Question

The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as [rholeft( r right) = A + Br.] Find its moment of inertia about the line perpendicular to the plane of the disc thorough its centre.

Answer 1

Consider a ring of thickness dx at a distance r from the centre of the disc.

Mass of the ring,

\[dm = \left( A + Br \right) \times 2\pi r \times dr\]

Therefore, moment of inertia about the centre,

\[I =    \int_0^a  r^2 dm\]

\[=  \int_o^a d\left( A + Br \right)  2\pi r . dr \times  r^2 \] 

\[ =  \int_o^a 2\pi A r^3   dr +  \int_o^a   2\pi B r^4   dr\] 

\[ =  \left[ 2\pi A  \left( \frac{r^4}{4} \right) + 2\pi B  \left( \frac{r^5}{5} \right) \right]_0^a \] 

\[ = 2\pi\left( \frac{A a^4}{4} + \frac{B a^5}{5} \right)\]