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)
• 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:
• 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:
• 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, . . . ,Pn
• 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 90o
• 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
• 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
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 |
♦ 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 |
♦ 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 |
• 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, . . . ,Pn
• 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 90o
• 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
19. We have considered only one particle (P1) in the body
• 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
Once we obtain the work done, we can discuss about power. We will see it in the next section
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