In the previous section, we saw the kinematics of rotational motion about a fixed axis. In this section we will see dynamics of rotational motion about a fixed axis
• For our present discussion, we are considering ‘rotation about a fixed axis’
• A door is an example of this type of rotation. We will write the steps:
1. Consider fig.7.128(a) below:
• A force $\mathbf\small{\vec{F}}$ is applied at the door handle
2. If $\mathbf\small{\vec{F}}$ is perpendicular to the door, it is well and good
• But in the fig.a, the $\mathbf\small{\vec{F}}$ is inclined to the door
3. Let us resolve $\mathbf\small{\vec{F}}$ into it's rectangular components. This is shown in fig.b
• We know that, the handle will be moving in a circular path
♦ The red component in fig.b is tangential to that circular path
♦ The green component is perpendicular to that tangent
♦ The blue component is parallel to the hinge
4. What are the effects of the three components?
• The red component causes the door to rotate
• The green component tries to detach the door from the hinge
♦ And it tries to move the door (in a horizontal direction) away from the hinge
• The blue component also tries to detach the door from the hinge
♦ And it tries to move the door (in a vertical direction towards the roof) away from the hinge
5. But in or day to day life, we do not see doors getting detached
• So we conclude that, the green and blue components will not accomplish their objectives
• This is because, the hinge is firmly fixed
• The fixity of the hinge helps to develop forces of constraint which are required to cancel the effects of green and blue components
6. So we see that, green and blue components can be discarded. We will first discuss about the blue component
• We can discard any forces which are parallel to the blue component
• Because they have no effect on the rotation about the hinge
(If they are so large that, they do cause some effect, it means that, the hinge comes of from the wall, and thus it is not a case of ‘rotation about a fixed axis’ any more)
7. Discarding ‘forces parallel to blue component’ means that, discarding forces which are perpendicular to the ‘horizontal plane in which the door handle moves’
• That means, we discard all forces perpendicular to the red plane shown in fig.7.129 below:
• We see that the red circle is the path followed by the door handle
• The red plane is the plane of that red circle
8. So we are left with forces which lie exactly on the red plane in fig.7.129
• Those forces can be resolved into two components:
(i) Those which are tangential to the circle
(ii) Those which are perpendicular to the tangents (the green component in fig.7.128.b)
9. We can discard the item 8(ii) above
• But for that, we will have to resolve the forces at every point on the circle
■ So in general, we take into account all forces that lie exactly on that red plane in fig.7.129
9. Thus for the general case, in rotation about a fixed axis, we take all forces which lie on the planes perpendicular to the axis
• Forces which are perpendicular to those planes are discarded
• Next we will see which position vectors are to be taken into account:
• In fig.7.130 below, we see a force $\mathbf\small{\vec{F}}$ acting on the door handle
11. This $\mathbf\small{\vec{F}}$ is eligible to be taken into account because, it lies in a plane perpendicular to the axis
• To obtain the torque created about the hinge (the z-axis) by $\mathbf\small{\vec{F}}$, we take the cross product: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• But $\mathbf\small{\vec{r}=\vec{OC}+\vec{CP}}$
• So we get: $\mathbf\small{\vec{\tau}=\left(\vec{OC}+\vec{CP}\right)\times \vec{F}}$
• $\mathbf\small{\Rightarrow \vec{\tau}=\left(\vec{OC}\times \vec{F}\right)+\left(\vec{CP}\times \vec{F}\right)}$
12. That means, the torque created by $\mathbf\small{\vec{F}}$ is in fact, the vector sum of two torques. They are:
(i) $\mathbf\small{\vec{\tau}_1=\left(\vec{OC}\times \vec{F}\right)}$
(ii) $\mathbf\small{\vec{\tau}_2=\left(\vec{CP}\times \vec{F}\right)}$
13. Let us now closely examine each of those torques:
• First we consider $\mathbf\small{\vec{\tau}_1=\left(\vec{OC}\times \vec{F}\right)}$
• We know this:
The vector obtained as the result of cross product will be perpendicular to both the vectors which are being multiplied
• So in our present case, we can write this:
The torque $\mathbf\small{\vec{\tau}_1}$ will be perpendicular to both $\mathbf\small{\vec{OC}}$ and $\mathbf\small{\vec{F}}$
14. Next we consider $\mathbf\small{\vec{\tau}_2=\left(\vec{CP}\times \vec{F}\right)}$
• In our present case, we can write this:
The torque $\mathbf\small{\vec{\tau}_2}$ will be perpendicular to both $\mathbf\small{\vec{CP}}$ and $\mathbf\small{\vec{F}}$
• 'Perpendicular to $\mathbf\small{\vec{CP}}$' is same as 'parallel to $\mathbf\small{\vec{OC}}$
15. So we get two interesting results:
(i) $\mathbf\small{\vec{\tau}_1}$ is perpendicular to $\mathbf\small{\vec{OC}}$
(ii) $\mathbf\small{\vec{\tau}_2}$ is parallel to $\mathbf\small{\vec{OC}}$
16. The first result 15(i) is not desirable. Let us see the reason:
• $\mathbf\small{\vec{OC}}$ is aligned with the axis
• So the torque $\mathbf\small{\vec{\tau}_1}$ is perpendicular to the axis
• A torque which is perpendicular to the axis will be trying to tilt that axis
17. In our present case, $\mathbf\small{\vec{\tau}_1}$ will be trying to tilt the 'line of hinge' of the door
• As a result, the hinges will have a tendency to come off from the wall
• But in normal doors, we do not see the hinges coming off. This is because, the hinges have sufficient fixity
• So we can discard $\mathbf\small{\vec{\tau}_1}$
• In general, for rotation about a fixed axis, we can discard $\mathbf\small{\vec{\tau}_1}$
18. Now, $\mathbf\small{\vec{\tau}_1}$ is caused by the perpendicular component $\mathbf\small{\vec{OC}}$ of $\mathbf\small{\vec{r}}$
• So for the general case of rotation about a fixed axis, we can discard the perpendicular component of the position vectors
19. Can we discard $\mathbf\small{\vec{\tau}_2}$?
Let us find out:
• We have seen in 15(ii) that, $\mathbf\small{\vec{\tau}_2}$ is parallel to $\mathbf\small{\vec{OC}}$
• That means, $\mathbf\small{\vec{\tau}_2}$ is parallel to the axis
• So $\mathbf\small{\vec{\tau}_2}$ is in fact the torque which causes the door to rotate. We cannot discard it
• In general, for rotation about a fixed axis, we cannot discard $\mathbf\small{\vec{\tau}_2}$
20. Now, $\mathbf\small{\vec{\tau}_2}$ is caused by the parallel component $\mathbf\small{\vec{CP}}$ of $\mathbf\small{\vec{r}}$
• So for the general case of rotation about a fixed axis, we cannot discard the parallel component of the position vectors
When a body rotates about a fixed axis:
1. Consider the planes perpendicular to the axis
• All forces lying in those planes should be taken into account
• Discard all forces which are perpendicular to those planes
2. Consider the position vectors of all particles in the body
• Components of the position vectors which are parallel to the planes in (1) should be taken into account
• Discard all components which are perpendicular to those planes
■ The red plane in fig.7.129 above is perpendicular to the axis of rotation
■ All the forces that we need, are parallel to that red plane
■ All the 'position vector components' that we need, are parallel to that red plane
• In other words, the 3D problem of 'rotation about a fixed axis' is now reduced to a 2D problem
• Let us see how to represent it as a 2D problem. We will write it in steps:
1. Consider the fig.7.130 that we saw above. The path of the door handle is shown as a pink circle
• Since it is now a 2D problem, the 'plane of that circle' is the plane which are are going to work on
2. If we look at the 'plane of that circle' from above, we will get a 2D view
• In that view, the z-axis will be seen as a small blue circle. This is shown in fig.7.131 below:
• The entire path of the handle is not shown in fig.7.131. Rather, a portion of it is shown as the pink arc
♦ The center of this arc is C
• The door handle is shown as a small yellow sphere
x'-axis is parallel to the x-axis of the reference frame
♦ x'-axis passes through C
• y'-axis is parallel to the y-axis of the reference frame
♦ y'-axis passes through C
3. A force $\mathbf\small{\vec{F}}$ acts on the door handle. This force lies on the x'y' plane. So we must take it into account
4. $\mathbf\small{\vec{r}_{\bot}}$ is the 'perpendicular component' of the 'position vector of the handle'. It lies on the x'y' plane. So we must take it into account
• For our present discussion, we are considering ‘rotation about a fixed axis’
• A door is an example of this type of rotation. We will write the steps:
1. Consider fig.7.128(a) below:
 |
| Fig.7.128 |
2. If $\mathbf\small{\vec{F}}$ is perpendicular to the door, it is well and good
• But in the fig.a, the $\mathbf\small{\vec{F}}$ is inclined to the door
3. Let us resolve $\mathbf\small{\vec{F}}$ into it's rectangular components. This is shown in fig.b
• We know that, the handle will be moving in a circular path
♦ The red component in fig.b is tangential to that circular path
♦ The green component is perpendicular to that tangent
♦ The blue component is parallel to the hinge
4. What are the effects of the three components?
• The red component causes the door to rotate
• The green component tries to detach the door from the hinge
♦ And it tries to move the door (in a horizontal direction) away from the hinge
• The blue component also tries to detach the door from the hinge
♦ And it tries to move the door (in a vertical direction towards the roof) away from the hinge
5. But in or day to day life, we do not see doors getting detached
• So we conclude that, the green and blue components will not accomplish their objectives
• This is because, the hinge is firmly fixed
• The fixity of the hinge helps to develop forces of constraint which are required to cancel the effects of green and blue components
6. So we see that, green and blue components can be discarded. We will first discuss about the blue component
• We can discard any forces which are parallel to the blue component
• Because they have no effect on the rotation about the hinge
(If they are so large that, they do cause some effect, it means that, the hinge comes of from the wall, and thus it is not a case of ‘rotation about a fixed axis’ any more)
7. Discarding ‘forces parallel to blue component’ means that, discarding forces which are perpendicular to the ‘horizontal plane in which the door handle moves’
• That means, we discard all forces perpendicular to the red plane shown in fig.7.129 below:
 |
| Fig.7.129 |
• The red plane is the plane of that red circle
8. So we are left with forces which lie exactly on the red plane in fig.7.129
• Those forces can be resolved into two components:
(i) Those which are tangential to the circle
(ii) Those which are perpendicular to the tangents (the green component in fig.7.128.b)
9. We can discard the item 8(ii) above
• But for that, we will have to resolve the forces at every point on the circle
■ So in general, we take into account all forces that lie exactly on that red plane in fig.7.129
9. Thus for the general case, in rotation about a fixed axis, we take all forces which lie on the planes perpendicular to the axis
• Forces which are perpendicular to those planes are discarded
10. So now we know which forces are to be taken into account
• In fig.7.130 below, we see a force $\mathbf\small{\vec{F}}$ acting on the door handle
 |
| Fig.7.130 |
• To obtain the torque created about the hinge (the z-axis) by $\mathbf\small{\vec{F}}$, we take the cross product: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• But $\mathbf\small{\vec{r}=\vec{OC}+\vec{CP}}$
• So we get: $\mathbf\small{\vec{\tau}=\left(\vec{OC}+\vec{CP}\right)\times \vec{F}}$
• $\mathbf\small{\Rightarrow \vec{\tau}=\left(\vec{OC}\times \vec{F}\right)+\left(\vec{CP}\times \vec{F}\right)}$
12. That means, the torque created by $\mathbf\small{\vec{F}}$ is in fact, the vector sum of two torques. They are:
(i) $\mathbf\small{\vec{\tau}_1=\left(\vec{OC}\times \vec{F}\right)}$
(ii) $\mathbf\small{\vec{\tau}_2=\left(\vec{CP}\times \vec{F}\right)}$
13. Let us now closely examine each of those torques:
• First we consider $\mathbf\small{\vec{\tau}_1=\left(\vec{OC}\times \vec{F}\right)}$
• We know this:
The vector obtained as the result of cross product will be perpendicular to both the vectors which are being multiplied
• So in our present case, we can write this:
The torque $\mathbf\small{\vec{\tau}_1}$ will be perpendicular to both $\mathbf\small{\vec{OC}}$ and $\mathbf\small{\vec{F}}$
14. Next we consider $\mathbf\small{\vec{\tau}_2=\left(\vec{CP}\times \vec{F}\right)}$
