First slide

Moment of intertia

Question

If the moment of inertia of a disc about al axis tangentially and parallel to its surface be I, then what will be the moment of inertia about the axis tangential but Perpendicular to the surface.

Moderate

A

$\frac{6}{5} I$
B

$\frac{3}{4} I$
C

$\frac{3}{2} I$
D

$\frac{5}{4} I$

Solution

According to the theorem of parallel axes, the moment of inertia of the disc about an axis tangentially and parallel to the surface is given by
$I = I_{11} = \frac{{MR}^{2}}{4} + {MR}^{2} = \frac{5}{4} {MR}^{2}$
The moment of inertia of the disc about an axis tangential but perpendicular to the surface is given by
$\begin{array}{l} I^{'} = I ⊥= \frac{{MR}^{2}}{2} + {MR}^{2} = \frac{3}{2} {MR}^{2} \\ = \frac{6}{5} [\frac{5}{4} {MR}^{2}] = \frac{6}{5} I \end{array}$

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