = Work done during segments one and two
   ♦ Work done on the gas = Work done during segments three and four
• Thus we get:
W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}\right )- \left ( n R T_2\,\,ln \frac{V_C}{V_D}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1} \right)}}$
⇒ W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}- n R T_2\,\,ln \frac{V_C}{V_D} \right)}}$
2. We know that, efficiency of a heat engine is given by:
$\mathbf\small{\rm{\eta = \frac{Net \; Work}{Heat \; Absorbed}}}$ (see Eq.12.9 in section 12.5)
3. Also, we have derived another form of the above equation:
$\mathbf\small{\rm{\eta = 1-\frac{Heat \;Rejected}{Heat \; Absorbed}}}$ (see Eq.12.10 in section 12.5)
• It is easier to find efficiency using this equation
4. So our next task is to find the 'Heat absorbed' and the 'Heat rejected'
• It can be done in steps:
(i) We know that, heat absorption takes place only during the first segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(ii) We know that, heat rejection takes place only during the third segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat rejected = $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$
5. So the result in (3) becomes: $\mathbf\small{\rm{\eta = 1-\frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_B}{V_A}}}}$
⇒ $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right ) \left (\frac{ln \frac{V_C}{V_D}}{ln \frac{V_B}{V_A}}\right )}}$
6. We see that, the right side of the above equation involves temperatures and volumes. So our next task is to find the relations between them
• For that, we consider the two adiabatic processes in the cycle. It can be written in two steps:
(i) For the  adiabatic process BC, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(see Eq.12.7 of section 12.3)
(ii) Similarly, for the adiabatic process DA, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_D}{V_A} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(iii) The right sides in (i) and (ii) are the same. So we get:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{V_D}{V_A} \right )^{\gamma - 1}}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )=\left (\frac{V_D}{V_A} \right )}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_D} \right )=\left (\frac{V_B}{V_A} \right )}}$
7. We see the above result in (5)
So the result in (5) becomes:
Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
8. Two expressions for efficiency:
   ♦ The above Eq.12.14 gives us an expression for efficiency
   ♦ Step (3) also gives us an expression for efficiency
• So we can equate the two:
$\mathbf\small{\rm{1-\frac{Heat \;Rejected}{Heat \; Absorbed} = 1- \left (\frac{T_2}{T_1} \right )}}$
• Thus we get Eq.12.15:
$\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{T_2}{T_1} \right )}}$

⇒ $\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{Temperature \; of \; cold \; reservoir}{Temperature \; of \; hot \; reservoir} \right )}}$
9. Let us use try the results in (1) and (2) to find efficiency. it can be written in 3 steps:
(i) Using the result in (6), the result in (1) becomes:
W = $\mathbf\small{\rm{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}}$
(ii) Using the result in 4(i), we have:
Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(iii) Applying the above two results into (2), we get:
$\mathbf\small{\rm{\eta = \frac{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}{n R T_1\,\,ln \frac{V_B}{V_A}}=\frac{T_1 - T_2}{T_1}}}$
⇒ $\mathbf\small{\rm{\eta =1-\left (\frac{T_2}{T_1} \right )}}$
This is the same result as in Eq.12.14

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Let us see a solved example
Solved example 12.10
Three heat engines operate between the following temperature differences
    ♦ Engine A: T₁ = 1000 K, T₂ = 700 K
    ♦ Engine A: T₁ = 800 K, T₂ = 500 K
    ♦ Engine A: T₁ = 600 K, T₂ = 300 K
• Which is the most efficient engine?
• Which is the least efficient engine?
Solution:
1. Efficiency of a heat engine is given by Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
• So we get:
    ♦ Efficiency of Engine A = $\mathbf\small{\rm{\eta = 1- \left (\frac{700}{1000} \right )}}$ = 0.3
    ♦ Efficiency of Engine B = $\mathbf\small{\rm{\eta = 1- \left (\frac{500}{800} \right )}}$ = 0.375
    ♦ Efficiency of Engine C = $\mathbf\small{\rm{\eta = 1- \left (\frac{300}{600} \right )}}$ = 0.5
2. We can write:
    ♦ Engine C is the most efficient
    ♦ Engine A is the least efficient
3. Dependence of efficiency on temperature:
    ♦ We see that, the temperature difference is the same for all three engines
    ♦ In such a situation, the engine with the smallest T₂ has the most efficiency 

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In the next section, we will see Carnot refrigerator


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Chapter 8.12 - Gravitational Potential Energy

In the previous section, we completed a discussion on the basics of gravitational force, gravity and acceleration due to gravity. We saw some solved examples and graphs also. In this section, we will see gravitational potential energy

• We know that, potential energy is that energy which is stored in a body at it’s given position
• If the body moves to a new position, it’s potential energy will change
• The ‘change in potential energy’ is equal to the amount of work done by the ‘force which causes the change in position’ on the body (Details here)
• We saw the definition of conservative forces also
    ♦ If the work done by a force is independent of the path, it is called a conservative force
    ♦ Gravitational force is a conservative force

We will now write a detailed analysis of the potential energy. In this analysis, we will include Newton's laws of gravitation also. We will write it in steps:
1. Consider a point mass m_A situated on the exterior of the earth
• Let it be at a distance of r from the center of the earth
2. We know that, the force of attraction experienced by that body will be given by: $\mathbf\small{|\vec{F}_G|=\frac{G\,M_E\,m_A}{r^2}}$
• Since r is in the denominator, it is clear that, the force will decrease with increase in the distance r
    ♦ We will see this same fact if we draw the graph of $\mathbf\small{|\vec{F}_G|}$
    ♦ This graph will be similar to the graph in fig.8.32 that we saw in the previous section 
• A demonstration of the 'decreasing force' can be given in 3 steps:
(i) In fig.8.37(a) below, m_A is at a distance of r₁ from the center O of the earth
• In fig.b, the same m_A is at a distance of r₂ from O

Fig.8.37

(ii) In fig.a, the force experienced by m_A will be $\mathbf\small{\frac{G\,M_E\,m_A}{r_1^2}}$
• In fig.b, the force experienced by m_A will be $\mathbf\small{\frac{G\,M_E\,m_A}{r_2^2}}$  
(iii) We see that, r₂ is greater than r₁
• Since r₁ and r₂ are in the denominators, the force in fig.b will be less than the force in fig.a
■ So we can write:
If the same body is taken further away from the earth, it will be experiencing a lesser force of attraction
3. Is there any possibility for the forces in both figs.8.36 (a) and (b) to be the same?
Let us check:
• We know that r can be split up as (R_E+h)
• So we get two more figs.: (c) and (d)
    ♦ In fig.c, m_A is at a height h₁ above the surface of the earth
    ♦ In fig.d, m_A is at a height h₂ above the surface of the earth
• Note that:
    ♦ Fig.a is same as fig.c. The only difference is that, the distance r₁ is split into R_E and h₁
    ♦ Fig.b is same as fig.d. The only difference is that, the distance r₂ is split into R_E and h₂
4. We know that, the force in fig.d will be less than that in fig.c
• Force acting on m_A in fig.c = $\mathbf\small{\frac{G\,M_E\,m_A}{(R_E+h_1)^2}}$
• Force acting on m_A in fig.d = $\mathbf\small{\frac{G\,M_E\,m_A}{(R_E+h_2)^2}}$
5. Suppose that, both h₁ and h₂ are very small when compared to R_E. We have seen that, in such cases, those heights can be ignored
We can write:
• Force acting on m_A in fig.c = $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^2}}$
• Force acting on m_A in fig.d = $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^2}}$
■ Both forces are the same
6. So we can write:
• If the heights attained by a body are very close to the surface of the earth, the same force will be acting on it
• This force is given by: $\mathbf\small{|\vec{F}_G|=\frac{G\,M_E\,m_A}{R_E^2}}$ 
7. In the right side of the above expression, all quantities (except m_A) are constants
• So $\mathbf\small{\frac{G\,M_E}{R_E^2}}$ is a constant
• We can calculate the value of this constant as:
$\mathbf\small{\frac{G\,M_E}{R_E^2}=\frac{6.6742 \times 10^{-11}\rm{(m^3\,kg^{-1}\,s^{-2})}\times5.972 \times 10^{24}\rm{(kg)}}{(6.3781 \times 10^{6})^2\rm{(m^2)}}=9.7979\;\rm{m\,s^{-2}}}$
8. So the result in (6) becomes:
• If the heights attained by a body are very close to the surface of the earth, the same force will be acting on it
• This force is given by: $\mathbf\small{|\vec{F}_G|=9.7979\;\rm{(m\,s^{-2}})}\times m_A\,\rm{(kg)}$
9. We see that, a constant force is acting at smaller heights
• If we divide this force by mass, we will get the acceleration experienced by m_A 
• We can write: $\mathbf\small{|\vec{a}|=\frac{|\vec{F}_G|}{m_A}=9.7979\;\rm{(m\,s^{-2}})}$
• This $\mathbf\small{|\vec{a}|}$ is the same 'acceleration due to gravity' that we denote as $\mathbf\small{|\vec{g}|}$
• So we can write: $\mathbf\small{|\vec{g}|=9.7979\;\rm{(m\,s^{-2})}}$
■ We see that:
When the mass m_A is at smaller heights, it experiences the same acceleration of 9.7979 ms^-2 at all those heights
10. The reader must be able to appreciate the following fact:
■ If the heights attained are not very small (when compared to R_E), we will not get the same acceleration 9.7979 ms^-2 at those heights
• If the reader is not able to appreciate it, he/she must examine the steps (1) to (9) again
• Also recall that, when the heights attained are not very small (when compared to R_E), the graph of $\mathbf\small{|\vec{g}_h|}$ is a horizontal line. 
    ♦ A horizontal line indicates a constant value
    ♦ We saw this graph in the fig.8.33 of the previous section  
11. Now we can write every thing in terms of acceleration:
• At smaller heights:
    ♦ A constant acceleration $\mathbf\small{|\vec{g}|}$ is acting on the body
    ♦ Hence, a constant force $\mathbf\small{m_A|\vec{g}|}$ is acting on the body
12. Since the force is constant, we can easily calculate the work done
Work done in lifting the body from h₁ to h₂ = Force × displacement = $\mathbf\small{m|\vec{g}|(h_2-h_1)}$
13. Now is a good time to write a note about datum
• We can consider any horizontal line as the 'datum line'
• Let us consider the surface of the earth as the datum line. This is shown in the fig.8.38 below:

Fig.8.38

■ We cannot say this:
At the surface of the earth, m_A has zero potential energy
• Why we cannot say that?
The answer can be written in 4 steps:
(i) Consider m_A to be resting on the surface of the earth
(ii) Let a pit of depth d be made on the surface
(iii) If m_A falls freely into that pit, some energy will be released
(iv) So m_A has some potential energy even when it is at the datum
14. So what is the 'amount of potential energy that m_A possess when it is at the datum'?
• The 'amount of potential energy that m_A possess when it is at the datum' does not really matter
• We can write the reason in 5 steps:
(i) Let the 'amount of potential energy that m_A possesses when it is at the datum' be W₀
(ii) When m_A is lifted from the datum to P₁, a work of mgh₁ will be done on it
• Then we can write:
Amount of potential energy that m_A possesses when it is at P₁ = mgh₁ +W₀
(iii) When m_A is lifted from the datum to P₂, a work of mgh₂ will be done on it
• Then we can write:
Amount of potential energy that m_A possesses when it is at P₂ = mgh₂ + W₀
(iv) Difference in the potential energies
= (Potential energy at P₂ - Potential energy at P₁)
= [(mgh₂ + W0) - (mgh₁ + W₀)]
= [mgh₂ - mgh₁] = mg(h2-h₁)
(v) This is the same result obtained in (12)
• W₀ gets cancelled
• The difference in potential energies at two points P₁ and P₂ is equal to the work done in moving the body from P₁ to P₂ 
• So it is the initial and final energies that matter. It is not necessary to know the potential energy at the datum
14. Let us make a graphical representation:
• We know that, the work done is equal to the area enclosed in the force-displacement graph (Details here) The fig.8.39 below shows such a graph:

Fig.8.39

The following points may be noted:
(i) In our present case, the force is a constant. So the graph representing force will be a horizontal line
• This is indicated by the yellow horizontal line
(ii) The distance between this yellow line and the x-axis will be the magnitude of the force
• Also, this distance is the width of the blue rectangle
(iii) The green vertical lines indicate the two heights h₁ and h₂
    ♦ The first green line is at a distance of h₁ from the origin
    ♦ The second green line is at a distance of h₂ from the origin
• So the distance between the green lines is (h₂-h₁)
• Also this distance is the length of the blue rectangle
(iv) So area of the blue rectangle = $\mathbf\small{m_A|\vec{g}|(h_2-h_1)}$
• This area is the work done by gravity in moving the mass from P₁ to P₂

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• If the force is not constant, we will not get a horizontal yellow line. So we will not get a rectangle
• In such cases, the area cannot be calculated easily
• We will now write a detailed analysis on such cases:

1. Let h₁ and h₂ in the fig.8.38 above, be very large when compared to R_E
This can be represented as in fig.8.40(a) below:

Fug.8.40

2. We have: r₁ = (R_E+h₁) and r₂ = (R_E + h₂)
• Since h₁ and h₂ are very large, we cannot ignore them. That is., we cannot write: r₁ = r₂ = R_E
3. Also, since h₁ and h₂ are very large when compared to R_E, there is no need for splitting the distances as (R_E+h₁) and (R_E+h₂)
• We can simply write:
    ♦ P₁ is at a distance of r₁ from the center of the earth
    ♦ P₂ is at a distance of r₂ from the center of the earth
4. We want the work done against gravity when m_A is moved from P₁ to P₂
• As usual, we need to use the formula: Work = Force × displacement
5. But the force is not a constant. It changes as we move from P₁ to P₂
• So what 'value of force' will we use?
6. Let us try the graphical method. Fig.8.41 below shows the force-displacement graph:

Fig.8.41

• The graph (the yellow curve) representing force is not a horizontal line
• It is not even a line. It is a curve
• How do we plot this curve on the graph paper?
The answer can be written in 3 steps:
(i) We have: $\mathbf\small{|\vec{F}|=\frac{G\,M_E\,m_A}{r^2}}$
(ii) This is same as: $\mathbf\small{y=K\left(\frac{1}{x^2}\right)}$
Where:
    ♦ $\mathbf\small{y=|\vec{F}|}$
    ♦ $\mathbf\small{K=G\,M_E\,m_A}$
    ♦ $\mathbf\small{x=r}$
(iii) That means, the equation of the yellow curve is: $\mathbf\small{y=K\left(\frac{1}{x^2}\right)}$
• Putting, various values of x, we will get the corresponding values of y
• Thus the curve can be plotted
(However, we do not need the actual plotting at present. We just need to confirm that, the 'graph of the force' is a curve)
7. Consider the blue region. It is bounded by four items:
(i) The x-axis
(ii) The vertical green line at r₁
(iii) The vertical green line at r₂
(iv) The yellow curve
8. Using calculus, we can find the area of such complicated shapes
■ For this particular case, calculus gives us this result:
Area of the blue region = $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$
9. But area of the blue region is equal to the work done in moving m_A from P1 to P2
So we can write:
Eq.8.14:
Work done in moving m_A from P₁ to P₂ = $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$
10. How do we verify that the right side of Eq.8.14, is indeed a 'work'?
It can be done in 3 steps:
(i) $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$ can be written as: $\mathbf\small{G\,M_E\,m_A\left(\frac{r_1-r_2}{r_1r_2}\right)}$
(ii) Inputting the units, we get: $\mathbf\small{(N\,m^2\,kg^{-2})(kg)(kg)\left(\frac{(m)}{(m)(m)}\right)=N\,m}$
(iii) $\mathbf\small{N\,m}$ corresponds to 'force × displacement', which is 'work done'  
11. Using Eq.8.14, we can calculate the work done for moving the body between any two points P₁ and P₂
• Let us consider two important points (see fig.8.40.b above):
    ♦ First point P₁ is at a distance of r from the center of the earth
    ♦ Second point P₂ is at a distance of infinity (∞) from the center of the earth
• Substituting these values in Eq.8.14, we get:
Work done in moving m_A from ∞ to P₁
= $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{\infty }-\frac{1}{r}\right)}$ = $\mathbf\small{G\,M_E\,m_A\left(0-\frac{1}{r}\right)}$ = $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$
(∵Any quantity divided by ∞ gives zero)
12. ∞ is such a very large distance. At  that distance, there will not be any action of earth's gravity
• Gravitational potential energy comes into effect only if there is an action of gravity
• But at ∞ distance from earth, there is no action of earth's gravity
■ So the gravitational potential energy at ∞ is zero
13. We moved m_A from a 'point of zero potential' to P₁
• For that, we did a work of $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$ joules
• So the potential energy stored in the body when it is at P₁ is $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$ joules
14. This is the case for a body A of mass m_A. We can write the general case:
Eq.8.15: $\mathbf\small{U_r=-\frac{G\,M_E\,m}{r}}$
• Where $\mathbf\small{U_r}$ is the gravitational potential energy stored in a body of mass m, when it is at a distance of r from the center of the earth
15. In Eq.8.15, the quantity on the right side is given a negative sign
• The reason can be written in steps:
(i) In the expression, since r is in the denominator, it is clear that when the distance r from the center of the earth increases, the potential energy decreases
(ii) Consider the 3 conditions:
• When r approaches ∞, the potential energy must approach zero
• When r becomes equal to ∞, the potential energy must become equal to zero
• Also, when r increases, the potential energy must increase. This is because, when height increases, the potential energy naturally increases
(iii) So we must satisfy 3 conditions mentioned in (ii)
■ This is possible only if a negative sign is given to the quantity on the right side
(iv) This case is similar to a number line shown in fig.8.42 below:

Fig.8.42

• In this number line, the values of U_r always lies on the left of zero
• As we move from far left towards the right, the absolute value of U_r decreases. But the actual value of U_r is increases
15. We have to note an important point here. It can be written in steps:
(i) The body moves from infinity to P₁
(ii) We get the impression that:
• The body falls from infinity to P₁, and so, the work done for the motion will be provided by gravity
(iii) But this is not the case
• If the body is allowed to fall freely, some of the energy will be converted to kinetic energy
• We do not want such a 'conversion to kinetic energy'. Because, we want to find the potential energy only
(iv) So, we must ensure that, the body moves (from ∞ to P₁) with out acceleration
• For that, we must do work against gravity. This work is stored as potential energy

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• So using Eq.8.15, we can calculate the potential energy of a mass m at any point (situated at a distance r) from the center of the earth
• How is all this related to the potential energy mgh that we learned in our previous classes?
• We will see it in the next section

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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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