Question

A uniform disc of mass $5$ kg has a radius of $0.5$m. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is?

• A. $1.25$ kg-$m^2$
• B. $0.5$ kg-$m^2$
• C. $4$ kg-$m^2$
• D. $1.875$ kg-$m^2$

Answer 1

The correct option is D $1.875$ kg-$m^2$

Moment of inertia of the disc along a line

Has to its plane and passing through centre = $\frac{M{R}^2}{2}.$

By using perpendicular axis therom.

moment of inertia through axis perpendicular

to plane and passing through point on circumference

= moment of inertia has to plane passing through centre

+ mass $\times$ distance between centre and point

$=\frac{M{R}^2}{2}+M{R}^2$

$=\frac{3M{R}^2}{2}=\frac{3\times 5\times 0.25}{2}$

$=1.875kg{m}^2$