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Question Number 46299 by rahul 19 last updated on 23/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

	$a r e a r e m a i n i n g p o r t i o n = π R^{2} - π {(\frac{R}{3})}^{2}$ $= π R^{2} (1 - \frac{1}{9}) = \frac{8 π R^{2}}{9}$ $m a s s p e r u n i t a r e a (ρ) = \frac{M}{\frac{8 π R^{2}}{9}} = \frac{9 M}{8 π R^{2}}$ $s o m a s s f o r c u t d i s c = π \frac{R^{2}}{9} \times \frac{9 M}{8 π R^{2}} = \frac{M}{8}$ $m a s s o f u n c u t c o m p l e t e d i s c = M + \frac{M}{8} = \frac{9 M}{8}$ $m o m e n t o f i n e r t i a o f u n c u t c o m p l e t e d i s c$ $a b o u t t h e a x i s ⊥ t o t h e p l a n e o f d i s c t b r o u g h$ $c e n t r e$ $= \frac{9 M}{8} \times \frac{R^{2}}{2} = \frac{9 {M R}^{2}}{16}$ $\frac{9 {M R}^{2}}{16} \leftarrow I_{o}$ $m o m e n g o f i n e r t i a c u t s m a l l d i s c a b o u t$ $m e n t i o n e d a x i x = \frac{M}{8} \times {(R - \frac{R}{3})}^{2} + \frac{M}{8} \times {(\frac{R}{3})}^{2} \times \frac{1}{2}$ $= \frac{M}{8} \times \frac{4 R^{2}}{9} + \frac{M}{8} \times \frac{R^{2}}{18} = \frac{{M R}^{2}}{8} (\frac{4}{9} + \frac{1}{18}) = \frac{{M R}^{2}}{8} \times \frac{1}{2}$ $= \frac{{M R}^{2}}{16}$ $s l r e q u i r e d a n s i s = \frac{9 {M R}^{2}}{16} - \frac{{M R}^{2}}{16} = \frac{{M R}^{2}}{2}$ $p l s c h e c k . . .$
	Commented by rahul 19 last updated on 24/Oct/18

	$S i r, i^{'} m g e t t i n g s a m e a n s w e r .$ $B u t i w o n d e r m o m e n t$ $o f i n e r t i a o f d i s c o f m a s s m = \frac{{m R}^{2}}{2} .$ $(n o r m a l d i s c)$ $i f w e t a k e a n o t h e r d i s c w h i c h h a s a h o l e,$ $o f m a s s m, M . I i s s t i l l s a m e!$
	Commented by MrW3 last updated on 24/Oct/18

	$y o u r c o m p a r a t i o n i s n o t r e a l l y c o r r e c t .$ $t h e d i s c w i t h o u t h o l e a n d t h e d i s c w i t h$ $h o l e d o h a v e i n f a c t n o t t h e s a m e M o I$ $w . r . t . t h e i r C o M .$ $I_{d i s c w i t h o u t h o l e} = \frac{{M R}^{2}}{2}$ $I_{d i s c w i t h h o l e} < \frac{{M R}^{2}}{2}$ $t h e M o I o f a o b j e c t w . r . t . i t s C o M i s$ $t h e s m a l l e s t . i t s M o I w . r . t . t o a n y$ $o t h e r p o i n t i s b i g g e r . f o r t h e d i s c w i t h$ $h o l e i f w e c a l c u l a t e i t s M o I n o t w . r . t . i t s$ $C o M, b u t w . r . t . t o a n o t h e r p o i n t, w e$ $w i l l g e t a b i g g e r v a l u e . i n f a c t w e c a n$ $g e t a n y v a l u e i f w e c h o o s e t h i s p o i n t$ $c o r r e s p o n d i n g l y . f o r t h e d i s c w i t h h o l e$ $i f w e t h e p o i n t O w h i c h i s n o t i t s C o M,$ ${w e}^{'} l l g e t t h e s a m e v a l u e a s t h e M o I$ $o f a d i s c w i t h o u t h o l e w . r . t . i t s C o M .$
	Answered by ajfour last updated on 24/Oct/18

	$I_{r e q d} = \frac{1}{2} {M R}^{2} .$
	Commented by rahul 19 last updated on 24/Oct/18

	$S i r, i s t h i s m e t h o d c o r r e c t ?$ $M . I = \frac{1}{2} \times M \times R^{2} w h e r e M = m a s s o f$ $r e m a i n i n g d i s c . . . . . . o r w e n e e d t o$ $c a l c u l a t e t h e w a y T a n m a y s i r h a s d o n e .$
	Commented by ajfour last updated on 24/Oct/18

	$y e s i h a d m y s e l f s o l v e d t h e s a m e$ $w a y; o n l y d i n t p o s t; b u t i t c o u l d$ $b e r e d u c e d a l i t t l e;$ $\frac{M + m}{m} = \frac{R^{2}}{{(R / 3)}^{2}} = 9 \Rightarrow M = 8 m$ $I = \frac{(M + m) R^{2}}{2} - [\frac{m}{2} {(\frac{R}{3})}^{2} + m {(\frac{2 R}{3})}^{2}]$ $= \frac{9 {m R}^{2}}{2} - \frac{{m R}^{2}}{18} - \frac{4 {m R}^{2}}{9}$ $= \frac{{m R}^{2}}{18} (81 - 1 - 8) = 4 {m R}^{2}$ $= 4 (\frac{M}{8}) R^{2} = \frac{{M R}^{2}}{2} .$
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