In the previous section, we saw two-dimensional problems for finding 'C'. In this section, we will find the location of 'C' of some common objects.
1. Consider a uniform square rod. Some images can be seen here.
• We are given a piece of such a rod. We want to find it's 'C'
• For that, we place it along the x-axis as shown in fig.7.36(a) below
• It is placed at a distance of 's' from O
• A 3D view is shown in fig.7.37 below:
2. Now, let us divide the rod into 5 equal pieces
Let the length of each piece be 'l'. This is shown in fig.7.36(b)
• Let x1 be the distance of the 'center of the first piece' from O
♦ Clearly, x1 = s + 0.5l
♦ This is shown in fig.7.36(b)
• Let x2 be the distance of the 'center of the second piece' from O
♦ Clearly, x2 = s + l + 0.5l = s + 1.5l
♦ This is shown in fig.7.36(b)
• The distances for the next 3 pieces can also be written in this way:
• Let x3 be the distance of the 'center of the third piece' from O
♦ Clearly, x3 = s + 2l + 0.5l = s + 2.5l
• Let x4 be the distance of the 'center of the fourth piece' from O
♦ Clearly, x4 = s + 3l + 0.5l = s + 3.5l
• Let x5 be the distance of the 'center of the fifth piece' from O
♦ Clearly, x5 = s + 4l + 0.5l = s + 4.5l
3. We have: $\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}}$
• We need to know the masses. Remember that, it is a 'uniform square rod'
• So where ever we make a cut, the cross section will be a square
• The size of that square will be the same at every section
• So, since the lengths of the pieces are the same, we can write:
• The masses are all equal. That is:
m1 = m2 = m3 = m4 = m5 = m
4. So the numerator in (3) will be:
m(s + 0.5l) + m(s + 1.5l) + m(s + 2.5l) + m(s + 3.5l) + m(s + 4.5l)
= ms + 0.5ml + ms + 1.5ml + ms + 2.5ml + ms + 3.5ml + ms + 4.5ml
= 5ms + 12.5 ml
• The denominator will be 5m
• So we get:
$\mathbf\small{X=\frac{5ms+12.5ml}{5m}=\frac{5ms+2.5\times 5ml}{5m}=(s+2.5l)}$
• That means, the 'C' of the rod is situated at a distance of (s+2.5l) from O
5. Now, the total length of the rod is '5l'
• So '2.5l' indicates the geometric center of the rod
• Thus it is clear that, 'C' is at the geometric center of the rod. This is shown in fig.7.36(b)
6. If we arrange the axes in such a way that, 'O' coincides with the left end of the rod, calculations will become a little more easier
• This is because, in such an arrangement, 's' will be absent
Some examples are:
1. Lamina
(i) A lamina is a plane object with a 'very small uniform thickness'
• In fig.7.38(a) below, a rectangular lamina is shown
(ii) Since it is rectangular, it's opposite sides are equal.
• Also since it is a lamina, the thickness is uniform
(iii) The geometric center can be easily calculated in such cases
• The 'C' will be same as the geometric center. it is shown as a small white sphere
2. Cube
(i) Fig.b shows a cube. In the fig., the material of the cube is assumed to be transparent. So we can see the interior
(ii) The geometric center of the cube can be determined using diagonals from opposite corners
(iii) If the 'distribution of particles in the cube' is uniform, the geometric center will be the 'C'
3. Cylinder
(i) Fig.c shows a cylinder. In the fig., the material of the cylinder is assumed to be transparent. So we can see the interior
(ii) The definition of the cylinder tells us that, the top and bottom must be 'circles of the same radius'. Then only we can call it a cylinder
(iii) The geometric center will be on the axis of the cylinder. Also it will be equi-distant from top and bottom
(iv) If the 'distribution of particles in the cylinder' is uniform, the geometric center will be the 'C'
4. Circular disc
(i) Fig.d shows a circular disc. It is thicker than a lamina. In the fig., the material of the disc cylinder is assumed to be transparent. So we can see the interior
(ii) A circular disc is same as a cylinder of small thickness
(iii) The geometric center will be on the axis of the disc. Also it will be equi-distant from top and bottom
(iv) If the 'distribution of particles in the disc' is uniform, the geometric center will be the 'C'
5. Circular ring
(i) Fig.e shows a circular ring
(ii) It is clear that, the geometric center will be on the axis. And also equi-distant from top and bottom
(iii) If the 'distribution of particles in the ring' is uniform, the geometric center will be the 'C'
• Note that, in this case, the material need not be transparent for us to see the interior
1. Consider a uniform square rod. Some images can be seen here.
• We are given a piece of such a rod. We want to find it's 'C'
• For that, we place it along the x-axis as shown in fig.7.36(a) below
• It is placed at a distance of 's' from O
 |
| Fig.7.36 |
 |
| Fig.7.37 |
Let the length of each piece be 'l'. This is shown in fig.7.36(b)
• Let x1 be the distance of the 'center of the first piece' from O
♦ Clearly, x1 = s + 0.5l
♦ This is shown in fig.7.36(b)
• Let x2 be the distance of the 'center of the second piece' from O
♦ Clearly, x2 = s + l + 0.5l = s + 1.5l
♦ This is shown in fig.7.36(b)
• The distances for the next 3 pieces can also be written in this way:
• Let x3 be the distance of the 'center of the third piece' from O
♦ Clearly, x3 = s + 2l + 0.5l = s + 2.5l
• Let x4 be the distance of the 'center of the fourth piece' from O
♦ Clearly, x4 = s + 3l + 0.5l = s + 3.5l
• Let x5 be the distance of the 'center of the fifth piece' from O
♦ Clearly, x5 = s + 4l + 0.5l = s + 4.5l
3. We have: $\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}}$
• We need to know the masses. Remember that, it is a 'uniform square rod'
• So where ever we make a cut, the cross section will be a square
• The size of that square will be the same at every section
• So, since the lengths of the pieces are the same, we can write:
• The masses are all equal. That is:
m1 = m2 = m3 = m4 = m5 = m
4. So the numerator in (3) will be:
m(s + 0.5l) + m(s + 1.5l) + m(s + 2.5l) + m(s + 3.5l) + m(s + 4.5l)
= ms + 0.5ml + ms + 1.5ml + ms + 2.5ml + ms + 3.5ml + ms + 4.5ml
= 5ms + 12.5 ml
• The denominator will be 5m
• So we get:
$\mathbf\small{X=\frac{5ms+12.5ml}{5m}=\frac{5ms+2.5\times 5ml}{5m}=(s+2.5l)}$
• That means, the 'C' of the rod is situated at a distance of (s+2.5l) from O
5. Now, the total length of the rod is '5l'
• So '2.5l' indicates the geometric center of the rod
• Thus it is clear that, 'C' is at the geometric center of the rod. This is shown in fig.7.36(b)
6. If we arrange the axes in such a way that, 'O' coincides with the left end of the rod, calculations will become a little more easier
• This is because, in such an arrangement, 's' will be absent
■ For all uniform bodies, the 'C' will be at the geometric center
1. Lamina
(i) A lamina is a plane object with a 'very small uniform thickness'
• In fig.7.38(a) below, a rectangular lamina is shown
 |
| Fig.7.38 |
• Also since it is a lamina, the thickness is uniform
(iii) The geometric center can be easily calculated in such cases
• The 'C' will be same as the geometric center. it is shown as a small white sphere
2. Cube
(i) Fig.b shows a cube. In the fig., the material of the cube is assumed to be transparent. So we can see the interior
(ii) The geometric center of the cube can be determined using diagonals from opposite corners
(iii) If the 'distribution of particles in the cube' is uniform, the geometric center will be the 'C'
3. Cylinder
(i) Fig.c shows a cylinder. In the fig., the material of the cylinder is assumed to be transparent. So we can see the interior
(ii) The definition of the cylinder tells us that, the top and bottom must be 'circles of the same radius'. Then only we can call it a cylinder
(iii) The geometric center will be on the axis of the cylinder. Also it will be equi-distant from top and bottom
(iv) If the 'distribution of particles in the cylinder' is uniform, the geometric center will be the 'C'
4. Circular disc
(i) Fig.d shows a circular disc. It is thicker than a lamina. In the fig., the material of the disc cylinder is assumed to be transparent. So we can see the interior
(ii) A circular disc is same as a cylinder of small thickness
(iii) The geometric center will be on the axis of the disc. Also it will be equi-distant from top and bottom
(iv) If the 'distribution of particles in the disc' is uniform, the geometric center will be the 'C'
5. Circular ring
(i) Fig.e shows a circular ring
(ii) It is clear that, the geometric center will be on the axis. And also equi-distant from top and bottom
(iii) If the 'distribution of particles in the ring' is uniform, the geometric center will be the 'C'
• Note that, in this case, the material need not be transparent for us to see the interior
Note: The five bodies that we saw above have uniform shape. Now consider the body shown in fig.7.39 below:
• The cross section of this body will be different at different points along it's length
• We can cut it into pieces of the same length 'l'. But since the cross sections are different, the masses of those pieces will not be the same
• In such cases, the location of 'C' can be easily calculated using principles of calculus
Solved example 7.4
Find the center of mass of a L-shaped lamina with dimensions as shown in fig.7.40(a) below. The mass of the lamina is 3 kg
Solution:
1. First we need to fix up a reference frame
• It will be most convenient if the x and y axes coincide with the edges of the lamina
• Such an arrangement is shown in fig.b
2. Based on this arrangement, the coordinates of the vertices (O, P, Q, R, S and T) can be easily written
• They are shown in fig.b
3. The measurements of the given lamina are in such a way that, it can be effectively divided into 3 equal squares. This is shown in fig.c
• Based on the coordinates written in fig.b, the coordinates of the center points (C1, C2 and C3) of the 3 squares can also be written. They are shown in fig.c
• Those coordinates will directly give us the distances of the squares from the two axes
4. Next, we want the masses
• Given that, the total mass of the lamina is 3 kg
• Since the lamina is uniform, each of the 3 squares will be having same mass
• Thus we get: Mass of each square = $\mathbf\small{Z=\frac{3\,(\text{kg})}{3}= 1\, \text{kg}}$
5. Now we can form the table:
• The first 3 columns can be filled up using the given data
• From the table, we get:
$\mathbf\small{\sum{m_i}=3}$
$\mathbf\small{\sum{m_ix_i}=2.5}$
$\mathbf\small{\sum{m_iy_i}=2.5}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{2.5}{3}=0.83\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{2.5}{3}=0.83\, \text{m}}$
Solved example 7.5
Find the center of mass of a T-shaped lamina with dimensions as shown in fig.7.40(a) below
Solution:
1. First we need to fix up a reference frame
• It will be most convenient if the x and y axes coincide with the edges of the lamina
• Also the T-shaped lamina can be divided into simple rectangles
• Such an arrangement is shown in fig.b
2. There are 3 rectangles and 1 void circle
• Let the centers of these components be C1, C2, C3, and C4
• We can easily write the coordinates of those centers. We have:
C1(6.5,18), C2(6.5,12), C3(6.5,8.75), and C4(6.5,4)
3. Next we want the masses of those rectangles and the circle
• Given that, the lamina is uniform. So every 'one square meter area' of the lamina will be having the same mass
• Let the mass of 'one square meter area' of the lamina be 'k' kg
• Then we get:
♦ Mass of 1 = (Area of 1) × k = (13×4) × k = 52k
♦ Mass of 2 = (Area of 2) × k = (6×8) × k = 48k
♦ Mass of 3 = (Area of 3) × k = (7.1) × k = 7.1k
♦ Mass of 4 = (Area of 4) × k = (2×8) × k = 16k
4. Now we can prepare the table:
• From the table, we get:
$\mathbf\small{\sum{m_i}=108.9k}$
$\mathbf\small{\sum{m_ix_i}=707.9k}$
$\mathbf\small{\sum{m_iy_i}=1513.9k}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{707.9k}{108.9k}=6.5\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{1513.9k}{108.9k}=13.9\, \text{m}}$
This is shown in fig.c
Solved example 7.6
Find the center of mass of a triangular lamina
Solution:
1. Fig.7.41(a) below shows a triangular lamina PQR
• Let us divide it into strips parallel to the base PQ
• We want the center of each of those strips
2. Three of those strips are shown in fig.b
• They are given three different colors: green, red and yellow
• Clearly, they are trapeziums. Because, top and bottom sides are parallel
