Five particles of mass $2 \ kg$ each are attached to the rim of a circular disc of radius $0.1 \ m$ and negligible mass. The moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is ........ $kg \ m^2$.

The moment of inertia of a circular ring of radius $r$ and mass $M$ about its diameter is:

The analogue of mass in rotational motion is:

$A$ thin metal rod of mass $M$ and length $L$ is cut into $4$ equal parts by cutting it perpendicular to its length. If the moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis is $I$,then what is the moment of inertia of each part about a similar axis?

The mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma(r) = A + Br$. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is

$A$ circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$,and another disc $Y$ of radius $4R$ is made from an iron plate of thickness $\frac{t}{4}$. The ratio of the moment of inertia $\frac{I_Y}{I_X}$ is: