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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Chapter 7.29 - Work done by a Torque

In the previous section, we saw how the 3D problem of 'rotation about a fixed axis' can be represented using a 2D plane. Based on that, in this section we will see work done by a torque

1. Consider a rigid body rotating about a fixed axis
• Let the z-axis be the fixed axis (see fig.7.132.a below)

Fig.7.132

• A horizontal plane is also shown in the fig.a
    ♦ It is shown in red color
    ♦ It is parallel to the xy plane
2. Cut the body by that horizontal plane. This is shown in fig.b
    ♦ Now the body is separated into an upper part and a lower part
• Remove the upper part 
• Remove the plane also. This is shown in fig.7.133(a) below:

Fig.7.133

• A new x' axis, which is parallel to the original x-axis of the reference frame can be drawn
    ♦ This x' axis lies on the 'top surface of lower part' 
• A new y' axis, which is parallel to the original y-axis of the reference frame can also be drawn
    ♦ This y' axis also lies on the 'top surface of lower part' 
• They are shown in fig.7.133(b) above
3. Look at the lower part from above
    ♦ We will get a 2D view of the 'top surface of lower part'. It is shown in fig.7.134(a) below:

Fig.7.134

• This 2d view of the 'top surface of lower part' is called the cross section of the body   
• The 'top surface of lower part' coincides with the new x’y’ plane
• All particles on that top surface will be moving in circular paths
• We can assume that the x’y’ plane coincides with the plane of the computer screen
• The z-axis appears as a small blue circle
    ♦ The z-axis is perpendicular to the computer screen 
4. The body consists of a large number of particles: P₁, P₂, P₃, . . . ,P_n
• Let us isolate one particle on the x'y' plane. We will call it P₁
    ♦ It is shown as a small yellow circle. It's path is shown in pink color
• In the fig.7.134(a), the full path of P₁ is not shown. Rather, a portion of the path is shown as the pink arc
5. Let a force $\mathbf\small{\vec{F}_1}$ (lying on the x'y' plane) act on the particle P₁ when it is at A
• This $\mathbf\small{\vec{F}_1}$ lies on the plane x’y’ so we must take it into account
6. As a result of the force, the particle reaches B
• The angle turned by P is θ
• The distance traveled by P is the arc length AB
• We want this arc length
• We have: $\mathbf\small{\text{Angle}=\frac{\text{Arc}}{\text{Radius}}}$
$\mathbf\small{\Rightarrow \text{Arc}=\text{Radius}\times \text{Angle}}$
7. The position vector of P₁ (with respect to origin O which is not shown in fig.7.134) when it is at A is $\mathbf\small{\vec{r}_1}$
• It’s perpendicular component $\mathbf\small{\vec{r}_{(\bot)1}}$ will lie on the x’y’ plane
• So radius of the arc = $\mathbf\small{|\vec{r}_{(\bot)1}|}$ 
• Thus we get: Arc length AB = $\mathbf\small{|\vec{r}_{(\bot)1}|\;\theta}$
8. Now consider the same situation when θ is very small. We will indicate that 'small value of  θ' as dθ. This is shown in fig.b
• If dθ is very small, the arc length AB will be equal to the straight line distance between A and B
• Using the result in (7), we get:
Straight line distance AB = Arc length AB = $\mathbf\small{|\vec{r}_{(\bot)1}|\;d\theta}$
9. Let us denote the straight line distance between A and B as $\mathbf\small{ds_1}$
• Then the result in (8) becomes:
$\mathbf\small{ds_1}$ = Arc length AB = $\mathbf\small{|\vec{r}_{(\bot)1}|\;d\theta}$
10. But straight line distance AB is the displacement from A to B
• Displacement is a vector. We must denote it as $\mathbf\small{\vec{ds}_1}$
• So magnitude of the displacement will be denoted as: $\mathbf\small{|\vec{ds}_1|}$ 
• Thus the result in (9) becomes:
$\mathbf\small{|\vec{ds}_1|=|\vec{r}_{(\bot)1}|\;d\theta}$
11. We have two items:
(i) Force acting on the particle P₁. It is $\mathbf\small{\vec{F}_1}$ 
(ii) Displacement suffered by the particle P₁. It is $\mathbf\small{\vec{ds}_1}$ 
• With those two items, we can calculate the work done by the $\mathbf\small{\vec{F}_1}$ on the particle
• We will denote this work as $\mathbf\small{dW_1}$
• 'Work done' is obtained as a dot product. The result is a scalar
• So we have: $\mathbf\small{dW_1=\vec{F}_1.\vec{ds}_1}$
12. To evaluate this dot product, we need the angle between $\mathbf\small{\vec{F}_1}$ and $\mathbf\small{\vec{ds}_1}$
• In fig.b, we know that AB can be considered as a straight line
    ♦ We know the reason: dθ is very small
• The direction of AB is in fact the direction of $\mathbf\small{\vec{ds}_1}$
• AB is extended upwards. This is the cyan line    
• Angle between the cyan line and $\mathbf\small{\vec{F}_1}$ is our required angle. It is denoted as Φ₁
■So the result in (11) becomes: $\mathbf\small{dW_1=|\vec{F}_1|\times |\vec{ds}_1|\times \cos \phi_1}$
13. In fig.b, we know that B is very close to A. 
    ♦ We know the reason: dθ is very small  
• So, the straight line AB is in fact, the tangent at A
• The cyan line is the extension of AB. So cyan line is the tangent at A
• OA is the radius drawn through A
• OA is extended upto A'. So OA' is an extension of radius
14. So we have two items:
(i) The tangent at A   
(ii) The radius through A
• These two will be perpendicular to each other. That means, the angle between OA' and the cyan line is 90^o
• So we get: $\mathbf\small{(\phi_1+\alpha_1)=90^o}$
$\mathbf\small{\Rightarrow \phi_1=(90-\alpha_1)}$
$\mathbf\small{\Rightarrow \cos \phi_1=\cos (90-\alpha_1)=\sin \alpha_1}$
    ♦ Where $\mathbf\small{\alpha_1}$ is the angle between $\mathbf\small{\vec{F}_1}$ and OA'
• So the result in (12) becomes: $\mathbf\small{dW_1=|\vec{F}_1|\times |\vec{ds}_1|\times \sin \alpha_1}$
15. Substituting for $\mathbf\small{|\vec{ds}_1|}$ from (10), we get:
$\mathbf\small{dW_1=|\vec{F}_1|\times \left(|\vec{r}_{(\bot)1}|\;d\theta\right)\times \sin \alpha_1}$
• Rearranging this, we get:
$\mathbf\small{dW_1=\left[|\vec{F}_1|\times |\vec{r}_{(\bot)1}|\times \sin \alpha_1 \right]\;d\theta}$
16. Inside the square brackets, there are 3 items:
(i) Magnitude of $\mathbf\small{\vec{F}_1}$
(ii) Magnitude of $\mathbf\small{\vec{r}_{(\bot)1}}$
(iii) sine of the angle between $\mathbf\small{\vec{F}_1}$ and $\mathbf\small{\vec{r}_{(\bot)1}}$
• When those 3 items are multiplied, obviously, we get the magnitude of the cross product: ($\mathbf\small{\vec{F}_1 \times\vec{r}_{(\bot)1}}$)
• That means, what we have inside the square brackets is: $\mathbf\small{|(\vec{F}_1 \times\vec{r}_{(\bot)1})|}$
So the result in (15) becomes: $\mathbf\small{dW_1=\left[|(\vec{F}_1 \times\vec{r}_{(\bot)1})|\right]\;d\theta}$
17. But $\mathbf\small{(\vec{F}_1 \times\vec{r}_{(\bot)1})}$ is the torque created by $\mathbf\small{\vec{F}_1}$ about the axis
• We will denote this torque as $\mathbf\small{\vec{\tau}_1}$
• Thus we can write: $\mathbf\small{(\vec{F}_1 \times\vec{r}_{(\bot)1})=\vec{\tau}_1}$
$\mathbf\small{\Rightarrow |(\vec{F}_1 \times\vec{r}_{(\bot)1})|=|\vec{\tau}_1|}$
• So the result in (16) becomes: $\mathbf\small{dW_1=\left[|\vec{\tau}_1|\right]\;d\theta}$
■ On the left side of the above equation, we have:
• work done on the particle P₁
    ♦ Work done is a scalar
■ On the right side we have:
(i) Magnitude of the torque
    ♦ It is a scalar
(ii) Angle turned (with respect to the axis) by the particle
    ♦ It is also a scalar
• Product of two scalars is a scalar. So we have scalars on either side of the equation
18. So we can write:
• Work done on a particle at any instant is equal to the product of two items:
(i) The torque experienced by the particle at that instant
(ii) The angle through which the particle turned at that instant

