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'
■ ‘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
• 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.
• 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'
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:
• 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
■ 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
1. Consider a rigid block shown in fig.7.3(a) below:
• 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
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']
• 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'
■ ‘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
• 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 |
• 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 |
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
Such a difficulty can be overcome by assuming the object to be ‘rigid’
• Consider any one 'pair of points' in that body
♦ The distance between the points in that pair, will not change
• 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
• 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
 |
| Fig.7.3 |
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
If at any instant, all particles of a body have the same velocity, then that body is said to be in pure translational motion
Now we will see another type of motion:
1. Consider the side view of a cylinder in fig.7.4(a) below:
• It is rolling towards the right, on a horizontal surface. This 'rolling' is indicated by the white curved arrow
1. Consider the side view of a cylinder in fig.7.4(a) below:
 |
| Fig.7.4 |
• 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']
In the next section, we will see what this 'something else' is.
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