First slide

Moment of inertia

Question

From a circular disc of radius R and mass 9M, a small disc of radius $\frac{R}{3}$ is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

Moderate

A

$4 {MR}^{2}$
B

$\frac{40}{9} {MR}^{2}$
C

$40 {MR}^{2}$
D

$\frac{37}{9} {MR}^{2}$

Solution

Mass per unit area of disc $= \frac{9 M}{{πR}^{2}}$
Mass of removed portion of disc
$= \frac{9 M}{{πR}^{2}} \times π {(\frac{R}{3})}^{2} = M$
Moment of inertia of removed portion about an axis passing through the centre of disc and perpendicular to the plane of disc, using theorem of parallel axes is
$I_{1} = \frac{M}{2} {(\frac{R}{3})}^{2} + M {(\frac{2 R}{3})}^{2} = \frac{1}{2} {MR}^{2}$
The moment of inertia of complete disc about the given axis is $I_{2} = \frac{9}{2} {MR}^{2}$
So moment of inertia of the disc with removed portion, about the given axis is $I = I_{2} - I_{1} = \frac{9}{2} {MR}^{2} - \frac{1}{2} {MR}^{2} = 4 {MR}^{2}$

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