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19. We have considered only one particle (P₁) in the body

• The force acting on that particle is $\mathbf\small{\vec{F}_1}$
• The radial distance of that particle from the axis is $\mathbf\small{\vec{r}_{(\bot)1}}$
• The magnitude of the torque experienced by that particle is $\mathbf\small{|\vec{\tau}_1|}$ 
• The angle through which that particle turned is $\mathbf\small{d\theta}$ 
• The work done ($\mathbf\small{dW_1}$) on that particle is given by: $\mathbf\small{dW_1=|\vec{\tau}_1|\;d\theta}$
20 There may be lots of other forces acting on the body:
    ♦ $\mathbf\small{\vec{F}_2}$ at P₂
    ♦ $\mathbf\small{\vec{F}_3}$ at P₃
    ♦ $\mathbf\small{\vec{F}_4}$ at P₄
    ♦ so on . . .
• Each of those particles will be having it's own radial distance from the axis:
    ♦ P₂ is at a radial distance of $\mathbf\small{|\vec{r}_{(\bot)2}|}$ 
    ♦ P₃ is at a radial distance of $\mathbf\small{|\vec{r}_{(\bot)3}|}$
    ♦ P₄ is at a radial distance of $\mathbf\small{|\vec{r}_{(\bot)4}|}$
    ♦ so on . . .
• Those particles will experience torques:
    ♦ P₂ experiences $\mathbf\small{\vec{\tau}_2}$ 
    ♦ P₃ experiences $\mathbf\small{\vec{\tau}_3}$
    ♦ P₄ experiences $\mathbf\small{\vec{\tau}_4}$
    ♦ so on . . .
• We can calculate work done on each particle:
    ♦ Work done by $\mathbf\small{\vec{F}_2}$ on P₂ is given by: $\mathbf\small{dW_2=|\vec{\tau}_2|\;d\theta}$
    ♦ Work done by $\mathbf\small{\vec{F}_3}$ on P3 is given by: $\mathbf\small{dW_3=|\vec{\tau}_3|\;d\theta}$
    ♦ Work done by $\mathbf\small{\vec{F}_4}$ on P₄ is given by: $\mathbf\small{dW_4=|\vec{\tau}_4|\;d\theta}$
    ♦ so on . . .
21. So total work done by all the forces 
= $\mathbf\small{dW_1+dW_2+dW_3+\;.\;.\;.\;+dW_n}$
= $\mathbf\small{|\vec{\tau}_1|\;d\theta+|\vec{\tau}_2|\;d\theta+|\vec{\tau}_3|\;d\theta\;+\;.\;.\;.\;+\;|\vec{\tau}_n|\;d\theta}$
= $\mathbf\small{\left(|\vec{\tau}_1|+|\vec{\tau}_2|+|\vec{\tau}_3|\;+\;.\;.\;.\;+\;|\vec{\tau}_n|\right)\;d\theta}$
Two points may be noted here:
(i) dθ is same for all particles because, all particles in the body will turn through the same angle. So it can be taken outside the brackets
(ii) Inside the brackets, we are simply adding the 'magnitudes of the torques' algebraically. This is possible because, all the torques are aligned along the axis. We saw this in the previous section
22. Let us denote the total work done by all the forces as $\mathbf\small{dW}$
• Also let us denote the algebraic sum of all the 'torque magnitudes' as $\mathbf\small{|\vec{\tau}|}$
• Then the result in (21) becomes: $\mathbf\small{dW=|\vec{\tau}|\;d\theta}$
    ♦ Where $\mathbf\small{|\vec{\tau}|}$ is the net external torque acting on the body
23. So we can write:
(i) Work done on a body in rotational motion about a fixed axis is given by:
Eq.7.27: $\mathbf\small{dW=|\vec{\tau}|\;d\theta}$
    ♦ In the right side, we have: Torque times angular displacement  
(ii) In the previous chapter we saw:
Work done on a body in translational motion = $\mathbf\small{dW=|\vec{F}|\;ds}$
    ♦ In the right side, we have: Force times linear displacement  
• The two expressions are similar

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Once we obtain the work done, we can discuss about power. We will see it in the next section

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Chapter 7.27 - Kinematics of Rotational motion about a fixed axis

In the previous section, we saw the theorem of parallel axes. In this section we will see kinematics of rotational motion about a fixed axis

• In the fig.7.126 below, a rigid body rotates about a fixed axis
• This axis is taken as the z-axis of the reference frame

Fig.7.126

• The body is given some transparency so that, particles inside it can be seen
• Three particles in the body are highlighted. They are shown as red, yellow and green spheres
    ♦ The yellow sphere is closer to the axis than the red sphere
    ♦ The green sphere is situated exactly on the axis
• We have seen this situation in an earlier section in fig.7.10
• There we noted the characteristics of rotational motion:
■ If a rigid body is in rotation about an axis, then:
• Every particle of the body, which lies on the axis will be stationary
• Each of the other particles will be rotating in it's own circular path
    ♦ The center of that circular path lies on the axis
    ♦ Radius of that circular path = Distance of the particle from the axis
• That circular path lies on a plane
    ♦ This plane is perpendicular to the axis
■ The following two points are also important for our present discussion:
1. At any instant:
• Linear velocity of red sphere > Linear velocity of yellow sphere
• That is., $\mathbf\small{\vec{v}_{Red}>\vec{v}_{Yellow}}$
2. At any instant:
• Angular velocity of red sphere
= Angular velocity of yellow sphere
= Angular velocity of the whole body
• That is., $\mathbf\small{\vec{\omega}_{Red}=\vec{\omega}_{Yellow}=\vec{\omega}_{(Whole\;Body)}}$

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Based on the above discussion, we can write the steps to find angular velocity:
1. The 'path of the red sphere', as well as the 'plane of the red sphere' are shown in fig.7.127 below:

Fig.7.127

• The red circle is the path

• The red plane is the plane
2. When the initial reading (t₁) in the stopwatch is zero, the particle is at A
• We need to specify the location A in terms of angles
• For that, we must have a reference direction
• The reference direction is denoted as x'
    ♦ x' is parallel to the x-axis
    ♦ x' lies in the red plane
3. Now we can specify the location A:
• It is at an angular distance of θ₀ from x'
■ So we can write:
• When the initial reading (t₁) in the stopwatch is zero, the angular displacement of the particle is θ₀
• In other words:
The initial angular displacement of the particle is θ₀
4. When the final reading (t₂) in the stopwatch is t, the particle is at B
• B is at an angular distance of θ from x’
■ So we can write:
• When the final reading (t₂) in the stopwatch is t, the angular displacement of the Particle is θ
• In other words:
The final angular displacement of the Particle is θ 
5. Now the change in angular displacement = Δθ = (θ - θ₀)
• This change occurred in a time interval of Δt = (t₂-t₁) = (t – 0) = t 
• So average angular velocity = $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ 
■ We call it ‘average angular velocity’ because, the angular velocity with which the particle travels from A to B may not be uniform
• This situation occurs especially when there is angular acceleration
6. If we want instantaneous angular velocity, the time interval (t₂-t₁) must be very small
• In such a situation, we use calculus 
• The instantaneous angular velocity is given by $\mathbf\small{\frac{d \theta}{d t}}$
7. Note that, the angular velocity (instantaneous or average) possessed by the particle is the 'same angular velocity possessed by the whole body'
8. We have seen that, the angular velocity (denoted as $\mathbf\small{\vec{\omega}}$ ) is a vector (details here)
• In our present case, we deal with rotation about a fixed axis
• That means the direction of $\mathbf\small{\vec{\omega}}$ is either upwards or downwards along that fixed axis
• 'upwards or downwards' can be distinguished by using either positive or negative sign 
• So for rotation about a fixed axis, we do not need to treat $\mathbf\small{\omega}$ as a vector

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■ The rotational motion about a fixed axis is analogous to linear motion

• The role played by linear velocity $\mathbf\small{v}$ is played by angular velocity $\mathbf\small{\omega}$
• The role played by linear acceleration $\mathbf\small{a}$ is played by angular acceleration $\mathbf\small{\alpha}$
• The role played by linear  displacement $\mathbf\small{x}$ is played by angular displacement $\mathbf\small{\theta}$
■ We have 3 kinematic equations in linear motion:
$\mathbf\small{v=v_0+at}$
$\mathbf\small{x=x_0+v_0 t+\frac{1}{2}at^2}$
$\mathbf\small{v^2=v_0^2+2ax}$
■ Correspondingly, we have 3 kinematic equations in rotational motion with uniform acceleration:
$\mathbf\small{\omega=\omega_0+\alpha t}$
$\mathbf\small{\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2}$
$\mathbf\small{\omega^2=\omega_0^2+2\alpha (\theta - \theta_0)}$

