Explanation
Moment of inertia of rod about its centre of mass is determined by the general form of moment of inertia which is given by:
$$I = \int\limits_0^M {{r^2}dm} $$
Now mass $$m$$ is termed into length element $$dr$$
$$I = \int\limits_{ - L/2}^{L/2} {{r^2}} \times \dfrac{M}{L}dr$$
$$I = \dfrac{{M{L^2}}}{{12}}$$
Now moment of inertia of rod about its one end is given by
$${I_R} = \dfrac{{M{L^2}}}{{12}} + M \times {\left( {\dfrac{L}{2}} \right)^2}$$
$${I_R} = \dfrac{{M{L^2}}}{{12}} + \dfrac{{M{L^2}}}{4}$$
$${I_R} = \dfrac{{M{L^2}}}{3}$$
Moment of inertia of another mass M which is attached at the other end of the rod
$$I = M{L^2}$$
Total moment of inertia
$${I_T} = {I_R} + I$$
$${I_T} = \dfrac{{4M{L^2}}}{3}$$
Thus moment of inertia of a rod and mass about its one of the end is$$\dfrac{{4M{L^2}}}{3}$$.
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