• In our present case, we can write this:
The torque $\mathbf\small{\vec{\tau}_2}$ will be perpendicular to both $\mathbf\small{\vec{CP}}$ and $\mathbf\small{\vec{F}}$
• 'Perpendicular to $\mathbf\small{\vec{CP}}$' is same as 'parallel to $\mathbf\small{\vec{OC}}$
15. So we get two interesting results:
(i) $\mathbf\small{\vec{\tau}_1}$ is perpendicular to $\mathbf\small{\vec{OC}}$
(ii) $\mathbf\small{\vec{\tau}_2}$ is parallel to $\mathbf\small{\vec{OC}}$
16. The first result 15(i) is not desirable. Let us see the reason:
• $\mathbf\small{\vec{OC}}$ is aligned with the axis
• So the torque $\mathbf\small{\vec{\tau}_1}$ is perpendicular to the axis
• A torque which is perpendicular to the axis will be trying to tilt that axis
17. In our present case, $\mathbf\small{\vec{\tau}_1}$ will be trying to tilt the 'line of hinge' of the door
• As a result, the hinges will have a tendency to come off from the wall
• But in normal doors, we do not see the hinges coming off. This is because, the hinges have sufficient fixity
• So we can discard $\mathbf\small{\vec{\tau}_1}$
• In general, for rotation about a fixed axis, we can discard $\mathbf\small{\vec{\tau}_1}$
18. Now, $\mathbf\small{\vec{\tau}_1}$ is caused by the perpendicular component $\mathbf\small{\vec{OC}}$ of $\mathbf\small{\vec{r}}$
• So for the general case of rotation about a fixed axis, we can discard the perpendicular component of the position vectors
19. Can we discard $\mathbf\small{\vec{\tau}_2}$?
Let us find out:
• We have seen in 15(ii) that, $\mathbf\small{\vec{\tau}_2}$ is parallel to $\mathbf\small{\vec{OC}}$
• That means, $\mathbf\small{\vec{\tau}_2}$ is parallel to the axis
• So $\mathbf\small{\vec{\tau}_2}$ is in fact the torque which causes the door to rotate. We cannot discard it
• In general, for rotation about a fixed axis, we cannot discard $\mathbf\small{\vec{\tau}_2}$
20. Now, $\mathbf\small{\vec{\tau}_2}$ is caused by the parallel component $\mathbf\small{\vec{CP}}$ of $\mathbf\small{\vec{r}}$
• So for the general case of rotation about a fixed axis, we cannot discard the parallel component of the position vectors
Now we can write a summary:
1. Consider the planes perpendicular to the axis
• All forces lying in those planes should be taken into account
• Discard all forces which are perpendicular to those planes
2. Consider the position vectors of all particles in the body
• Components of the position vectors which are parallel to the planes in (1) should be taken into account
• Discard all components which are perpendicular to those planes
This is a very interesting situation:
■ All the forces that we need, are parallel to that red plane
■ All the 'position vector components' that we need, are parallel to that red plane
• So the 3D problem is now reduced to a 'problem in a plane'
• Let us see how to represent it as a 2D problem. We will write it in steps:
1. Consider the fig.7.130 that we saw above. The path of the door handle is shown as a pink circle
• Since it is now a 2D problem, the 'plane of that circle' is the plane which are are going to work on
2. If we look at the 'plane of that circle' from above, we will get a 2D view
• In that view, the z-axis will be seen as a small blue circle. This is shown in fig.7.131 below:
 |
| Fig.7.131 |
♦ The center of this arc is C
• The door handle is shown as a small yellow sphere
x'-axis is parallel to the x-axis of the reference frame
♦ x'-axis passes through C
• y'-axis is parallel to the y-axis of the reference frame
♦ y'-axis passes through C
3. A force $\mathbf\small{\vec{F}}$ acts on the door handle. This force lies on the x'y' plane. So we must take it into account
4. $\mathbf\small{\vec{r}_{\bot}}$ is the 'perpendicular component' of the 'position vector of the handle'. It lies on the x'y' plane. So we must take it into account
Thus we have succeeded in representing the 'required items' in a 2D plane. Based on that, we can discuss the work done by a torque. We will see it in the next section
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