First slide

Moment of intertia

Question

If the moment of inertia of a disc about an axis tangentially and parallel to its surface be I, 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
$I = = \frac{{MR}^{2}}{4} + {MR}^{2} = \frac{5}{4} {MR}^{2} \Rightarrow M R^{2} = \frac{4 I}{5}$
$given by I^{'} = \frac{{MR}^{2}}{2} + {MR}^{2} = \frac{3}{2} {MR}^{2} = \frac{3}{2} [\frac{4 I}{5}] = \frac{6 I}{5}$

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