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Now we will see a solved example:

Solved example 7.32
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time?
Solution:
1. Initial angular velocity = 1200 rpm = 1200 revolutions in one minute
(This is the angular velocity at the instant when the stop watch is started)
• 1 revolution is 2π radians
• So 1200 revolutions = (1200 × 2π) radians = 2400π radians
• So the motor wheel travels an angular distance of 2400π radians in one minute
• That means, the motor wheel travels an angular distance of $\mathbf\small{\frac{2400 \pi}{60}=40 \pi}$ radians in one second
• So initial angular velocity ω₀= 40π radians/sec 
2.In the same way, final angular velocity ω = 3120 rpm = 104π radians/sec  
3. Given that, the time during which this change in angular velocity occurs = 16 s
4. We will use the equation: $\mathbf\small{\omega=\omega_0+\alpha t}$
• Substituting the values, we get: $\mathbf\small{104 \pi=40 \pi+\alpha \times 16}$
⇒ α = 4π radians/sec².
• This is the answer for part (i)
5. We want the angular distance traveled by the wheel during those 16 seconds
• We will use the equation: $\mathbf\small{\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2}$
• We do not want the 'angular distance traveled' before the stopwatch is started. So θ₀ = 0
• Substituting the values, we get: $\mathbf\small{\theta=0+40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2=1152 \pi}$
6. So we get:
• The angular distance traveled during those 16 seconds = 1152π radians
• 1 revolution is 2π radians
• So number of revolutions = $\mathbf\small{\frac{1152 \pi}{2\pi}}$ = 576 revolutions

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In the next section, we will see dynamics of rotational motion about a fixed axis

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Chapter 7.15 - Relation between Linear and Angular velocities

In the previous section, we completed a discussion on cross products. In this section we will see angular velocity

We have seen the characteristics of rotational motion in a previous section of this chapter (Details here)
There we saw this:

■ If a rigid body is in rotation about a fixed axis, then 3 facts can be written:

1. Every particle of the body, which lies on the axis will be stationary
2. Each of the other particles will be rotating in it's own circular path
    ♦ The center of that circular path lies on the axis
    ♦ Radius of that circular path = Perpendicular distance of the particle from the axis
3. That circular path lies on a plane
    ♦ This plane is perpendicular to the axis

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Now we will write the steps to find angular velocity:
1. Consider a rigid body in rotation about a fixed axis. Let us isolate a 'single particle' in that rigid body 

• That 'single particle' is shown as a small red sphere in fig.7.76 below:

Fig.7.76

• The direction of rotation is indicated by the yellow curved arrow 
2. At the instant when the reading in the stop watch is 't₁', the particle is at A
    ♦ The linear velocity of the particle at that instant is $\mathbf\small{\vec{v}_{(t1)}}$
    ♦ It is tangential to the circular path
    ♦ It is shown in magenta color   
• At the instant when the reading in the stop watch is 't₂', the particle is at B
    ♦ The linear velocity of the particle at that instant is $\mathbf\small{\vec{v}_{(t2)}}$
    ♦ It is tangential to the circular path
    ♦ It is shown in magenta color   
3. The time elapsed between the two instants = Δt = (t₂-t₁)
• During this time interval Δt, the particle turns through ∠AOB
• We will denote this ∠AOB as Δθ.
4. We know that $\mathbf\small{\text{Linear velocity}=\frac{\text{Displacement}}{time}}$ 
• In a similar way, $\mathbf\small{\text{Angular velocity}=\frac{\text{Angular displacement}}{time}}$  
• So we can write:
The angular velocity with which the particle travels from A to B = $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$
5. But the rotation may not be uniform. That is., angular velocity may be different at different instances
• In such a situation, the the result obtained in (4) will be the 'average angular velocity'
6. If we make the time interval Δt smaller and smaller, it will tend to become zero
• If Δt is very close to zero, we can call it an instant
• Then the result obtained in (4) will be the 'instantaneous angular velocity' 
(Note that, Δt must not become exact zero. Then we do not have an instant. Besides, division by zero is undefined)
• The evaluation of 'instantaneous angular velocity' when Δt is very small, can be easily done using calculus
The formula is: $\mathbf\small{\text{Instantaneous angular velocity}=\frac{d\theta}{dt}}$
We will see more details when we learn calculus
7. We denote the 'instantaneous angular velocity' by ω (the Greek letter omega) 
• When we learned about circular motion, we saw that, the relation between this ω and the linear velocity 'v' is given by: v = rω
    ♦ Where r is the radius of the circle
8. Using this relation, we can find the instantaneous linear velocity of any particle in the rigid body
• For example, let there be n particles in the rigid body
• We can represent them as: 1^st particle, 2^nd particle, 3^rd particle, . . . , n^th particle
• For generality, consider the i^th particle. 
• We want it's linear velocity at the instant when the reading in the stop watch is 't'
• We can obtain it as: $\mathbf\small{v_{i(t)}=r_{\bot i} \times \omega_{(t)}}$
• Where:
    ♦ $\mathbf\small{\omega_{(t)}}$ is the instantaneous angular velocity of the rigid body, at the instant when the reading in the stop watch is 't'
    ♦ $\mathbf\small{r_{\bot i}}$ is the perpendicular distance of that i^th particle from the axis
■ Why do we specifically write 'perpendicular distance'?
Answer can be written in 4 steps: 
(i) Let the particle be at any random position that we select
(ii) From that position, there are 'infinite number of distances' possible to the axis
(iii) Some of those possible distances are indicated by the red lines in fig.7.77 below:

