question_answer

The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:                                    [AIPMT (S) 2005]

A) \[\frac{1}{2}\,M{{R}^{2}}\]

B) \[M{{R}^{2}}\]

C) \[\frac{7}{2}M{{R}^{2}}\]

D) \[\frac{3}{2}M{{R}^{2}}\]

Correct Answer: D

Solution :

Key Idea: We should use parallel axis theorem.

Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane is

\[{{I}_{AB}}=\frac{1}{2}M{{R}^{2}}\]

Using theorem of parallel axes, we have,

\[{{I}_{CD}}={{I}_{AB}}+M{{R}^{2}}\]

\[=\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}\]

\[=\frac{3}{2}M{{R}^{2}}\]

Note:    The role of moment of inertia in the study of rotational motion is analogous to that or mass in study of linear motion.