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Question Number 35732 by ajfour last updated on 22/May/18

Commented by ajfour last updated on 22/May/18

$F i n d m o m e n t o f i n e r t i a o f d i s c$ $a b o u t a n a x i s, a s i n f i g u r e a b o v e,$ $i f d e n s i t y o f d i s c b e p r o p o r t i o n a l$ $t o \sin θ .$
	Answered by ajfour last updated on 23/May/18

	$c h o r d l e n g t h a t a n g l e θ i s$ $ρ = 2 R \sin θ$ $l e t a r e a l d e n s i t y b e δ = k \sin θ$ $d I = {(r \sin θ)}^{2} d m; r b e i n g d i s t a n c e o f$ $a m a s s e l e m e n t f r o m p o i n t o f$ $t a n g e n c y o f d i s c w i t h a x i s o f$ $r o t a t i o n .$ $\int d I = \int_{0}^{π} \int_{0}^{ρ} {(r \sin θ)}^{2} (k \sin θ) (r d θ) d r$ $= k \int_{0}^{π} [(\sin^{3} θ) \int_{0}^{ρ} r^{3} d r] d θ$ $= k \int_{0}^{π} (\sin^{3} θ) (\frac{ρ^{4}}{4}) d θ$ $a s ρ = 2 R \sin θ, w e h a v e$ $I = 4 {k R}^{4} \int_{0}^{π} \sin^{7} θ d θ$ $= - 4 {k R}^{4} \int_{0}^{π} {(1 - \cos^{2} θ)}^{3} (- \sin θ d θ)$ $l e t \cos θ = t \Rightarrow - \sin θ d θ = d t$ $I = - 4 {k R}^{4} \int_{1}^{- 1} {(1 - t^{2})}^{3} d t$ $= 8 R^{2} k \int_{0}^{1} (1 - 3 t^{2} + 3 t^{4} - t^{6}) d t$ $= 8 {k R}^{4} [t - t^{3} + \frac{3 t^{5}}{5} - \frac{t^{7}}{7}] ∣_{0}^{1}$ $\Rightarrow I = 8 {k R}^{4} (\frac{16}{35})$ $o r I = \frac{128 {k R}^{4}}{35} .$
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