Find the moment of inertia
(a) of a thin uniform rod relative to the axis which is perpendicular to the rod and passes through its end, if the mass of the rod is m and its lengh l,
(b) of a thin uniform rectangular plate relative to the axis passing perpendicular to the plane of the plate through one of its vertices, if the sides of the plate are equal to a and b, and its mass is m.

Explanation

(a) Consider a strip of length `dx` at a perpendicular distance x from the axis about which we have to find the moment of inertia of the rod. The elemental mass of the rod equals
`dm=(m)/(l)dx`
Moment of inertia of this element about the axis
`dI=dmx^2=m/ldx*x^2`
Thus, moment of inertia of the rod, as a whole about the given axis
`I=underset(0)overset(l)intm/lx^2dx=(ml^2)/(3)`
(b) Let us imagine the plane of plate as xy plane taking the origin at the intersection point of the sides of the plate(Figure).
Obviously `I_x=intdmy^2`
`=underset(0)overset(a)int((m)/(ab)bdy)y^2`
`=(ma^2)/(3)`
Similarly `I_y=(mb^2)/(3)`
Hence from perpendicular axis theorem
`I_x=I_x+I_y=m/3(a^2+b^2)`, which is the sought moment of inertia.