Chapter 7.1 - Rotation About An Axis

In the previous section, we saw that, motion of a rigid body can be: [Translation + 'something else']. In this section, we will see what this 'something else' is.

1. To study about this situation, we first need to constrain the rigid body
• We have to constrain it in such a way that, it cannot have translational motion
2. The most effective way to achieve this, is to fix it along a straight line
• The animation in fig.7.5 below shows an example:

Fig.7.5

• The small yellow rigid block is attached to the larger red rigid block
3. The attachment is along an edge. This edge is highlighted by a blue line
• This blue line is the 'straight line' along which the yellow block is fixed
• Because of this 'fixity', the yellow block cannot have translational motion
■ When a rigid body is fixed along a line, only one type of motion is possible for that body:
The rotational motion
(If in the above fig.7.5, the large red block was absent, the yellow block can complete a full 360^o rotation)
■ The line along which the rigid body is fixed is called it's axis of rotation
• Note that, the axis of rotation need not pass through the geometric center of the body 

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In our day to day life, we come across many examples where rigid bodies rotate about an axis. let us see some of them:

Example 1: The ceiling fan
An animation of a 'rotating ceiling fan' is shown in fig.7.6 below

Fig.7.6

• The axis of rotation passes through the center of the ceiling fan. it is shown in fig.7.7 below:

Fig.7.7

• The axis is indicated by the blue straight line  
• The blue curved arrow indicates the direction of rotation
• The fan is fixed along the axis
■ If during rotation, the 'mechanism which fixes the axis' fails, the fan will move with both rotation and translation, causing serious accidents
• So we see that, fixity along the axis is important to ensure pure rotation 

Example 2: The potter's wheel
An animation of a rotating potter's wheel is shown in fig.7.8 below:

Fig.7.8

• The axis of rotation passes through the center of the wheel. It is shown in fig.7.9 below:

Fig.7.9

• The axis is indicated by the blue straight line  
• The curved arrow indicates the direction of rotation
■ If the 'mechanism which fixes the axis' fails, the wheel will move with both rotation and translation
• So we see that, fixity along the axis is important to ensure pure rotation 

Characteristics of rotational motion

1. The animation in fig.7.10 below, shows a rotating body

Fig.7.10

• The axis of rotation is shown in blue color
■ For convenience of mathematical calculations, we will consider this axis to be the 'z-axis' of the 'frame of reference'
    ♦ The x-axis is shown in red color
    ♦ The y-axis is shown in green color
2. We are inclined to think that: Every particle in the body will be in rotation
• But that is not entirely true. Let us see the reason:
• We will consider 3 particles inside the body
    ♦ They are indicated by small spheres
    ♦ The red, yellow and green spheres
3. If we trace the path of the red sphere, that path will be a circle
• It is the red circle that we see in the animation
• What is the radius of this circle?
• Obviously, radius = distance of the red sphere from the axis
• We will denote it as r_R
4. If we trace the path of the yellow sphere, it will be a circle
• It is the yellow circle that we see in the animation
• What is the radius of this circle?
• Obviously, radius = distance of the yellow sphere from the axis
• We will denote it as r_Y
5. If we try to trace the path of the green sphere, we will not succeed
• This is because there is no such path
• In other words, there is no green circle. That is., r_G = 0
• 'No path' means that, there is no movement. That is., the green sphere is stationary
■ Like the green sphere, all the particles which lie on the axis are stationary. This is the answer to the doubt mentioned in (2) above
6. So we have two circles: The red circle and the yellow circle
• We know that, any circle will be having it's own plane
• The plane in which the red circle lies, is shown in red color in fig.7.11(a) below

Fig.7.11

• The plane in which the yellow circle lies, is shown blue color in fig.b.
• The plane of the red circle is perpendicular to the axis of rotation
    ♦ This is indicated by the white set-square in fig.a
• The plane of the yellow circle is also perpendicular to the axis
    ♦ This is indicated by the white set-square in fig.b

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Let us write a summary of 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

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Also it is worth mentioning that, in the animation in fig.7.10 above, the red sphere moves with a greater speed than the yellow sphere
This is because, r_R > r_Y
We saw the detailed explanation in a previous section.

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In the next section, we will see another type of rotational motion

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Chapter 7 - Systems of Particles And Rotational Motion

In the previous section, we completed a discussion on two-dimensional collision. In this chapter, we will see rotational motion.

• In the previous chapters, we analyzed many cases in which ‘objects’ are involved. Let us recall some examples that we have seen:
    ♦ An object acted upon by a net force moves from one position to another
    ♦ A projectile moving in a parabolic path is acted upon by various forces
    ♦ An object at a height possess potential energy
    ♦ An object moving with a certain velocity possess kinetic energy
    ♦ Collision between objects causes changes in their original velocities     
so on . . .
• In all those cases, we considered the objects to be point masses
    ♦ It is true that, we drew them not as points, but as squares, rectangles, spheres, cars, trucks, etc.,
    ♦ But we drew the vectors from their centers
    ♦ We assumed that, all the mass of an object is concentrated at the 'center point' 

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■ ‘Point mass’ can be explained as follows:
• If an object is said to have a mass ‘m’ kg, all that mass is assumed to be concentrated at a point
• That point is a ‘point mass’  
• Even though it is a ‘point’, it's mass is not zero
• Considering objects to be point masses helps to simplify calculations

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• So we can do calculations by two methods:
(i) Considering an object to be a point mass
(ii) Considering the actual distribution of mass in the object
• On many occasions, the results obtained by the two methods will not be appreciably different
■ If we consider the ‘actual distribution of mass’ in an object, then it is called an extended object
• While considering an extended object, it’s size becomes important
    ♦ It cannot be considered as a point
• So basically, an ‘extended object’ is a ‘system of particles’
• In some problems in physics, it is compulsory to consider objects as ‘extended objects’
• In fig.7.1(a) below, a cube is shown. 

Fig.7.1

• It is a uniform cube. That is., the mass is distributed uniformly through out the entire volume of the cube. 
• In such cases, we can consider the cube to be a point mass located at the 'geometric center' of the cube. 
• The geometric center of a cube is easy to locate. Recall that, in many diagrams that we saw in the previous chapters, we drew velocity vectors, force vectors etc., through the geometric center of the objects. 
• Now consider fig.b. Geometrically, it is a cube. But more mass is concentrated towards the left side.
• That is., the 'left side is heavier' and 'right side is lighter'. We cannot consider the mass to be concentrated at the geometric center. If we make such an assumption, the results will not be correct 
• In such cases, we need to define a new ‘point’
■ This point is called the ‘center of mass’
• So we have two points:
    ♦ A point which is the geometric center
    ♦ A point which is the center of mass
• Unlike the ‘geometric center’, the ‘center of mass’ is not easy to find
• However in later sections, we will see methods to find it for 'simple objects'

Significance of Rigid bodies

Following the above discussion, we encounter a new ‘situation’. It can be explained as follows:
Consider a scenario in steps:
1. We have to solve a problem involving an extended object
2. We start off by calculating the center of mass of that object
• An example is shown in fig.7.2(a) below:

Fig.7.2

• The center of mass is denoted by the red '❌' mark
3. But when a force act on the object, it gets deformed
• As a result, the center of mass shifts to a new position. This is shown in fig.7.2(b)
• We will have to recalculate the new position

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Such a difficulty can be overcome by assuming the object to be ‘rigid’

■ A rigid body is a body with a perfectly definite and unchanging shape
• Consider any one 'pair of points' in that body
    ♦ The distance between the points in that pair, will not change

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• However, no real body is truly rigid. Because, all of them deforms under the influence of forces
• But in many cases, those deformations are negligible
• We can confidently do problems by considering them to be rigid
• We will learn about advanced cases (where deformations also have to be considered) in higher classes

Motion of a rigid body

1. Consider a rigid block shown in fig.7.3(a) below:

Fig.7.3

• It moves with uniform velocity $\mathbf\small{\vec{v}}$ on a horizontal surface
2. Consider any two points P and Q in the body
We can write three detailed steps as follows:
(i) Let  $\mathbf\small{\vec{v}_P}$ be the velocity with which the point P moves
(ii) Let  $\mathbf\small{\vec{v}_Q}$ be the velocity with which the point Q moves
(iii) Then $\mathbf\small{\vec{v}_P=\vec{v}_Q=\vec{v}}$
• The result in (iii) is obvious. In fact we do not even need to write the three detailed steps. We know it is true
3. The same block is shown in fig.b
• But this time it moves with a non-uniform velocity
• For example, if it is moving with an acceleration 'a', we can say that, it is moving with a non-uniform velocity
• In such a situation, the velocity of the block will be different at different instances
4. A point 'A' is marked on the horizontal surface on which the block moves
• Let the velocity of the block at the instant when it just passes 'A' be $\mathbf\small{\vec{v}_{(A)}}$
• We can write three detailed steps as follows:
(i) Let $\mathbf\small{\vec{v}_{P(A)}}$ be the 'velocity with which the point P moves' at the instant when the block just passes 'A'
(ii) Let $\mathbf\small{\vec{v}_{Q(A)}}$ be the 'velocity with which the point Q moves' at the instant when the block just passes 'A'
(iii) Then $\mathbf\small{\vec{v}_{P(A)}=\vec{v}_{Q(A)}=\vec{v}_{(A)}}$
• The result in (iii) is obvious. In fact we do not even need to write the three detailed steps. We know it is true

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If at any instant, all particles of a body have the same velocity, then that body is said to be in pure translational motion

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Now we will see another type of motion:
1. Consider the side view of a cylinder in fig.7.4(a) below:

Fig.7.4

• It is rolling towards the right, on a horizontal surface. This 'rolling' is indicated by the white curved arrow

2. A point 'A' is marked on the horizontal surface on which the cylinder rolls
• Consider the instant at which the cylinder passes 'A' 
• A normal is drawn to the horizontal surface at A. It is shown in red color
• Also, this normal passes through the center O of the cylinder
3. Consider any two points P and Q on this normal
We can write three detailed steps as follows:
(i) Let $\mathbf\small{\vec{v}_{P(A)}}$ be the 'velocity with which the point P moves' at the instant when the cylinder just passes 'A'
(ii) Let $\mathbf\small{\vec{v}_{Q(A)}}$ be the 'velocity with which the point Q moves' at the instant when the cylinder just passes 'A'
(iii) Then $\mathbf\small{\vec{v}_{P(A)}\neq \vec{v}_{Q(A)}}$
4. In 3(iii), We have an inequality
• Unlike the previous cases that we saw about the block, this result is not so obvious. We need to give a bit of explanation
• The explanation can be written based on the second fig.b, as follows:
(i) In fig.b, a circle is drawn through P
• The center of this circle is 'O', which is the center of the cylinder
    ♦ $\mathbf\small{\vec{v}_{P(A)}}$ will be tangential to this circle
    ♦ $\mathbf\small{\vec{v}_{P(A)}}$ will be perpendicular to the red line
• Another circle is drawn through Q. The center of this circle also is 'O'
    ♦ $\mathbf\small{\vec{v}_{Q(A)}}$ will be tangential to this circle
    ♦ $\mathbf\small{\vec{v}_{Q(A)}}$ will be perpendicular to the red line
(ii) Since both the velocities are perpendicular to the red line, we can write:
• $\mathbf\small{\vec{v}_{P(A)}}$ and $\mathbf\small{\vec{v}_{Q(A)}}$ have the same direction
(iii) But their magnitudes will be different. Let us see the reason:
• Let $\mathbf\small{\omega_{(A)}}$ be the instantaneous angular velocity when the cylinder just passes 'A'
• This instantaneous angular velocity will be same for both P and Q
• It is the linear speeds $\mathbf\small{|\vec{v}_{P(A)}|}$ and $\mathbf\small{|\vec{v}_{Q(A)}|}$ which differ
• This difference is due to the 'greater radius' of the circle through P
• We saw a detailed explanation in a previous chapter
(v) For two vectors to be equal, both their magnitudes and directions should be equal
• In our present case, magnitudes are not equal
• This explains the result in 3(iii) that we saw above
5. Let us go back to fig.(a). Besides P and Q, one more point R is also marked
• It is on a violet line passing through O
6. In fig.c, a circle is drawn through R
• The center of this circle is O
• The instantaneous velocity $\mathbf\small{\vec{v}_{R(A)}}$ for R is tangential to this circle
• Also this $\mathbf\small{\vec{v}_{R(A)}}$ is perpendicular to the violet line
• Obviously, the direction of $\mathbf\small{\vec{v}_{R(A)}}$ is different from those of $\mathbf\small{\vec{v}_{P(A)}}$ and $\mathbf\small{\vec{v}_{Q(A)}}$
• Thus we can write: $\mathbf\small{\vec{v}_{P(A)}\neq \vec{v}_{Q(A)}\neq \vec{v}_{R(A)}}$
7. The rolling cylinder is at point 'A' that we marked on the horizontal surface
• If we mark another point 'B' to the right of 'A', after a short time, the cylinder will surely reach that 'B' 
• So we can say that the cylinder is in translational motion
• But all the particles in the cylinder are not moving with the same velocity
• So the cylinder is not in pure translational motion 
• It moves with: [Translation + ''something else']

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In the next section, we will see what this 'something else' is.

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