First slide

Moment of intertia

Question

From a disc of radius $R$ and mass $M$ a circular hole of diameter $R$ whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Moderate

A

$11 M R^{2} / 32$
B

$9 M R^{2} / 32$
C

$15 M R^{2} / 32$
D

$13 M R^{2} / 32$

Solution

$Mass per unit area of disc = \frac{M}{π R^{2}}$
Mass of removed portion of disc,
$M^{'} = \frac{M}{π R^{2}} \times π {(\frac{R}{2})}^{2} = \frac{M}{4}$
Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,
$I_{O}^{'} = I_{O^{'}} + M^{'} d^{2}$
$= \frac{1}{2} \times \frac{M}{4} \times {(\frac{R}{2})}^{2} + \frac{M}{4} \times {(\frac{R}{2})}^{2} = \frac{M R^{2}}{32} + \frac{M R^{2}}{16} = \frac{3 M R^{2}}{32}$
When portion of disc would not have been removed, the moment of inertia of complete disc about centre O is
$I_{O} = \frac{1}{2} M R^{2}$
So, moment of inertia of the disc with removed portion is
$I = I_{O} - I_{O}^{'} = \frac{1}{2} M R^{2} - \frac{3 M R^{2}}{32} = \frac{13 M R^{2}}{32}$

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