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?
Answer & explanation
Correct answer: option 2
I = \(\frac{MR^2}{2} - \frac{3\sigma}{2} \pi (\frac{R}{2})^2 (\frac{R}{2})^2\)
where \(\sigma\) = \(\frac{M}{\pi R^2}\)
I = \(\frac{MR^2}{2} - \frac{3\sigma}{2} MR^2\)
= 13 MR2 /32