Question

if lemda = alphax² + betax , M ,L . Find c.o.m of rod of mass M and length L​

Answer 1

The linear mass density of the rod is given by,

$\sf{\longrightarrow\lambda=\alpha x^2+\beta x}$

or,

$\sf{\longrightarrow\dfrac{dm}{dx}=\alpha x^2+\beta x}$

$\sf{\longrightarrow dm=(\alpha x^2+\beta x)\ dx}$

Then, position of center of mass of the rod,

$\sf{\longrightarrow\bar x=\dfrac{\displaystyle\int\limits_0^Mx\ dm}{\displaystyle\int\limits_0^Mdm}}$

$\sf{\longrightarrow\bar x=\dfrac{\displaystyle\int\limits_0^Lx(\alpha x^2+\beta x)\ dx}{\displaystyle\int\limits_0^L(\alpha x^2+\beta x)\ dx}}$

$\sf{\longrightarrow\bar x=\dfrac{\displaystyle\int\limits_0^L\left(\alpha x^3+\beta x^2\right)\ dx}{\displaystyle\int\limits_0^L\left(\alpha x^2+\beta x\right)\ dx}}$

$\sf{\longrightarrow\bar x=\dfrac{\dfrac{\alpha}{4}\left[x^4\right]_0^L+\dfrac{\beta}{3}\left[x^3\right]_0^L}{\dfrac{\alpha}{3}\left[x^3\right]_0^L+\dfrac{\beta}{2}\left[x^2\right]_0^L}}$

$\sf{\longrightarrow\bar x=\dfrac{\dfrac{\alpha L^4}{4}+\dfrac{\beta L^3}{3}}{\dfrac{\alpha L^3}{3}+\dfrac{\beta L^2}{2}}}$

$\sf{\longrightarrow\underline{\underline{\bar x=\dfrac{3\alpha L^2+4\beta L}{4\alpha L+6\beta}}}}$