A 1 mlong rod having a constant cross sectional area is made of four materials. The first 0.2 m is made of iron, the next 0.3 m of lead, the following 0.2m of aluminium and the remaining part is made of copper. Find the centre of mass of the rod. The densities of iron, lead, aluminium and copper are 7.9 xx 10^3kg//m^3, 11.4 xx 10^3 kg//m^3, 2.7 xx 10^3kg//m^3 and 8.9 xx 10^3 kg//m^3 respectively.

A 1 mlong rod having a constant cross sectional area is made of four m

1.	A 1 mlong rod having a constant cross sectional area is made of four materials. The first 0.2 m is made of iron, the next 0.3 m of lead, the following 0.2m of aluminium and the remaining part is made of copper. Find the centre of mass of the rod. The densities of iron, lead, aluminium and copper are 7.9 xx 10^3kg//m^3, 11.4 xx 10^3 kg//m^3, 2.7 xx 10^3kg//m^3 and 8.9 xx 10^3 kg//m^3 respectively.
Answer» Solution :According to the definition of density we can write, `d=m/v` `rArr m = v xx d` `rArr m=A xx l xx d [therefore V = A xx l]` In case of uniform rod its area of cross section is constant. Hence `m PROP l xx d [ therefore " A is constant "]` MASS of IRON part, `m_(1) = 0.2 xx 7.9 xx 10^(3)=1.58 xx 10^(3) kg` Mass of lead part, `m_(2) prop 0.3 xx 11.4 xx 10^(3) = 3.42 xx 10^(3)` kg Mass of ALUMINIUM part, `m_(3) prop 0.2 xx 2.7 xx 10^(3) = 0.54 xx 10^(3) kg` Mass of copper part, `m_(4) prop 0.3 xx 8.9 xx 10^(3) = 2.67 xx 10^(3)` kg CM of iron part from end "A" of the rod, `x_(1)=0.1`m CM of lead part from end "A" of the rod, `x_(2)= 0.35 m` CM of aluminium part from end "A" of the rod, `x_(3) = 0.6 m` CM of copper part from end "A" of the rod, `x_4= 0.85m` `therefore`Centre of mass of the rod, `X_( c)=(m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3) + m_(4)x_(4))/(m_(1) + m_(2) + m_(3) + m_(4))` `rArr X_(cm) = 0.481 m`from the end "A" of the rod.