Chapter 7.3 - Center of Mass

In the previous section, we saw that the 'center of mass' of a body has much significance. In this section, we will see methods to find the location of that center.

1. In fig.7.24 below, two points P and Q lie on the x-axis

Fig.7.24

• Point P is at a distance of 5 cm from O
• Point Q is at a distance of 9 cm from O
2. First we add the above distances
• Then we divide that sum by the 'number of distances'
• what do we get?
3. Let us try:
• Sum of the distances = (5+9) = 14
• Number of distances:
    ♦ '5' is one distance
    ♦ '9' is the other distance
    ♦ So number of distances = 2
• Thus we get:
$\mathbf\small{\frac{\text{Sum of distances}}{\text{No. of distances}}=\frac{14}{2}=7}$
4. This '7' is marked in the fig.b as 'C'
• Distance of C from P = (7-5) = 2 cm
• Distance of C from Q = (9-7) = 2 cm
■ So 'C' is equidistant from P and Q
■ That means, C is the midpoint of PQ
5. The 'midpoint' can be used if there are only two points P and Q
• It is more appropriate to use 'average'
    ♦ P is at a distance of 5 cm from O
    ♦ Q is at a distance of 9 cm from O
• We know that the average distance is calculated as:
$\mathbf\small{\frac{\text{dist. 1 + dist. 2 + . . . + dist n}}{\text{No. of distances}}=\frac{\text{dist. 1 + dist. 2 + . . . + dist n}}{\text{n}}}$
• So for our present case, 
Average distance = $\mathbf\small{\frac{5 + 9}{\text{2}}=\frac{\text{14}}{\text{2}}=7}$
(same as the midpoint)
6. The advantage of 'average' is that, it can be applied to more than 2 'values'
• Examples of 'values' are: distances, heights, speeds, velocities, masses etc.
7. Let us see a case involving 3 heights:
• A team has 3 players
• Their heights are 158 cm, 164 cm and 167 cm
• We know that average height = $\mathbf\small{\frac{158 + 164 + 167}{\text{3}}=\frac{\text{489}}{\text{3}}=163}$
• So 'height 163 cm' is an appropriate representation of the 'height of the team of 3 players'
8. What if there are more than 1 player having the same height ?
Consider the following data:
• Total number of players in the team = 9
    ♦ 3 players have height 158 cm 
    ♦ 4 players have height 164 cm 
    ♦ 2 players have height 167 cm 
9. Let us calculate the average height:
• The first height is 158 cm. 
    ♦ This height of '158' has to be given due weightage
    ♦ Three players have this height
    ♦ So to give the due weightage, we multiply '158' by '3'
(Here the word 'due' is used as a noun. It means 'claim'. The 'height 158' has a claim of '3')
• The second height is 164 cm. 
    ♦ This height of '164' has to be given due weightage
    ♦ Four players have this height
    ♦ So to give the due weightage, we multiply '164' by '4'
• The third height is 167 cm. 
    ♦ This height of '167' has to be given due weightage
    ♦ Two players have this height
    ♦ So to give the due weightage, we multiply '167' by '2'
10. In the denominator, of course, it should be the total number of 'values'
• In our present case, the total number of values is 9 
• So we get: Average height = $\mathbf\small{\frac{{(158\times 3)+(164 \times 4) + (167 \times 2)}}{9}=\frac{{1464}}{9}=162.67}$
■ So ' height 162.67 cm' is an appropriate representation of the 'height of the team of 9 players'

---

Note:

• Let us forget about 'giving due weightage' and find the average by the regular method. We get: Average height =
$\mathbf\small{\frac{158 + 158 + 158 + 164 + 164 + 164 + 164 + 167 + 167}{\text{9}}}$
In the numerator, there are:
• 3 No. 158
    ♦ This can be written as (3 × 158)
• 4 No. 164
    ♦ This can be written as (4 × 164)
• 2 No. 168
    ♦ This can be written as (2 × 168)
■ So 'giving due weightage' is a natural phenomenon

---

• In the above example, we wanted to find a 'representative height'

• We achieved it by giving 'due weightage' to each individual heights 
• In the next example, we want to find a 'representative distance'. Let us see the details:
• In fig.7.25(a) below, a sphere of mass 5 kg is situated at 'P' on the x-axis
    ♦ 'P' is at a distance of 7 m from O
• Another sphere of mass 9 kg is situated at 'Q' on the x-axis
    ♦ 'Q' is at a distance of 12 m from O

Fig.7.25

• We want to find a 'representative distance' which can be used to represent the two distances '7 m' and '12 m'. The masses also should be taken into account.
Solution:
1. The first distance is '7 m'
• This distance has to be given due weightage
• For that, we make the following 2 assumptions:
(i) The sphere at 'P' is a group of 5 'smaller spheres'
    ♦ This is shown in fig.7.25(b)
(ii) Each of these 'smaller spheres' has a mass of 1 kg
• So we can make a comparison with the previous example:
■ Previous example:
• 3 individuals have a height of 158 cm
    ♦ So the weightage due for 'height 158' is 3
■ Present example:
• 5 individual spheres (each with mass 1 kg) are at a distance of 7 m
    ♦ So the weightage due for 'distance 7' is 5
2. The second distance is '12 m'
• This distance has to be given due weightage
• For that, we make the following 2 assumptions:
(i) The sphere at 'Q' is a group of 9 'smaller spheres'
(ii) Each of these 'smaller spheres' has a mass of 1 kg
• So as in (1) above, we can write:
• 9 individual spheres (each with mass 1 kg) are at a distance of 12 m
    ♦ So the weightage due for 'distance 12' is 9
3. So in the numerator, we have: (5 × 7) + (9 × 12)
4. In the denominator, of course, it is the 'total number of individuals'
• In our present case, we have:
    ♦ 5 individual spheres at P
    ♦ 9 individual spheres at Q
• So 'total number of individuals' is 14
• Thus in the denominator, we have 14
5. So the 'representative distance' that we obtain from the two distances 7 m and 12 m in fig.7.25 is:
$\mathbf\small{\frac{{(5\times 7)+(9 \times 12)}}{14}=\frac{{143}}{14}=10.21}$
• A point 'C' is marked as 'C' at this distance in fig.7.25(b)
6. Consider the system given to us. It consists of the following two items:
(i) A 5 Kg mass situated at a distance of 7 m from the origin O     
(ii) A 9 Kg mass situated at a distance of 12 m from the origin O     
7. The system described in (6) can be replaced by:
■ A 14 kg mass situated at a distance of 10.21 m from the origin 
8. Recall that in the first section of this chapter, we saw the following 3 items:
(i) If the mass of a body is not distributed uniformly, we cannot use the geometric center
(ii) In such cases, we have to find the 'center of mass'
(iii) We can assume that, all the mass of a body is concentrated at the 'center of mass' 
9. So the result that we obtained in (5) above, is actually, the 'location of the center of mass'
• To locate the 'center of mass', all we need is it's distance from O
    ♦ The 'center of mass' is denoted by the letter 'C'
    ♦ The distance of 'C' from O is denoted by the letter 'X'
10. Since 'calculation of X' is to be done frequently, we need to write a general formula
• If we examine step (3), we can see that:
The numerator is: (Mass₁ × Distance₁) + (Mass₂ × Distance₂)
• If we examine step (4), we can see that:  
The denominator is: (Mass₁ + Mass₂) = Total mass (M)
11. So we can write:
Eq.7.1:
$\mathbf\small{X=\frac{m_1 x_1+m_2 x_2}{m_1 + m_2}}$
If there are more than 2 masses, we can write:
$\mathbf\small{X=\frac{m_1 x_1+m_2 x_2+\,.\, .\,.\,+m_n x_n}{m_1 + m_2+\,.\, .\,.\,+m_n}=\frac{\sum{m_ix_i} }{\sum{m_i} }=\frac{\sum{m_ix_i} }{M}}$

---

• In the fig.7.25 that we saw above, both the spheres lie on the x axis
• A 3D view of that system is shown in fig.7.26 below:

Fig.7.26

• The x-axis is shown in red color
• The y-axis is shown in green color
• The z-axis is shown in blue color
• The plane (shown in grey color) which is bounded by the x and y axes, is called the xy-plane 
• When we look at this plane from above, we get the 2D view that we saw earlier in fig.7.25
• In such a view, the z-axis will be pointing towards us
• In our present example, the z-axis has no role to play. So it is not shown in the fig.7.25

---

An interesting result:
1. We now know the position of 'C' of the system of two spheres P and Q in fig.7.25
• Let us rearrange the 'frame of reference' in such a way that, the y-axis passes through 'C'
• This is shown in fig.7.27(a) below:

Fig.7.27

2. The sphere P lies on the left side of y-axis
• So x₁ will be a negative value. We can write:
x₁ = -3.21 m, x₂ = +1.79 m
3. Thus we get: $\mathbf\small{\sum{m_ix_i}=m_1 x_1+m_2 x_2}$ 
= [(5×-3.21)+(9×1.79)] = [-16.1+16.1] = 0
4. If we connect P and Q by an uniform rod, the whole system can be balanced by placing a 'pointed support' at 'C'
• This is shown in fig.7.27(c)

---

Now we will see a solved example

Solved example 7.1
Three spheres P, Q and R are placed on the x-axis as shown in fig.7.28(a) below:

Fig.7.28

Find the location of 'C'. The masses of P, Q and R are 30, 45 and 25 kg respectively
Solution:
1. The calculations can be done easily if we arrange the data in a tabular form:

• The first two columns can be filled up using the given data
2. From the table, we get:
$\mathbf\small{\sum{m_i}=100}$
$\mathbf\small{\sum{m_ix_i}=460}$
3. Thus we get: $\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{460}{100}=4.6\, \text{m}}$
• This is shown in fig.b

An interesting result:
1. We now know the position of 'C' of the system of three spheres P, Q and R in fig.7.28
• Let us rearrange the 'frame of reference' in such a way that, the y-axis passes through 'C'
• This is shown in fig.7.27(b) above
2. The sphere P lies on the left side of y-axis
• So x₁ will be a negative value. We can write:
x₁ = -2.6 m, x₂ = +0.4 m, x₃ = +2.4 m 
3. Thus we get: $\mathbf\small{\sum{m_ix_i}=m_1 x_1+m_2 x_2+m_3 x_3}$ 
= [(30×-2.6)+(45×0.4)+(25×2.4)] = [-78+18+60] = 0
4. We get the same result quickly if we write the values in a tabular form:

5. If we connect P, Q and R by an uniform rod, the whole system can be balanced by placing a 'pointed support' at 'C'
• This is shown in fig.7.27(d) above

---

■ Based on figs.7.27 (a) & (b), we can write:

• If the y-axis passes through 'C', then $\mathbf\small{\sum{m_ix_i}=0}$

---

In the above discussion, we saw the cases where all the bodies lie on the x-axis. In the next section, we will see the more advanced case, where bodies are distributed in the xy-plane

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 7.2 - Rotation about a Moving Axis

In the previous section, we saw the characteristics of 'rotation about an axis'. In this section, we will see another type of rotation.
We will write the steps:
1. The fig.7.12 below shows an animation of a spinning top

Fig.7.12

• The top is moving around in a circle
• The axis for this circular motion is the blue vertical line shown in fig.7.13 below
    ♦ This blue axis passes through the 'point of contact of the top with the ground'

Fig.7.13

• The direction of rotation is indicated by the blue circular arrow
2. While rotating around the blue axis, the top is spinning around it's own axis
• This axis is the orange line shown in fig.7.13
    ♦ It passes through the center of the top
• The 'spin about this axis' is indicated by the orange circular arrow
3. We can write:
• The blue axis is the Global axis
• The orange axis is the Local axis
■ The top is spinning about it's local axis and at the same time, moving around the global axis
• This movement of the local axis of the top around the global axis is termed as precession 
4. In the the previous section,we saw that all the points on the 'axis of the rotating object' will be stationary
• Let us analyse our present case:
(i) The 'axis of the rotating object' in our present case is the orange axis
(ii) Only 'one point on this axis' is stationary
    ♦ It is the point at the intersection of the orange and blue axes
    ♦ That is., the point of contact of the top with the ground
(iii) At any instant, the local axis of the top passes through the point of contact 
(iv) The point of contact of the top with the ground should not change
• If that point changes, the steps that we wrote above from (1) to (4) become invalid
• Because, then it would mean that, the top has some translation also. The blue line will not be the global axis any more

---

Let us see another case belonging to this category:
1. The fig.7.14 below shows an animation of an 'oscillating table fan'

Fig.7.14

• When we consider the total system, the central portion has an oscillating motion
• The path traced by the central portion, during this oscillating motion, is indeed a 'portion of a circle'
• The axis for this circular motion is the blue vertical line shown in fig.7.15 below:

Fig.7.15

• The direction of rotation is indicated by the blue circular arrow
    ♦ This arrow is double-headed
    ♦ This indicates oscillating or to-and-fro motion
2. While the oscillation is taking place around the blue axis, the fan is spinning around it's own axis
• This axis is the orange line shown in fig.7.15
    ♦ It passes through the center of the fan
• The 'spin about this axis' is indicated by the orange circular arrow
3. We can write:
• The blue axis is the Global axis

• The orange axis is the Local axis
■ The fan is spinning about it's local axis and at the same time, moving around the global axis
4. In the the previous section,we saw that all the points on the 'axis of the rotating object' will be stationary
• Let us analyse our present case:
(i) The 'axis of the rotating object' in our present case is the orange axis
(ii) Only 'one point on this axis' is stationary
    ♦ It is the point at the intersection of the orange and blue axes
(iii) At any instant, the local axis of the fan passes through the point of intersection
(iv) The point of intersection of the two axes should not change
• If that point changes, the steps that we wrote above from (1) to (4) become invalid
• Because, then it would mean that, the fan has some 'translation' also. The blue line will not be the global axis any more

---

In the 3D animation in fig.7.16 below, a body is moving along a path

Fig.7.16

Let us write an analysis:
1. The path is a straight line. It is shown in magenta color
2. The motion of the body has one peculiarity:
■ At any instant, the 'center of mass' of the body lies on the magenta line
• This can be explained using the 7 steps given below:
(i) The body has it's own 3 local axes
(ii) The local z-axis of the body is shown to be straight up (the blue arrow) in the 3D view in fig.7.17 below:

Fig.7.17

• The local x (red arrow) and y (green arrow) axes are also shown
(iii) The direction of motion is indicated by the yellow arrow 
(iv)The 'center of mass' is the white sphere
(v) The 'center of mass' coincides with the origin O of the three axes
(vi) This 'center of mass' always lies on the magenta line
(vii) This is shown in the 2D side view in fig.7.18 below also:

