Question

An infinite number of bricks are placed one over the other as shown in the figure. Each succeeding brick having
half the length and breadth of its preceding brick and the mass of each succeeding bricks being
1/4th of the preceding one, take 'O' as the origin, the x-coordinate of centre of mass of the system of bricks is:-

:​

Answer 1

Answer:

$\frac{3a}{7}$

Explanation:

The figure is attached

The centre of mass of each blocks will lie in the middle

Let the mass of the lower most brick be M

The centre of mass will act in the middle of the blocks

Let the length of this block is a

The horizontal distance of centre of mass of M from O = $\frac{L}{2}$

Similarly the mass of the block just above it will be $\frac{M}{4}$

The horizontal distance of centre of mass of M/2 from O = $\frac{L}{4}$

Similarly for the other blocks mass and distance can be calculated

Sum of all the masses

$M'=M+\frac{M}{4}+\frac{M}{16}+\frac{M}{64}+.........$

$\implies M'=M(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.........$

$\implies M'=M(\frac{1}{1-(1/4)}$         (sum of infinite GP)

$\implies M'=M(\frac{4}{3}$  

Therefore,

$x_{com}=\frac{\Sigma Mx}{M'}$

or, $M'x_{com}=\Sigma Mx$

$\implies M'x_{com}=M\times \frac{a}{2}+\frac{M}{4}\times\frac{a}{4}+\frac{M}{4^2}\times\frac{a}{8}+.....$

$\implies M'x_{com}=\frac{Ma}{2}[1+\frac{1}{8}+\frac{1}{64}+.....$

$\implies M'x_{com}=\frac{Ma}{2}[1+\frac{1}{2^3}+\frac{1}{2^6}+\frac{1}{2^9}.....$

$\implies M'x_{com}=\frac{Ma}{2}(\frac{1}{1-(1/2^3)})$     (Sum of an infinite GP)

$\implies M'x_{com}=\frac{Ma}{2}(\frac{1}{1-(1/8)})$

$\implies M'x_{com}=\frac{Ma}{2}\times\frac{8}{7}$

$\implies M'x_{com}=\frac{4Ma}{7}$

$\implies x_{com}=\frac{4Ma}{7M'}$

$\implies x_{com}=\frac{4Ma}{7}\times\frac{3}{4M}$

$\implies x_{com}=\frac{3a}{7}$

Hope this helps.