In the previous section, we saw a 3D view in fig.7.26 where all the bodies in the system lie on the x-axis. In this section, we will see the more advanced case where the bodies are distributed in the xy-plane.
1. Consider the sphere P in the fig.7.29(a) below:
• It lies in the xy-plane
♦ It is at a distance of 4 m from the y-axis
♦ At the same time, it is at a distance of 3 m from the x-axis
• We can say: The coordinates of P are (4,3)
2. Similarly, sphere Q also lies in the xy-plane
• It's coordinates are (7,2)
• The 3D view is shown in fig.7.30 below:
3. P has a mass of 15 kg and Q has a mass of 19 kg
■ We want to find the location of the 'C' of this system of two spheres
4. In such cases, the 'C' lies at a point whose coordinates are (X,Y)
• We already know the method to find 'X'
5. We apply the same method to find 'Y'
• But to find 'Y', the method should be applied in the y-direction
• The following table illustrates the procedure:
• The first 3 columns can be filled up using the given data
6. From the table, we get:
$\mathbf\small{\sum{m_i}=34}$
$\mathbf\small{\sum{m_ix_i}=193}$
$\mathbf\small{\sum{m_iy_i}=83}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{193}{34}=5.68\, \text{m}}$
7. We can use a similar formula to calculate 'Y'
We can write a new equation:
Eq.7.2: $\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}}$
• Thus we get:
$\mathbf\small{Y=\frac{83}{34}=2.44\, \text{m}}$
• The coordinates (X,Y) = (5.68,2.44) will give the location of 'C'
• This is shown in fig.7.29(b)
8. If we connect the two spheres by an uniform rod and put a support below 'C', the system will balance. This is shown in the 3D view below:
(i) The bodies lie in the x-axis
♦ In this case we use one equation: Eq.7.1
♦ This is a one-dimensional problem
(ii) The bodies are distributed in the xy-plane
♦ In this case we use two equations: Eqs. 7.1 and 7.2
♦ This is a two-dimensional problem
Let us see some solved examples:
Solved example 7.2
Three spheres P, Q and R of masses m1, m2 and m3 are randomly placed on the xy-plane. Find the location of 'C' of that system of 3 spheres
Solution:
1. Any 'three points on a plane' will give a triangle
• It need not be an equilateral triangle
• It need not be an isosceles triangle
• It need not be a right triangle
• It need not be any special type of triangle
• What is important is that, we do get a triangle
2. Since they are randomly placed, we can assume their coordinates to be: (x1,y1), (x2,y2) and (x3,y3)
• A sample is shown in fig.7.34(a) below:
• A triangle is formed by joining the centers of the spheres
• Note that, since it is given that the spheres lie on the xy-plane, there will be no z-coordinates
3. From the coordinates, we get the distances from the x and y axes
• So we can directly write the answer:
■ 'C' of the system of 3 spheres lies on the xy-plane. The location of the 'C' is given by the coordinates (X,Y)
Where:
X = $\mathbf\small{\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}}$
Y = $\mathbf\small{\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3}}$
4. A special case:
• Suppose, the masses of the 3 spheres are equal, then we have: m1 = m2 = m3 = m
• Then the results in (3) become:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{m\, x_1+m\, x_2+m\, x_3}{m+m+m}=\frac{m(x_1+x_2+x_3)}{3m}=\frac{x_1+x_2+x_3}{3}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{m\, y_1+m\, y_2+m\, y_3}{m+m+m}=\frac{m(y_1+y_2+y_3)}{3m}=\frac{y_1+y_2+y_3}{3}}$
5. Let us connect the centers of the three equal spheres by straight lines. This is shown in fig.7.34(b)
• Now we have a triangle whose vertex coordinates are (x1,y1), (x2,y2) and (x3,y3)
• In our earlier maths classes, we have seen that, if the vertices of a triangle are (x1,y1), (x2,y2) and (x3,y3), then the coordinates of the centroid of that triangle are:
$\mathbf\small{\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}}$
6. Comparing the results in (4) and (5), we can write a summary about this special case:
• Three spheres lie on the xy-plane
• There masses are equal
• Imagine a triangle formed by joining the centers of those spheres
• That triangle will be having a centroid
• That centroid will be the location of the 'C' of the 'system of three spheres'
• But always remember that, this special case will work only if the spheres have the same mass
Solved example 7.3
Three particles P, Q and R are situated at the vertices of an equilateral triangle of side 0.5 m. Their masses are 100 g, 150 g and 200 g respectively. Find the location of the 'C' of this system of 3 particles
Solution:
1. We have three particles so situated in space that, they are at the vertices of an equilateral triangle
• For ease in calculations, we arrange the frame of reference in the following way:
♦ The plane of the triangle lies in the xy-plane
♦ One of the sides (say PQ), lies on the x-axis
♦ The left vertex (P) of that side coincides with O
• This arrangement is shown in fig.7.35 below:
2. Once this arrangement is fixed, we can easily write the coordinates of the bottom vertices
• The coordinates of P will be (0,0)
• The coordinates of Q will be (0.5,0)
• The x-coordinate of R will be 0.25
3. To find the y-coordinate of R, we use the following steps:
• Drop a perpendicular from R. This is shown in fig.b
• The foot of this perpendicular is R'
• In the right triangle PR'R, we have:
$\mathbf\small{\sin 60=\frac{RR'}{PR}=\frac{RR'}{0.5}}$
$\mathbf\small{\Rightarrow RR'=\sin 60 \times 0.5=0.433}$
So the y-coordinate of R is 0.433
4. Now we can prepare the table:
• The first 3 columns can be filled up using the given data
• From the table, we get:
$\mathbf\small{\sum{m_i}=0.45}$
$\mathbf\small{\sum{m_ix_i}=0.125}$
$\mathbf\small{\sum{m_iy_i}=0.0866}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{0.125}{0.45}=0.278\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{0.0866}{0.45}=0.192\, \text{m}}$
5. Note that, if the masses were equal, we could have used the result from the special case that we saw in solved example 7.2
• But since the masses are different, the 'C' obtained in (4) will be different from the 'centroid of the triangle'
1. Consider the sphere P in the fig.7.29(a) below:
 |
| Fig.7.29 |
♦ It is at a distance of 4 m from the y-axis
♦ At the same time, it is at a distance of 3 m from the x-axis
• We can say: The coordinates of P are (4,3)
2. Similarly, sphere Q also lies in the xy-plane
• It's coordinates are (7,2)
• The 3D view is shown in fig.7.30 below:
 |
| Fig.7.30 |
■ We want to find the location of the 'C' of this system of two spheres
4. In such cases, the 'C' lies at a point whose coordinates are (X,Y)
• We already know the method to find 'X'
5. We apply the same method to find 'Y'
• But to find 'Y', the method should be applied in the y-direction
• The following table illustrates the procedure:
6. From the table, we get:
$\mathbf\small{\sum{m_i}=34}$
$\mathbf\small{\sum{m_ix_i}=193}$
$\mathbf\small{\sum{m_iy_i}=83}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{193}{34}=5.68\, \text{m}}$
7. We can use a similar formula to calculate 'Y'
We can write a new equation:
Eq.7.2: $\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}}$
• Thus we get:
$\mathbf\small{Y=\frac{83}{34}=2.44\, \text{m}}$
• The coordinates (X,Y) = (5.68,2.44) will give the location of 'C'
• This is shown in fig.7.29(b)
8. If we connect the two spheres by an uniform rod and put a support below 'C', the system will balance. This is shown in the 3D view below:
 |
| Fig.7.31 |
An interesting result:
1. We now know the position of 'C' of the system of two spheres P and Q in fig.7.29
• Let us rearrange the 'frame of reference' in such a way that, both the x and y axes passes through 'C'
■ If both the x and y axes are to pass through 'C', the origin O must coincide with 'C'
• This is shown in fig.7.32 below:
• The new coordinates are also shown
• The 3D view of this situation is shown below:
2. The sphere P lies on the left side of y-axis
• Also it is above the x-axis
♦ So the x-coordinate of P will be negative
♦ And it's y-coordinate will be positive
3. The sphere Q lies on the right side of y-axis
• Also it is below the x-axis
♦ So the x-coordinate of Q will be positive
♦ And it's y-coordinate will be negative
4. Thus we get: $\mathbf\small{\sum{m_ix_i}=m_1 x_1+m_2 x_2}$
= [(15×-1.68)+(19×1.32)] = [-25.1+25.1] = 0
• Also we get: $\mathbf\small{\sum{m_iy_i}=m_1 y_1+m_2 y_2}$
= [(15×0.56)+(19×-0.44)] = [8.4-8.4] = 0
5. We get the same result quickly if we write the values in a tabular form:
• Let us rearrange the 'frame of reference' in such a way that, both the x and y axes passes through 'C'
■ If both the x and y axes are to pass through 'C', the origin O must coincide with 'C'
• This is shown in fig.7.32 below:
 |
| Fig.7.32 |
• The 3D view of this situation is shown below:
 |
| Fig.7.33 |
• Also it is above the x-axis
♦ So the x-coordinate of P will be negative
♦ And it's y-coordinate will be positive
3. The sphere Q lies on the right side of y-axis
• Also it is below the x-axis
♦ So the x-coordinate of Q will be positive
♦ And it's y-coordinate will be negative
4. Thus we get: $\mathbf\small{\sum{m_ix_i}=m_1 x_1+m_2 x_2}$
= [(15×-1.68)+(19×1.32)] = [-25.1+25.1] = 0