• But they are not isosceles-trapeziums
• Since they are not isosceles-trapeziums, it is not very easy to find the centers of those strips
3. So we adopt another method:
• We decrease the heights of those trapeziums
• We decrease the heights to such a low level that, they almost become lines
• But they cannot become exact lines. Because, if they do, then height is zero and the area becomes zero
■ So we can write:
• The lamina is divided into strips of very small heights
• The heights tends to zero, but never becomes zero
4. In such a situation, those strips looks like lines
• And the 'midpoint of the strips' can be considered to be same as 'the midpoints of those lines'
5. Now we join the midpoints of those thin strips (lines). This is shown in fig.c
• The white line joins the midpoints
• This white line passes through two important points:
♦ The apex
♦ The midpoint of the base
• So the white line is a median
6. Let us write a summary of the steps written so far. It can be written in 4 steps:
(i) The lamina is divided into thin strips parallel to base PQ
(ii) The 'C' of any one of those strips is at the 'midpoint of the line representing that strip'
(iii) The 'C' of all the strips falls on a median
(iv) So the 'C' of the total lamina must be somewhere on that median
7. But saying 'somewhere on the median' is not good enough. We want the exact point
• So we repeat the step by taking another vertex as the apex. This is shown in fig.d
• Q is chosen as the apex and PR is the base
• The median thus obtained is shown in black color
• The 'C' of the lamina must be somewhere on this black median
8. The two medians are shown together in fig.e
• The 'C' of the lamina must be on both the medians
• So it will be the point of intersection of the two medians
• But the point of intersection of the medians of a triangle is the centroid 'G' of that triangle
■ So we can write:
For a triangular lamina, 'C' coincides with 'G'
 |
| Fig.7.39 |
• We can cut it into pieces of the same length 'l'. But since the cross sections are different, the masses of those pieces will not be the same
• In such cases, the location of 'C' can be easily calculated using principles of calculus
We have seen the properties of lamina. We will now see some solved examples based on those properties
Solved example 7.4
Find the center of mass of a L-shaped lamina with dimensions as shown in fig.7.40(a) below. The mass of the lamina is 3 kg
 |
| Fig.7.40 |
1. First we need to fix up a reference frame
• It will be most convenient if the x and y axes coincide with the edges of the lamina
• Such an arrangement is shown in fig.b
2. Based on this arrangement, the coordinates of the vertices (O, P, Q, R, S and T) can be easily written
• They are shown in fig.b
3. The measurements of the given lamina are in such a way that, it can be effectively divided into 3 equal squares. This is shown in fig.c
• Based on the coordinates written in fig.b, the coordinates of the center points (C1, C2 and C3) of the 3 squares can also be written. They are shown in fig.c
• Those coordinates will directly give us the distances of the squares from the two axes
4. Next, we want the masses
• Given that, the total mass of the lamina is 3 kg
• Since the lamina is uniform, each of the 3 squares will be having same mass
• Thus we get: Mass of each square = $\mathbf\small{Z=\frac{3\,(\text{kg})}{3}= 1\, \text{kg}}$
5. Now we can form the table:
• From the table, we get:
$\mathbf\small{\sum{m_i}=3}$
$\mathbf\small{\sum{m_ix_i}=2.5}$
$\mathbf\small{\sum{m_iy_i}=2.5}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{2.5}{3}=0.83\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{2.5}{3}=0.83\, \text{m}}$
Solved example 7.5
Find the center of mass of a T-shaped lamina with dimensions as shown in fig.7.40(a) below
 |
| Fig.7.40 |
1. First we need to fix up a reference frame
• It will be most convenient if the x and y axes coincide with the edges of the lamina
• Also the T-shaped lamina can be divided into simple rectangles
• Such an arrangement is shown in fig.b
2. There are 3 rectangles and 1 void circle
• Let the centers of these components be C1, C2, C3, and C4
• We can easily write the coordinates of those centers. We have:
C1(6.5,18), C2(6.5,12), C3(6.5,8.75), and C4(6.5,4)
3. Next we want the masses of those rectangles and the circle
• Given that, the lamina is uniform. So every 'one square meter area' of the lamina will be having the same mass
• Let the mass of 'one square meter area' of the lamina be 'k' kg
• Then we get:
♦ Mass of 1 = (Area of 1) × k = (13×4) × k = 52k
♦ Mass of 2 = (Area of 2) × k = (6×8) × k = 48k
♦ Mass of 3 = (Area of 3) × k = (7.1) × k = 7.1k
♦ Mass of 4 = (Area of 4) × k = (2×8) × k = 16k
4. Now we can prepare the table:
$\mathbf\small{\sum{m_i}=108.9k}$
$\mathbf\small{\sum{m_ix_i}=707.9k}$
$\mathbf\small{\sum{m_iy_i}=1513.9k}$
• Thus we get:
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{707.9k}{108.9k}=6.5\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{1513.9k}{108.9k}=13.9\, \text{m}}$
This is shown in fig.c
Solved example 7.6
Find the center of mass of a triangular lamina
Solution:
1. Fig.7.41(a) below shows a triangular lamina PQR
 |
| Fig.7.41 |
• Let us divide it into strips parallel to the base PQ
• We want the center of each of those strips
2. Three of those strips are shown in fig.b
• They are given three different colors: green, red and yellow
• Clearly, they are trapeziums. Because, top and bottom sides are parallel
• But they are not isosceles-trapeziums
• Since they are not isosceles-trapeziums, it is not very easy to find the centers of those strips
3. So we adopt another method:
• We decrease the heights of those trapeziums
• We decrease the heights to such a low level that, they almost become lines
• But they cannot become exact lines. Because, if they do, then height is zero and the area becomes zero
■ So we can write:
• The lamina is divided into strips of very small heights
• The heights tends to zero, but never becomes zero
4. In such a situation, those strips looks like lines
• And the 'midpoint of the strips' can be considered to be same as 'the midpoints of those lines'
5. Now we join the midpoints of those thin strips (lines). This is shown in fig.c
• The white line joins the midpoints
• This white line passes through two important points:
♦ The apex
♦ The midpoint of the base
• So the white line is a median
6. Let us write a summary of the steps written so far. It can be written in 4 steps:
(i) The lamina is divided into thin strips parallel to base PQ
(ii) The 'C' of any one of those strips is at the 'midpoint of the line representing that strip'
(iii) The 'C' of all the strips falls on a median
(iv) So the 'C' of the total lamina must be somewhere on that median
7. But saying 'somewhere on the median' is not good enough. We want the exact point
• So we repeat the step by taking another vertex as the apex. This is shown in fig.d
• Q is chosen as the apex and PR is the base
• The median thus obtained is shown in black color
• The 'C' of the lamina must be somewhere on this black median
8. The two medians are shown together in fig.e
• The 'C' of the lamina must be on both the medians
• So it will be the point of intersection of the two medians
• But the point of intersection of the medians of a triangle is the centroid 'G' of that triangle
■ So we can write:
For a triangular lamina, 'C' coincides with 'G'
We completed a discussion on two cases:
(i) All the bodies lie on the x-axis
(ii) All the bodies are distributed in the xy-plane
In the next section, we will see the cases where the bodies are distributed in space
(i) All the bodies lie on the x-axis
(ii) All the bodies are distributed in the xy-plane
In the next section, we will see the cases where the bodies are distributed in space
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