From a circular disc of radius R and mass 9M , a small disc of radius R/3 is removed from the disc. The moment of inertia ….

Q: Free a circular disc of radius R and mass 9M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

Numerical

(a) 4 MR2

(b) $\frac{40}{9}MR^2 $

(c) $10 MR^2 $

(d) $\frac{37}{9}MR^2 $

Ans: (a)

Sol: Mass of removed disc , $M’ = \frac{9 M}{\pi R^2} \times \pi (\frac{R}{3})^2 = M $

Moment of Inertia of complete circular disc about O ,

$\displaystyle I = \frac{9 M R^2}{2}$

Moment of Inertia of removed disc about O will be ,

$\displaystyle I’ =  (\frac{M(R/3)^2}{2}) + M (\frac{2R}{3})^2$ (Applying Parallel Axes Theorem)

$\displaystyle I’ = \frac{M R^2}{18} + \frac{4 M R^2}{9}$

$\displaystyle I’ = \frac{M R^2}{2} $

Required Moment of Inertia ,

$\displaystyle I_{req} = I – I’ $

$\displaystyle I_{req} = \frac{9 M R^2}{2} – \frac{M R^2}{2} = 4 M R^2 $