Chapter 7.5 - Center of mass of some common objects

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

Fig.7.36

• A 3D view is shown in fig.7.37 below:

Fig.7.37

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

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■ For all uniform bodies, the 'C' will be at the geometric center

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

Fig.7.38

(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

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Note: The five bodies that we saw above have uniform shape. Now consider the body shown in fig.7.39 below:

Fig.7.39

• 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

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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

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

Fig.7.40

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

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'

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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

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Chapter 7.4 - Center of Mass in Two-Dimensional Problems

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:

Fig.7.29

• 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:

Fig.7.30

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:

Fig.7.31

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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:

Fig.7.32

• The new coordinates are also shown
• The 3D view of this situation is shown below:

Fig.7.33

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:

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Thus we have seen 2 cases:

(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:

Fig.7.34

• 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:

Fig.7.35

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'

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In the next section, we will see the location of 'C' for some common objects

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