• Also we get: $\mathbf\small{\sum{m_iy_i}=m_1 y_1+m_2 y_2}$
= [(15×0.56)+(19×-0.44)] = [8.4-8.4] = 0
5. We get the same result quickly if we write the values in a tabular form:
Thus we have seen 2 cases:
♦ In this case we use one equation: Eq.7.1
♦ This is a one-dimensional problem
(ii) The bodies are distributed in the xy-plane
♦ In this case we use two equations: Eqs. 7.1 and 7.2
♦ This is a two-dimensional problem
Let us see some solved examples:
Solved example 7.2
Three spheres P, Q and R of masses m1, m2 and m3 are randomly placed on the xy-plane. Find the location of 'C' of that system of 3 spheres
Solution:
1. Any 'three points on a plane' will give a triangle
• It need not be an equilateral triangle
• It need not be an isosceles triangle
• It need not be a right triangle
• It need not be any special type of triangle
• What is important is that, we do get a triangle
2. Since they are randomly placed, we can assume their coordinates to be: (x1,y1), (x2,y2) and (x3,y3)
• A sample is shown in fig.7.34(a) below:
 |
| Fig.7.34 |
• Note that, since it is given that the spheres lie on the xy-plane, there will be no z-coordinates
3. From the coordinates, we get the distances from the x and y axes
• So we can directly write the answer:
■ 'C' of the system of 3 spheres lies on the xy-plane. The location of the 'C' is given by the coordinates (X,Y)
Where:
X = $\mathbf\small{\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}}$
Y = $\mathbf\small{\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3}}$
4. A special case:
• Suppose, the masses of the 3 spheres are equal, then we have: m1 = m2 = m3 = m
• Then the results in (3) become:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{m\, x_1+m\, x_2+m\, x_3}{m+m+m}=\frac{m(x_1+x_2+x_3)}{3m}=\frac{x_1+x_2+x_3}{3}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{m\, y_1+m\, y_2+m\, y_3}{m+m+m}=\frac{m(y_1+y_2+y_3)}{3m}=\frac{y_1+y_2+y_3}{3}}$
5. Let us connect the centers of the three equal spheres by straight lines. This is shown in fig.7.34(b)
• Now we have a triangle whose vertex coordinates are (x1,y1), (x2,y2) and (x3,y3)
• In our earlier maths classes, we have seen that, if the vertices of a triangle are (x1,y1), (x2,y2) and (x3,y3), then the coordinates of the centroid of that triangle are:
$\mathbf\small{\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}}$
6. Comparing the results in (4) and (5), we can write a summary about this special case:
• Three spheres lie on the xy-plane
• There masses are equal
• Imagine a triangle formed by joining the centers of those spheres
• That triangle will be having a centroid
• That centroid will be the location of the 'C' of the 'system of three spheres'
• But always remember that, this special case will work only if the spheres have the same mass
Solved example 7.3
Three particles P, Q and R are situated at the vertices of an equilateral triangle of side 0.5 m. Their masses are 100 g, 150 g and 200 g respectively. Find the location of the 'C' of this system of 3 particles
Solution:
1. We have three particles so situated in space that, they are at the vertices of an equilateral triangle
• For ease in calculations, we arrange the frame of reference in the following way:
♦ The plane of the triangle lies in the xy-plane
♦ One of the sides (say PQ), lies on the x-axis
♦ The left vertex (P) of that side coincides with O
• This arrangement is shown in fig.7.35 below:
 |
| Fig.7.35 |
• The coordinates of P will be (0,0)
• The coordinates of Q will be (0.5,0)
• The x-coordinate of R will be 0.25
3. To find the y-coordinate of R, we use the following steps:
• Drop a perpendicular from R. This is shown in fig.b
• The foot of this perpendicular is R'
• In the right triangle PR'R, we have:
$\mathbf\small{\sin 60=\frac{RR'}{PR}=\frac{RR'}{0.5}}$
$\mathbf\small{\Rightarrow RR'=\sin 60 \times 0.5=0.433}$
So the y-coordinate of R is 0.433
4. Now we can prepare the table:
• From the table, we get:
$\mathbf\small{\sum{m_i}=0.45}$
$\mathbf\small{\sum{m_ix_i}=0.125}$
$\mathbf\small{\sum{m_iy_i}=0.0866}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{0.125}{0.45}=0.278\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{0.0866}{0.45}=0.192\, \text{m}}$
5. Note that, if the masses were equal, we could have used the result from the special case that we saw in solved example 7.2
• But since the masses are different, the 'C' obtained in (4) will be different from the 'centroid of the triangle'
In the next section, we will see the location of 'C' for some common objects
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