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?

Difficult

A

13MR²/32
B

11MR²/32
C

9MR²/32
D

15MR²/32

Solution

$\large I_{Total}=I_{remain}+I_{R/2}$
$\large I_{remain}=I_{Total}-I{_{R/2}$
$\large =\frac {mR^2}{2}-\left [ \frac {\frac m4(R/2)^2 }{2}+\frac m4\left ( \frac {R}{2} \right )^2 \right ]$
$\large = \frac{{m{R^2}}}{2} - \frac{{3m{R^2}}}{{32}}\therefore I = \frac{{13m{R^2}}}{{32}}$

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