Question 150: Centre of Mass and Rotational Motion

Question

A rod of length L has non-uniform linear mass
density given by $\rho $(x) = $a + b{\left( {{x \over L}} \right)^2}$
, where a
and b are constants and 0 $ \le $ x $ \le $ L. The value
of x for the centre of mass of the rod is at :

Accepted Answer

$\rho $ = a + b${\left( {{x \over L}} \right)^2}$
dm = $\rho $dx = $\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx$
M = $\int {dm} $ = $\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx} $
X
com
= ${{\int {xdm} } \over {\int {dm} }}$
= ${{\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)xdx} } \over {\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx} }}$
= ${{{{a{L^2}} \over 2} + {b \over {{L^2}}}.{{{L^4}} \over 4}} \over {aL + {b \over {{L^2}}}.{{{L^3}} \over 3}}}$
= ${{\left( {{{4a + 2a} \over 8}} \right)L} \over {\left( {{{3a + b} \over 3}} \right)}}$
= ${3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L$

Answer

${3 \over 2}\left( {{{a + b} \over {2a + b}}} \right)L$

Answer

${4 \over 3}\left( {{{a + b} \over {2a + 3b}}} \right)L$

Answer

${3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L$

Answer

${3 \over 2}\left( {{{2a + b} \over {3a + b}}} \right)L$

Centre of Mass and Rotational Motion

A rod of length L has non-uniform linear mass density given by \rho (x) = a + b{\left( {{x \over L}} \right)^2} , where a and b are constants and 0 \le x \le L. The value of x for the centre of mass of the rod is at :

Choose the correct answer: