Answer:
[tex]x_c= \dfrac{5}{9}L[/tex]
[tex]I=\dfrac {7}{12}\lambda_ 0 L^3[/tex]
Explanation:
Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.
At any distance x from point A mass density
[tex]\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x[/tex]
[tex]\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x[/tex]
Lets take element mass at distance x
dm =λ dx
mass moment of inertia
[tex]dI=\lambda x^2dx[/tex]
So total moment of inertia
[tex]I=\int_{0}^{L}\lambda x^2dx[/tex]
By putting the values
[tex]I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx[/tex]
By integrating above we can find that
[tex]I=\dfrac {7}{12}\lambda_ 0 L^3[/tex]
Now to find location of center mass
[tex]x_c = \dfrac{\int xdm}{dm}[/tex]
[tex]x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}[/tex]
Now by integrating the above
[tex]x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}[/tex]
[tex]x_c= \dfrac{5}{9}L[/tex]
So mass moment of inertia [tex]I=\dfrac {7}{12}\lambda_ 0 L^3[/tex] and location of center of mass [tex]x_c= \dfrac{5}{9}L[/tex]