Fig.7.18

3. In the fig.7.18, the different positions of 'center of mass' O are denoted as O_A, O_B, O_C so on . . .
• P is an arbitrary point on the local z-axis
    ♦ The different positions of P are denoted as P_A, P_B, P_C so on . . .
• We can write the following 3 facts:
(i) In fig.a, the body is at 'Position A'
■ Consider the instant at which the body just passes 'Position A'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(A)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(A)}}}$
    ♦ This velocity will be parallel to the path
• Since the two velocities are parallel, they have the same direction
• Since the body is rigid, the magnitudes will also be the same
• So we can write: $\mathbf\small{\vec{v}_{\text{P(A)}}=\vec{v}_{\text{O(A)}}}$
■ Now we need a prominent direction to fix up the orientation of the body
• Let us choose the 'horizontal'
• The angle which the local z-axis of the body (line O_AP_A) makes with the horizontal is denoted as α_(A) 
    ♦ This is shown separately in fig.a
(ii) In fig.b, the body is at 'Position B'
■ Consider the instant at which the body just passes 'Position B'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(B)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(B)}}}$
    ♦ This velocity will be parallel to the path
• Since the two velocities are parallel, they have the same direction
• Since the body is rigid, the magnitudes will also be the same
• So we can write: $\mathbf\small{\vec{v}_{\text{P(B)}}=\vec{v}_{\text{O(B)}}}$
■ The angle which local z-axis of the body (line O_BP_B) makes with the horizontal is denoted as α_(B) 
    ♦ This is shown separately in fig.b
(iii) In fig.c, the body is at 'Position C'
■ Consider the instant at which the body just passes 'Position C'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(C)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(C)}}}$
    ♦ This velocity will be parallel to the path
• Since the two velocities are parallel, they have the same direction
• Since the body is rigid, the magnitudes will also be the same
• So we can write: $\mathbf\small{\vec{v}_{\text{P(C)}}=\vec{v}_{\text{O(C)}}}$
■ The angle which local z-axis of the body (line O_CP_C) makes with the horizontal is denoted as α_(C) 
    ♦ This is shown separately in fig.c
Summary of the 3 facts can be written:
• At position A:
    ♦ Velocities of both O_A and P_A are the same
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(A) with the horizontal
• At position B:
    ♦ Velocities of both O_B and P_B are the same
    ♦ The local z-axis of the body (line O_BP_B) makes an angle of α_(B) with the horizontal
• At position C:
    ♦ Velocities of both O_C and P_C are the same
    ♦ The local z-axis of the body (line O_CP_C) makes an angle of α_(C) with the horizontal
• We see that: α_(A) = α_(B) = α_(C)
■ That means: The orientation of the body always remains the same

---

Let us check whether this is true when the path is curved:
Consider fig.7.19 below:

Fig.7.19

• Just as in the case of the straight line path, here also we can write the 3 facts
• Steps are the same. However, we will write them again
• The reader is advised to read all 3 steps carefully and become convinced that, they are all valid with respect to fig.7.19
(i) In fig.7.19(a), the body is at 'Position A'
■ Consider the instant at which the body just passes 'Position A'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(A)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(A)}}}$
    ♦ This velocity will be parallel to the path
• Since the two velocities are parallel, they have the same direction
• Since the body is rigid, the magnitudes will also be the same
• So we can write: $\mathbf\small{\vec{v}_{\text{P(A)}}=\vec{v}_{\text{O(A)}}}$
■ Now we need a prominent direction to fix up the orientation of the body
• Let us choose the 'horizontal'
• The angle which the local z-axis of the body (line O_AP_A) makes with the horizontal is denoted as α_(A) 
    ♦ This is shown separately in fig.a
(ii) In fig.b, the body is at 'Position B'
■ Consider the instant at which the body just passes 'Position B'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(B)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(B)}}}$
    ♦ This velocity will be parallel to the path
• Since the two velocities are parallel, they have the same direction
• Since the body is rigid, the magnitudes will also be the same
• So we can write: $\mathbf\small{\vec{v}_{\text{P(B)}}=\vec{v}_{\text{O(B)}}}$
■ The angle which local z-axis of the body (line O_BP_B) makes with the horizontal is denoted as α_(B) 
    ♦ This is shown separately in fig.b
(iii) In fig.c, the body is at 'Position C'
■ Consider the instant at which the body just passes 'Position C'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(C)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(C)}}}$
    ♦ This velocity will be parallel to the path
• Since the two velocities are parallel, they have the same direction
• Since the body is rigid, the magnitudes will also be the same
• So we can write: $\mathbf\small{\vec{v}_{\text{P(C)}}=\vec{v}_{\text{O(C)}}}$
■ The angle which local z-axis of the body (line O_CP_C) makes with the horizontal is denoted as α_(C) 
    ♦ This is shown separately in fig.c
Summary of the 3 facts can be written:
• At position A:
    ♦ Velocities of both O_A and P_A are the same
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(A) with the horizontal
• At position B:
    ♦ Velocities of both O_B and P_B are the same
    ♦ The local z-axis of the body (line O_BP_B) makes an angle of α_(B) with the horizontal
• At position C:
    ♦ Velocities of both O_C and P_C are the same
    ♦ The local z-axis of the body (line O_CP_C) makes an angle of α_(C) with the horizontal
• We see that: α_(A) = α_(B) = α_(C)

