Question

A 1 m long rod having a constant cross sectional area is made of four materials. The first $0.2$ m are made of iron, the next $0.3$ m of lead, and the next $0.2$ m of aluminium, and the remaining is made of copper. Find the COM of the rod. The densities of iron, lead, aluminium and copper are $7.9 \times {10^3}kg/{m^3}$, $11.4 \times {10^3}kg/{m^3}$, $2.7 \times {10^3}kg/{m^3}$ and $8.9 \times {10^3}kg/{m^3}$, respectively.

Answer 1

The centre of mass of a system is a point where the whole mass of the body can be assumed to be concentrated. It is calculated depending on the masses and the positions of all the objects in a system.
Formula used: $X = \dfrac{{\sum {mx} }}{{\sum m }}$, where X is the position of the centre of mass, and m is the mass of the individual bodies at position x.

Complete step by step answer:
We have a rod made of different materials. To find the centre of mass of the rod, we can find the position of the centre of mass for all the different constituent materials as the point where all the mass of that section is concentrated. These individual centres of masses can then be used to calculate the centre of mass of the body as a whole.
The information provided to us includes:
Total length of the road $L = 1m$
Length of iron ${L_i} = 0.2m$
Length of lead ${L_l} = 0.3m$
Length of aluminium ${L_a} = 0.2m$
Length of copper ${L_c} = L - ({L_i} + {L_l} + {L_a}) = 1 - 0.7 = 0.3m$
Area of cross section is A
We know that the centre of mass for a uniform body lies at its centre. Hence, considering the four material blocks, their centre of masses will lie at their centre plus the distance from one end of the rod as:
Centre of mass of iron ${x_i} = 0.1m$
Centre of mass of lead ${x_l} = {L_i} + 0.15 = 0.35m$
Centre of mass of aluminium ${x_a} = {L_i} + {L_l} + 0.1 = 0.6m$
Centre of mass of copper ${x_c} = {L_i} + {L_l} + {L_a} + 0.15 = 0.85m$
Now, the centre of mass of the rod can be calculated as:
$X = \dfrac{{\sum {mx} }}{{\sum m }} = \dfrac{{{m_i}{x_i} + {m_l}{x_l} + {m_a}{x_a} + {m_c}{x_c}}}{{{m_l} + {m_l} + {m_a} + {m_c}}}$ [Eq. 1]
We know that the mass of an object depends upon its density as:
$m = {\text{density}} \times {\text{volume}} = {\text{density}} \times {\text{length}} \times {\text{area}}$
Since, the area of cross section is constant in our case, the mass will become:
$m = {\text{density}} \times {\text{length}}$
Substituting the respective density and length values in Eq. 1, gives us:
$X = \dfrac{{7.9 \times {{10}^3} \times 0.2{x_i} + 11.4 \times {{10}^3} \times 0.3{x_l} + 2.7 \times {{10}^3} \times 0.2{x_a} + 8.9 \times {{10}^3} \times 0.3{x_c}}}{{7.9 \times {{10}^3} \times 0.2 + 11.4 \times {{10}^3} \times 0.3 + 2.7 \times {{10}^3} \times 0.2 + 8.9 \times {{10}^3} \times 0.3}}$
Solving this further, we get:
$X = \dfrac{{1.58{x_i} + 3.42{x_l} + 0.54{x_a} + 2.67{x_c}}}{{1.58 + 3.42 + 0.54 + 2.67}}$
Now, we input the values of individual centre of masses to get:
$X = \dfrac{{1.58 \times 0.1 + 3.42 \times 0.35 + 0.54 \times 0.6 + 2.67 \times 0.85}}{{8.21}}$
$\Rightarrow X = \dfrac{{0.158 + 1.197 + 0.324 + 2.27}}{{8.21}} = 0.48m$
Hence, this is the position of the centre of mass from the left end of the rod.

Note:
As we saw in this question, the centre of mass makes it easier by simplifying the whole body as one mass acts at one point. Centre of mass is crucial in helping us balance systems. For humans, the centre of mass lies a bit below the navel, and hence when we stoop a little, we tend to lose balance as the centre of mass gets shifted.