Fig.7.77

(iv) But there will be one and only one 'perpendicular distance'. This is indicated by the white line
• The white line is perpendicular to the axis
• It is the 'shortest possible distance' between the particle and the axis
• The white lines (of length $\mathbf\small{r_{\bot}}$) in the previous fig.7.76 are perpendicular distances
    ♦ One is the perpendicular distant at 'A'
    ♦ The other is the perpendicular distant at 'B'
9. For any particle situated on the axis of rotation, $\mathbf\small{r_{\bot}}$ = 0
• So the linear velocity of those particles = 0 × ω = 0   
• That means, all particles on the axis are stationary
• This proves that the axis is fixed
10. For particles not on the axis, the $\mathbf\small{r_{\bot}}$ will be different for different particles
• So for all particles not on the axis, $\mathbf\small{\vec{v}}$ will be different for different particles
■ However, we must always remember that, ω is the same for all particles at any instant
11. Since the ω is the same for all the particles, we can say that, the 'rigid body as a whole' has a particular ω at any instant
■ We defined pure translation earlier:
All particles of the body will be having the same linear velocity at any instant
■ In a similar way, we can now define pure rotation:
All particles of the body will be having the same angular velocity ω at any instant

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• From the above discussion, we get the impression that, the angular velocity is a scalar. But in fact, it is a vector. We will see the proof in higher classes. 

• If it is a vector, there will be both magnitude and direction
• We saw that the magnitude is given by: $\mathbf\small{\text{Instantaneous angular velocity}=\frac{d\theta}{dt}}$
• Now we want the direction
It can be explained by the following steps
1. Consider a rotating disc as shown in the fig.7.78(a) below:

Fig.7.78

• The direction of rotation is indicated by the orange curved arrow
• The blue line indicates the axis of rotation
2. Now, the $\mathbf\small{\vec{\omega}}$ will always lie along the axis of rotation
• So, in our present case, it will lie along the blue line
3. But we want more information:
• Is it from A to B?
• OR, Is it from B to A?
4. For that, we use the right hand screw rule
• It is applied as follows:
(i) Place a right handed screw along the axis of rotation
(ii) Turn the screw in the same direction in which the body is rotating
(Note that, in the fig.7.78(a) above, the brown curved arrow has the same direction as the orange curved arrow)
(iii) The ‘direction in which the screw moves’ gives the direction of omega vector
• The screw in fig.a will be moving towards the disc
• So the direction of $\mathbf\small{\vec{\omega}}$ in fig.a, is from A to B
• In fig.b, the disc is rotating in the opposite direction. So the brown curved arrow is also reversed
• The screw in this case will be moving away from the disc
• So in this case, the direction of $\mathbf\small{\vec{\omega}}$ is from B to A

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Now we will find the relation between linear and angular velocities of the particle:

