question_answer

If linear density of a rod of length 3 m varies as \[\lambda =2+x,\] then the position of the centre of gravity of the rod is:

A)  \[\frac{7}{3}m\]                              

B)         \[\frac{12}{7}m\]

C)         \[\frac{10}{7}m\]                           

D)         \[\frac{9}{7}m\]

Correct Answer: B

Solution :

Let rod is placed along \[x-\] axis. Mass of element  PQ of length \[dx\]situated at \[x=x\]is \[dm=\lambda dx=(2+x)dx\] The COM of the element has coordinates \[(x,0,0).\] Therefore, \[x-\]coordinate of COM of the rod will be \[{{x}_{COM}}=\frac{\int_{0}^{3}{xdm}}{\int_{0}^{3}{dm}}\] \[=\frac{\int_{0}^{3}{x(2+x)dx}}{\int_{0}^{3}{(2+x)dx}}\] \[=\frac{\int_{0}^{3}{(2x+{{x}^{2}})dx}}{\int_{0}^{3}{(2+x)dx}}\] \[=\frac{\left[ \frac{2{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ 2x+\frac{{{x}^{2}}}{2} \right]_{0}^{3}}\] \[=\frac{\left[ {{(3)}^{2}}+\frac{{{(3)}^{3}}}{3} \right]}{\left[ 2\times 3+\frac{{{(3)}^{2}}}{2} \right]}\] \[=\frac{9+9}{6+9/2}\] \[=\frac{18\times 2}{21}\] \[=\frac{12}{7}m\]