question_answer 38)

                  The density of a non-uniform rod of length 1m is given by \[\rho \,(x)\,=\,a(1+\,b{{x}^{2}})\] where a and b are constants and \[0\le \,x\le \,1.\] The centre of mass of the rod will be at                 (a) \[\frac{3(2+b)}{4(3+\,b)}\]               (b) \[\frac{4(2+b)}{3(3+\,b)}\]                 (c) \[\frac{3(3+b)}{4(2+\,b)}\]               (d) \[\frac{4(3+b)}{3(2+\,b)}\]                

Answer:

                  (a) Mass of small element at a distance x of width dx from one end of the rod                 \[dm=\rho \,dx=\,a\,(1+\,b{{x}^{2}})dx\]                 \[\therefore \]\[{{x}_{c}}\,=\frac{\int_{{}}^{{}}{dm\,x}}{\int_{{}}^{{}}{dm}}\,=\,\frac{\int\limits_{0}^{1}{a\,(1+\,3{{x}^{2}})\,x\,dx}}{\int\limits_{0}^{1}{\,(1+\,b{{x}^{2}})\,\,dx}}\]                 \[(a)=\,\frac{\left. \frac{{{x}^{2}}}{2} \right|_{0}^{1}+\,b\left. \frac{{{x}^{4}}}{4} \right|_{0}^{1}\,}{\left. \frac{{{x}^{2}}}{2}x \right|_{0}^{1}+\,b\left. \frac{{{x}^{3}}}{3} \right|_{0}^{1}}=\,\frac{\frac{1}{2}+\,\frac{b}{4}}{1+\,\frac{b}{3}}\]                 \[=\,\frac{(2+b)\,/4}{(3+\,b)/3}\,\,=\frac{3(2+b)}{4(3+\,b)}\]