Question

The coordinates of the centre of mass of the following system are,

$\begin{aligned} & a)\dfrac{4}{3},\dfrac{8}{3} \\\ & b)\dfrac{8}{3},\dfrac{4}{3} \\\ & c)0,0 \\\ & d)\dfrac{3}{4},\dfrac{3}{8} \\\ \end{aligned}$

Answer 1

When two or more masses are present in a coordinate system, we need to find the respective x-coordinates and y-coordinates of the centre of mass. The distance of the masses from the origin can be calculated easily. Therefore, we can find the centre of mass coordinates in such a way.

Formula used:
$(x,y)=(\dfrac{\sum\limits_{x}^{{{x}_{2}}}{mx}}{\sum\limits_{m}^{{{m}_{2}}}{m}},\dfrac{\sum\limits_{y}^{{{y}_{2}}}{my}}{\sum\limits_{m}^{{{m}_{2}}}{m}})$

Complete step by step answer:
We were given the positions of the three masses and we need to find the centre of mass of the three masses. The distance of the three masses in x-coordinate and y-coordinate can be calculated from the graph. So now,
Let’s calculate the x coordinate of the centre of mass,

& {{x}_{com}}=\dfrac{\sum\limits_{x}^{{{x}_{2}}}{mx}}{\sum\limits_{m}^{{{m}_{2}}}{m}} \\\ & \Rightarrow {{x}_{com}}=\dfrac{(1\times 0m)+(2\times 1m)+(3\times 2m)}{1kg+2kg+3kg} \\\ & \therefore {{x}_{com}}=\dfrac{8}{6}m \\\ \end{aligned}$$ Similarly, let's find the y-coordinate of the centre of mass, $\begin{aligned} & {{y}_{com}}=\dfrac{\sum\limits_{y}^{{{y}_{2}}}{my}}{\sum\limits_{m}^{{{m}_{2}}}{m}} \\\ & \Rightarrow {{y}_{com}}=\dfrac{(1\times 0m)+(2\times 2m)+(3\times 4m)}{6kg} \\\ & \Rightarrow {{y}_{com}}=\dfrac{16}{6}m \\\ & \therefore {{y}_{com}}=\dfrac{8}{3}m \\\ \end{aligned}$ **Therefore, the centre of mass will be $(\dfrac{4}{3},\dfrac{8}{3})$.** **The correct option is option a.** **Additional information:** Centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particles appears to be concentrated. When we are studying the dynamics of the motion of the system of a particle as a whole, then we need not bother about the dynamics of individual particles of the system. But only focus on the dynamic of a unique point corresponding to that system. Motion of this unique point is identical to the motion of a single particle whose mass is equal to the sum of all individual particles of the system and the resultant of all the forces exerted on all the particles of the system by surrounding bodies (or) action of a field of force is exerted directly to that particle. This point is called the centre of mass of the system of particles. **Note:** In the above question, the masses given are point masses and the masses are assumed to be evenly distributed. If the masses are uneven, we have to integrate the mass and find the centre of mass. Also, the distance of the masses is all considered from the origin but not the nearest mass for them. Centre of mass position is calculated in the same way for all problems.