Question 145: 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: