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A thin uniform rod $50 \:\rm cm$ long has a mass of $100\:\rm g$. A metal hemisphere with a diameter of $10\:\rm cm$ and having a mass of $50\:\rm g$ is affixed to one end.

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Now I know that the centre of the mass (along the horizontal direction) will be at A, and the centre of mass of the hemisphere is at B.

Can I now find the centre of mass of the body by considering the system to be made of two particles: One of mass $100\:\rm g$ at A and the other of mass $50\:\rm g$ at B, since the centre of mass moves as if the entire mass of the system were concentrated there?

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Yes.

Think of it this way, if one isn't working with rigid bodies, then the center of mass can be considered as a point mass.

mathematically speaking:

$$x_{c.m,\,total}=\frac{M_{1}\,\frac{\sum_n(x_{i,1}\,m_{i,1})}{\sum_n(m_{i,1})}+M_{2}\,\frac{\sum_n(x_{i,2}\,m_{i,2})}{\sum_n(m_{i,2})}}{M_{total}}$$
where index $\square_{i,1}$ is for the rod and $\square_{i,2}$ is for the hemisphere, and $\square_i$ is for the entire body (a point or mass belonging to either components).

Notice that the denominator in both fractions is equal to the fraction's coefficient. Thus:

$$x_{c.m,\,total}=\frac{\sum_n(x_{i,1}\,m_{i,1}+x_{i,2}\,m_{i,2})}{\sum_n(m_i)}=\frac{\sum_nx_im_i}{\sum_nm_i}$$

So there you have it, the center of mass is equal to the one you'd get if you'd take the hemisphere and the rod to be point masses, located at their respective centers of mass.

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