■ That means: The orientation of the body always remains the same

---

• The cases in both figs.7.18 and 7.19 that we saw above are pure translation. Now we will see another type:
• In the 3D animation in fig.7.20 below, a body is moving along a path

Fig.7.20

Let us write an analysis:
1. The path is a straight line. It is shown in magenta color
2. The motion of the body has two peculiarities:
(i) At any instant, the 'center of mass' of the body lies on the magenta line
(ii) The body is spinning about the local y-axis passing through the 'center of mass'
• These can be explained using the 9 steps given below:
(i) The body has it's own 3 local axes
(ii) The local z-axis of the body is shown to be straight up in the 3D view in fig.7.21 below:

Fig.7.21

• The local x (red arrow) and y (green arrow) axes are also shown 
(iii) The direction of motion is indicated by the yellow arrow
(iv) The direction of spin is indicated by the green curved arrow
(v) The body is spinning about the local y-axis 
(vi) The 'center of mass' is the white sphere
(vii) The 'center of mass' coincides with the origin O of the three axes
(viii) This 'center of mass' always lies on the magenta line
(ix) This is shown in the 2D side view in fig.7.22 below also:

Fig.7.22

3. The different positions of O are denoted as O_A, O_B, O_C so on . . .
• P is an arbitrary point in the body. It lies on the local z-axis
    ♦ The different positions of P are denoted as P_A, P_B, P_C so on . . .
• We can write the following 3 facts:
(i) In fig.a, the body is at 'Position A'
■ Consider the instant at which the body just passes 'Position A'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(A)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(A)}}}$
    ♦ This velocity will be different from $\mathbf\small{\vec{v}_{\text{O(A)}}}$
    ♦ This is because, the body has both translation and rotation
• So we can write: $\mathbf\small{\vec{v}_{\text{P(A)}}\neq\vec{v}_{\text{O(A)}}}$
■ Now we need a prominent direction to fix up the orientation of the body
• Let us choose the 'horizontal'
• The angle which the local z-axis of the body (line OP) makes with the horizontal is denoted as α_(A) 
    ♦ This is shown separately in fig.a
(ii) In fig.b, the body is at 'Position B'
■ Consider the instant at which the body just passes 'Position B'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(B)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(B)}}}$
    ♦ This velocity will be different from $\mathbf\small{\vec{v}_{\text{O(B)}}}$
    ♦ This is because, the body has both translation and rotation
• So we can write: $\mathbf\small{\vec{v}_{\text{P(B)}}\neq\vec{v}_{\text{O(B)}}}$
■ The angle which local z-axis of the body (line OP) makes with the horizontal is denoted as α_(B) 
    ♦ This is shown separately in fig.b
(iii) In fig.c, the body is at 'Position C'
■ Consider the instant at which the body just passes 'Position C'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(C)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(C)}}}$
    ♦ This velocity will be different from $\mathbf\small{\vec{v}_{\text{O(C)}}}$
    ♦ This is because, the body has both translation and rotation
• So we can write: $\mathbf\small{\vec{v}_{\text{P(C)}}\neq\vec{v}_{\text{O(C)}}}$
■ The angle which local z-axis of the body (line OP) makes with the horizontal is denoted as α_(C) 
    ♦ This is shown separately in fig.c
Summary of the 3 facts can be written:
• At position A:
    ♦ Velocities of O_A and P_A are different
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(A) with the horizontal
• At position B:
    ♦ Velocities of O_B and P_B are different
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(B) with the horizontal
• At position C:
    ♦ Velocities of O_C and P_C are different
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(C) with the horizontal
• We see that: α_(A) ≠ α_(B) ≠ α_(C)
■ That means: The orientation of the body always changes

