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
• 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'
• 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
• 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
• 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:
$\mathbf\small{X=\frac{m_1 x_1+m_2 x_2+\,.\, .\,.\,+m_n x_n}{m_1 + m_2+\,.\, .\,.\,+m_n}=\frac{\sum{m_ix_i} }{\sum{m_i} }=\frac{\sum{m_ix_i} }{M}}$
• 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
Solved example 7.1
Three spheres P, Q and R are placed on the x-axis as shown in fig.7.28(a) below:
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
• If the y-axis passes through 'C', then $\mathbf\small{\sum{m_ix_i}=0}$
1. In fig.7.24 below, two points P and Q lie on the x-axis
 |
| Fig.7.24 |
• 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:
$\mathbf\small{\frac{158 + 158 + 158 + 164 + 164 + 164 + 164 + 167 + 167}{\text{9}}}$
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'
• 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 |
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:
• A 3D view of that system is shown in fig.7.26 below:
 |
| Fig.7.26 |
• 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:
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)
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 |
• 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
Three spheres P, Q and R are placed on the x-axis as shown in fig.7.28(a) below:
 |
| Fig.7.28 |
Solution:
1. The calculations can be done easily if we arrange the data in a tabular form:
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:
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
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