1. In fig.7.79 below, a particle rotates about the z-axis
• So z-axis is the axis of rotation

Fig.7.79

• The direction of rotation is indicated by the yellow curved arrow
2. We know that, direction of $\mathbf\small{\vec{\omega}}$ of the particle will be along the axis of rotation
• So here it is along the z-axis
• Is $\mathbf\small{\vec{\omega}}$ directed towards the +ve side of z-axis ?
• OR, is it directed towards the -ve side of z-axis ?
3. To find the answer, we apply the right hand screw rule:
(i) The screw is shown at the bottom most portion, below the z-axis
(ii) We turn it in the same direction as the yellow curved arrow
(iii) Due to such a turning, the screw will move towards the +ve side of z-axis
(iv) So we write: $\mathbf\small{\vec{\omega}}$ of the particle is directed towards the +ve side of z-axis
• $\mathbf\small{\vec{\omega}}$ is indicated by the yellow vector at the top most point above the z-axis 
4. In the fig.7.79, the position vector $\mathbf\small{\vec{r}}$ of the particle is also shown
• It is shown in brown color
• For defining a position vector, we need a frame of reference. Note that in this case, the origin 'O' of the frame of reference is placed on the axis of rotation
5. Let us take the cross product of $\mathbf\small{\vec{\omega}}$ and $\mathbf\small{\vec{r}}$     
We have: $\mathbf\small{\vec{\omega}\times \vec{r}=\vec{\omega}\times \vec{OP}}$
6. But $\mathbf\small{\vec{OP}}$ is the resultant of $\mathbf\small{\vec{OC}}$ and $\mathbf\small{\vec{CP}}$ 
That is: $\mathbf\small{\vec{OP}=(\vec{OC}+\vec{CP})}$
7. So the result in (5) becomes: $\mathbf\small{\vec{\omega}\times \vec{r}=\vec{\omega}\times (\vec{OC}+\vec{CP})}$
• We have seen that, cross products obey distributive property. So we can write:
$\mathbf\small{\vec{\omega}\times \vec{r}=(\vec{\omega}\times \vec{OC})+(\vec{\omega}\times\vec{CP})}$
8. Consider the first term on the right side: $\mathbf\small{(\vec{\omega}\times \vec{OC})}$
• $\mathbf\small{\vec{\omega}}$ lies along the z-axis
• $\mathbf\small{\vec{OC}}$ also lies along the z-axis
• So we get: $\mathbf\small{(\vec{\omega}\times \vec{OC})=|\vec{\omega}|\times |\vec{OC}| \times \sin 0=\vec{0}}$
9. Thus we need not consider the first term. The result in (7) becomes:
$\mathbf\small{\vec{\omega}\times \vec{r}=(\vec{\omega}\times\vec{CP})}$
10. Consider the vector multiplication on the right side
• The resulting vector from this multiplication should be perpendicular to both $\mathbf\small{\vec{\omega}}$ and $\mathbf\small{\vec{CP}}$
• That is., the product vector must be perpendicular to the plane containing $\mathbf\small{\vec{\omega}}$ and $\mathbf\small{\vec{CP}}$
• This plane is shown in cyan color in fig.7.80 below:

Fig.7.80

11. So the product vector that we are seeking, is perpendicular to the cyan plane. But we want more information:
• Is that vector directed towards us ?
• OR, Is it directed away from us?
• To find the answer, we apply the right hand screw rule:
(i) We place the screw perpendicular to the cyan plane
• This screw is shown in violet color
(ii) Now, in which direction do we turn that screw?
• To find the direction of $\mathbf\small{(\vec{\omega}\times \vec{CP})}$, we turn it in the direction from $\mathbf\small{\vec{\omega}}$ to $\mathbf\small{\vec{CP}}$
• To find the direction of $\mathbf\small{(\vec{CP}\times \vec{\omega})}$, we turn it in the direction from $\mathbf\small{\vec{CP}}$ to $\mathbf\small{\vec{\omega}}$   
• So in our present case, we turn it from $\mathbf\small{\vec{\omega}}$ to $\mathbf\small{\vec{CP}}$
• This is indicated by the violet curved arrow
(Note that, to find the 'direction of turning', the two vectors must be placed such that, their tail ends coincide. In our present case, we see that, if we bring the $\mathbf\small{\vec{\omega}}$ downwards, the tail ends will coincide. Then the 'direction of turning' is indeed the one shown by the violet curved arrow)    
(iii) When we turn it in the direction of the violet curved arrow, the screw will move away from us
• So we can write: The cross product $\mathbf\small{(\vec{\omega}\times \vec{CP})}$ is perpendicular to the cyan plane and is directed away from us
12. If $\mathbf\small{(\vec{\omega}\times \vec{CP})}$ is perpendicular to the cyan plane, it will be perpendicular to $\mathbf\small{\vec{CP}}$ also
• This is because, $\mathbf\small{\vec{CP}}$ lies in the cyan plane
■ If $\mathbf\small{(\vec{\omega}\times \vec{CP})}$ is perpendicular to $\mathbf\small{\vec{CP}}$, it will be tangential (at P) to the red circular path    
• This is because, tangents are always perpendicular to the radius at the point of tangency
13. So now we have all the information about the direction of the 'cross product vector' that we are seeking 
• The cross product $\mathbf\small{(\vec{\omega}\times \vec{CP})}$ that we are seeking, is indicated by the magenta vector in fig.7.81 below:

Fig.7.81

14. Next we try to find the magnitude of that vector
We can write: $\mathbf\small{|(\vec{\omega}\times \vec{CP})|=|\vec{\omega}|\times r_{\bot}\times \sin 90=\omega \times r_{\bot}}$
15. Thus we can write the following 3 points:
(i) $\mathbf\small{(\vec{\omega}\times \vec{CP})}$ is the magenta vector shown in fig.7.81
(ii) It's direction is tangential (at P) to the red circular path 
(iii) It's magnitude is $\mathbf\small{\omega \times r_{\bot}}$ 
16. Now consider the result in (9) that we saw earlier. We will write it again:
$\mathbf\small{\vec{\omega}\times \vec{r}=(\vec{\omega}\times\vec{CP})}$
17. Comparing (15) and (16), we can write the 3 points again:
(i) $\mathbf\small{(\vec{\omega}\times \vec{r})}$ is the magenta vector shown in fig.7.81
(ii) It's direction is tangential (at P) to the red circular path 
(iii) It's magnitude is $\mathbf\small{\omega \times r_{\bot}}$
18. Now, based on the discussion that we had about circular motion in chapter 4 (Details here), we can write some details about our present particle in fig.7.81
• If at the instant when the particle is at P, the angular velocity is ω, then:
    ♦ The magnitude of the linear velocity $\mathbf\small{\vec{v}}$ at that instant is $\mathbf\small{\omega \times r_{\bot}}$
    ♦ The direction of the $\mathbf\small{\vec{v}}$ at that instant is tangential (at P) to the circular path
19. Comparing (17) and (18), we can write:
• The vector represented by $\mathbf\small{(\vec{\omega}\times \vec{r})}$, is none other than $\mathbf\small{\vec{v}}$
• That is:
Eq.7.16: $\mathbf\small{(\vec{\omega}\times \vec{r})=\vec{v}}$
■ This is a very useful result. From now on, we can calculate the tangential velocity of any particle in a rigid body. All we need to know are the following two items:
(i) Angular velocity vector $\mathbf\small{(\vec{\omega})}$ of the rigid body at that instant
(ii) The position vector $\mathbf\small{(\vec{r})}$ of that particle
• This is particularly helpful when the vectors are given in component forms. We do not need to find $\mathbf\small{r_{\bot}}$ 
20. If $\mathbf\small{\vec{\omega}}$ and $\mathbf\small{\vec{r}}$ are given in component forms, we can find the cross product by writing the table or the determinant
• If they are not in component form, we will be given the magnitudes and the angle θ between them. This is shown in fig.7.81
• In that case we find the cross product as: $\mathbf\small{(\vec{\omega}\times \vec{r})=|\vec{\omega}|\times |\vec{r}|\times \sin \theta}$
• Note that, in the right triangle OPC in fig.7.81, $\mathbf\small{|\vec{r}|\times \sin \theta=r_{\bot}}$

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Angular acceleration

From our discussion so far, it is clear that, translational motion and rotational motion are analogous. We can write the following points:
1. In translational motion, we have linear displacement x
• In rotational motion, we have angular displacement θ 
2. In translational motion, we have linear velocity v
    ♦ It is given by: $\mathbf\small{v = \frac{\text{displacement}}{time}=\frac{dx}{dt}}$ 
• In rotational motion, we have angular velocity ω
    ♦ It is given by: $\mathbf\small{\omega = \frac{\text{angular displacement}}{time}=\frac{d\theta}{dt}}$
3. In the same way, we can write about acceleration also:
• In translational motion, we have linear acceleration a
    ♦ It is given by: $\mathbf\small{a = \frac{\text{change in linear velocity}}{time}=\frac{dv}{dt}}$ 
• In rotational motion, we have angular acceleration
    ♦ It is denoted as α 
    ♦ It is given by: $\mathbf\small{\alpha = \frac{\text{change in angular velocity}}{time}=\frac{d\omega}{dt}}$

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In the next section, we will see torque

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