---

Let us check whether this is true when the path is curved:

Consider fig.7.23 below:

Fig.7.23

• Just as in the case of the straight line path, here also we can write the 3 facts
• Steps are the same. However, we will write them again
• The reader is advised to read all those steps carefully and become convinced that, they are all valid with respect to fig.7.23
(i) In fig.7.23(a), the body is at 'Position A'
■ Consider the instant at which the body just passes 'Position A'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(A)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(A)}}}$
    ♦ This velocity will be different from $\mathbf\small{\vec{v}_{\text{O(A)}}}$
    ♦ This is because, the body has both translation and rotation
• So we can write: $\mathbf\small{\vec{v}_{\text{P(A)}}\neq\vec{v}_{\text{O(A)}}}$
■ Now we need a prominent direction to fix up the orientation of the body
• Let us choose the 'horizontal'
• The angle which the local z-axis of the body (line OP) makes with the horizontal is denoted as α_(A) 
    ♦ This is shown separately in fig.a
(ii) In fig.b, the body is at 'Position B'
■ Consider the instant at which the body just passes 'Position B'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(B)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(B)}}}$
    ♦ This velocity will be different from $\mathbf\small{\vec{v}_{\text{O(B)}}}$
    ♦ This is because, the body has both translation and rotation
• So we can write: $\mathbf\small{\vec{v}_{\text{P(B)}}\neq\vec{v}_{\text{O(B)}}}$
■ The angle which local z-axis of the body (line OP) makes with the horizontal is denoted as α_(B) 
    ♦ This is shown separately in fig.b
(iii) In fig.c, the body is at 'Position C'
■ Consider the instant at which the body just passes 'Position C'
• The instantaneous velocity of the 'particle at O' at that instant is $\mathbf\small{\vec{v}_{\text{O(C)}}}$
    ♦ This velocity will be parallel to the path
    ♦ Further, the 'particle at O' lies in the path
    ♦ So we can say: This velocity will be tangential to the path
• The instantaneous velocity of the 'particle at P' at that instant is $\mathbf\small{\vec{v}_{\text{P(C)}}}$
    ♦ This velocity will be different from $\mathbf\small{\vec{v}_{\text{O(C)}}}$
    ♦ This is because, the body has both translation and rotation
• So we can write: $\mathbf\small{\vec{v}_{\text{P(C)}}\neq\vec{v}_{\text{O(C)}}}$
■ The angle which local z-axis of the body (line OP) makes with the horizontal is denoted as α_(C) 
    ♦ This is shown separately in fig.c
Summary of the 3 facts can be written:
• At position A:
    ♦ Velocities of O_A and P_A are different
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(A) with the horizontal
• At position B:
    ♦ Velocities of O_B and P_B are different
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(B) with the horizontal
• At position C:
    ♦ Velocities of O_C and P_C are different
    ♦ The local z-axis of the body (line O_AP_A) makes an angle of α_(C) with the horizontal
• We see that: α_(A) ≠ α_(B) ≠ α_(C)
■ That means: The orientation of the body always changes

---

Thus we have two cases:
Case 1:
• This case is explained based on figs.7.16, 7.17, 7.18 and 7.19
• At any instant, all the particles of the body have the same velocity
• The orientation of the body remains the same
• This motion is pure translation
Case 2:
• This case is explained based on figs.7.20, 7.21, 7.22 and 7.23
• At any instant, various particles of the body have different velocities
• The orientation of the body changes continuously
• This motion is [Translation + Something else] 
• We see that, the difference between the two cases is:
• In case 2, there is rotational motion about a local axis
• So the 'something else' is: Rotational motion

---

In rotational motion, we saw two cases:
Case 1:
• Rotational motion in which every point in the axis is fixed
Examples: Ceiling fan, Potter's wheel
Case 2:
• Rotational motion in which only one point in the axis is fixed
Examples: Spinning top, Oscillating fan
• In this chapter, we will be dealing with case 1 only  

---

In the above discussion, we see that,'center of mass' has an important role to play. In the next section, we will see the methods to find it

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved