Q.
The moment of inertia of a circular disc of radius $2\ m$ and mass $1\ kg$ about an axis passing through the centre of mass but perpendicular to the plane of the disc is $2\ kg\ m^{2}$. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is ………… (see the given figure)
A) $8\ kgm^{2}$
B) $4\ kgm^{2}$
C) $10\ kgm^{2}$
D) $6\ kgm^{2}$
Explanation
(d) Here, mass of the disc, $M = 1\ kg$
Radius of the disc, $R = 2m$
Moment of inertia of the circular disc about $XY$ is
[image:7_image147.emf]
$I_{XY} = \frac{MR^{2}}{2} = 2kg\ m^{2}$ [Given]
According to theorem of parallel axes the moment of inertia of the circular disc about $X_{1}Y_{1}$ is
$I_{X_{1}Y_{1}} = I_{XY} + MR^{2} = \frac{MR^{2}}{2} + MR^{2}$
$= \frac{3}{2}MR^{2} = \frac{3}{2} \times (1kg) \times (2m)^{2} = 6kgm^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.