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Question Number 162949 by ajfour last updated on 02/Jan/22

Commented by ajfour last updated on 02/Jan/22

$I f t h e s e m i r i n g p l u s r o d {c o m b i n a}^{n},$ $h a v i n g t h e s a m e l i n e a r m a s s d e n s -$ $i t y i s r e l e a s e d a s s h o w n, h o w l o n g$ $d o e s i t t a k e t o t u r n a r o u n d t h e p i v o t$ $a n d h i t t h e g r o u n d ?$
	Answered by mr W last updated on 02/Jan/22

	Commented by mr W last updated on 02/Jan/22

	$ρ = m a s s o f u n i t l e n g t h$ $m = t o t a l m a s s$ $m = (π R + 3 R) ρ = (π + 3) R ρ$ $x = R \cos ϕ$ $y = \frac{3 R}{2} + R \sin ϕ$ $d m = ρ R d ϕ$ $m e = \int x d m = \int_{- \frac{π}{2}}^{\frac{π}{2}} R \cos ϕ ρ R d ϕ$ $m e = ρ R^{2} {[\sin ϕ]}_{- \frac{π}{2}}^{\frac{π}{2}} = 2 ρ R^{2}$ $(π + 3) R ρ e = 2 ρ R^{2}$ $\Rightarrow e = \frac{2 R}{π + 3}$ $I_{x} = \frac{(3 R ρ) {(3 R)}^{2}}{3} + \int y^{2} d m$ $I_{x} = 9 ρ R^{3} + \int_{- \frac{π}{2}}^{\frac{π}{2}} {(\frac{3 R}{2} + R \sin ϕ)}^{2} ρ R d ϕ$ $I_{x} = 9 ρ R^{3} + ρ R^{3} \int_{- \frac{π}{2}}^{\frac{π}{2}} {(\frac{3}{2} + \sin ϕ)}^{2} d ϕ$ $I_{x} = 9 ρ R^{3} + ρ R^{3} \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{9}{4} + 3 \sin ϕ + \sin^{2} ϕ) d ϕ$ $I_{x} = 9 ρ R^{3} + ρ R^{3} \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{11}{4} + 3 \sin ϕ - \frac{\cos 2 ϕ}{2}) d ϕ$ $I_{x} = 9 ρ R^{3} + ρ R^{3} {[\frac{11 ϕ}{4} - 3 \cos ϕ - \frac{\sin 2 ϕ}{4}]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $I_{x} = 9 ρ R^{3} + \frac{11 π ρ R^{3}}{4} = (9 + \frac{11 π}{4}) ρ R^{3}$ $\Rightarrow I_{x} = \frac{(36 + 11 π) {m R}^{2}}{4 (3 + π)}$ $I_{y} = \int x^{2} d m$ $I_{y} = \int_{- \frac{π}{2}}^{\frac{π}{2}} R^{2} \cos^{2} ϕ ρ R d ϕ$ $I_{y} = \frac{ρ R^{3}}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} (\cos 2 ϕ + 1) d ϕ$ $I_{y} = \frac{ρ R^{3}}{2} {[\frac{\sin 2 ϕ}{2} + ϕ]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $I_{y} = \frac{π ρ R^{3}}{2}$ $\Rightarrow I_{y} = \frac{π {m R}^{2}}{2 (3 + π)}$ $I_{0} = I_{x} + I_{y} = \frac{(36 + 11 π) {m R}^{2}}{4 (3 + π)} + \frac{π {m R}^{2}}{2 (3 + π)}$ $\Rightarrow I_{0} = \frac{(36 + 13 π) {m R}^{2}}{4 (3 + π)}$
	Commented by mr W last updated on 02/Jan/22

	Commented by mr W last updated on 02/Jan/22

	$\cos φ = \frac{R}{1.5 R} = \frac{2}{3} \Rightarrow φ = \cos^{- 1} \frac{2}{3}$ $ω = \frac{d θ}{d t}$ $α = \frac{d ω}{d t} = ω \frac{d ω}{d θ}$ $I_{0} α = m g (1.5 R \sin θ + e \cos θ)$ $\frac{(36 + 13 π) {m R}^{2}}{4 (3 + π)} ω \frac{d ω}{d θ} = m g (\frac{3 R}{2} \sin θ + \frac{2 R}{3 + π} \cos θ)$ $ω \frac{d ω}{d θ} = \frac{2 g}{(36 + 13 π) R} [3 (3 + π) \sin θ + 4 \cos θ]$ $\int_{0}^{ω} ω d ω = \frac{2 g}{(36 + 13 π) R} \int_{0}^{θ} [3 (3 + π) \sin θ + 4 \cos θ] d θ$ $\frac{ω^{2}}{2} = \frac{2 g}{(36 + 13 π) R} [3 (3 + π) (1 - \cos θ) + 4 \sin θ]$ $ω = \sqrt{\frac{4 g}{(36 + 13 π) R}} \times \sqrt{3 (3 + π) (1 - \cos θ) + 4 \sin θ}$ $\sqrt{\frac{(36 + 13 π) R}{4 g}} \frac{d θ}{\sqrt{3 (3 + π) (1 - \cos θ) + 4 \sin θ}} = d t$ $\sqrt{\frac{(36 + 13 π) R}{4 g}} \int_{0}^{φ} \frac{d θ}{\sqrt{3 (3 + π) (1 - \cos θ) + 4 \sin θ}} = \int_{0}^{T} d t$ $T = \sqrt{\frac{(36 + 13 π) R}{4 g}} \int_{0}^{\cos^{- 1} \frac{2}{3}} \frac{d θ}{\sqrt{3 (3 + π) (1 - \cos θ) + 4 \sin θ}}$ $T \approx 0.751917 \sqrt{\frac{(36 + 13 π) R}{4 g}} \approx 3.2956 \sqrt{\frac{R}{g}}$
	Commented by ajfour last updated on 02/Jan/22

	$O m G! g i m m e t i m e, s i r . .$
	Commented by mr W last updated on 02/Jan/22

	$d o y o u h a v e t h e a n s w e r s i r ?$
	Commented by ajfour last updated on 02/Jan/22

	$n o, m y c r e a t i o n . ., s i r!$
	Commented by mr W last updated on 02/Jan/22
	��
	Commented by Ar Brandon last updated on 02/Jan/22
	Is this Lekh diagram, Sir?
	Commented by mr W last updated on 02/Jan/22

	$f o r t h i s o n e i {d i d n}^{'} t u s e L E K H, b u t$ $a n a n y i m a g e a p p i n m y s m a r t p h o n e .$ $c e r t a i n l y y o u c a n u s e L E K H .$
	Commented by Tawa11 last updated on 03/Jan/22

	$Great sir .$
	Commented by Ar Brandon last updated on 03/Jan/22
	OK
	Answered by ajfour last updated on 02/Jan/22

	Commented by ajfour last updated on 03/Jan/22

	$M g r \cos (θ + β) + m g L \cos θ = I (\frac{ω d ω}{d θ})$ $\frac{ω^{2}}{2 g} = \frac{M r \sin (θ + β) + m L \sin θ}{I}$ $\int_{0}^{T} d t = \sqrt{\frac{I}{2 g}} \int_{0}^{ϕ} \frac{d θ}{\sqrt{M r \sin (θ + β) + m L \sin θ}}$ $. . . .$ $T \sqrt{\frac{2 g}{I}} = \int_{0}^{ϕ} \frac{d θ}{\sqrt{M (e \sin θ + \frac{3 R}{2} \cos θ) + m (\frac{3 R}{2} \sin θ)}}$ $c o s ϕ = \frac{2}{3}$ $\frac{M}{m} = \frac{π R}{3 R} = \frac{π}{3}$ $M (\frac{R}{2}) + m (0) = (M + m) e$ $e = \frac{R / 2}{1 + \frac{3}{π}} = \frac{π R}{2 (π + 3)}$ $I = ? (i d o u b t, i f w e c a n e a s i l y f i n d . .)$ $T \sqrt{\frac{2 m g R}{I}} = J$ $J = \int_{0}^{ϕ} \frac{d θ}{\sqrt{[(\frac{π}{6}) (\frac{π}{π + 3}) + \frac{3}{2}] \sin θ + (\frac{π}{2}) \cos θ}}$ $i c a n g o t h i s f a r, S i r!$
	Commented by mr W last updated on 03/Jan/22

	$t h e C O M o f a s e m i r i n g i s n o t \frac{R}{2}$ $a p a r t f r o m t h e c e n t e r o f t h e c i r c l e,$ $b u t \frac{2 R}{π} .$ $t h e r e f o r e i t s h o u l d b e :$ $M (\frac{2 R}{π}) + m (0) = (M + m) e$ $e = \frac{2 R / π}{1 + \frac{3}{π}} = \frac{2 R}{π + 3}$ $i g o t t h i s t o o .$
	Commented by mr W last updated on 03/Jan/22

	Commented by mr W last updated on 03/Jan/22

	$I_{0} c a n a l s o b e ‘ ‘ {e a s i l y}^{″} o b t a i n e d i n$ $f o l l o w i n g w a y :$ $t h e M o I o f a c o m p l e t e r i n g a b o u t$ $i t s a x i s i s o b v i o u s l y I_{p} = {M R}^{2}, s o t h e$ $M o I a b o u t i t s d i a m e t e r i s o b v i o u s l y$ $\frac{{M R}^{2}}{2}, s i n c e I_{P} = I_{x} + I_{y} a n d I_{x} = I_{y} .$ $o n t h e o t h e r s i d e a c o m p l e t e r i n g$ $c o n s i s t s o f t w o s e m i r i n g s .$ $\frac{{M R}^{2}}{2} = 2 (I_{s e m i, y, 0} + m_{s e m i} e^{2})$ $I_{s e m i, y} = I_{s e m i, y, 0} + m_{s e m i} e^{2} = \frac{{M R}^{2}}{4} = \frac{m_{s e m i} R^{2}}{2}$ $\frac{{M R}^{2}}{2} = 2 I_{s e m i, x, 0}$ $I_{s e m i, x, 0} = \frac{{M R}^{2}}{4} = \frac{m_{s e m i} R^{2}}{2}$ $I_{s e m i, x} = I_{s e m i, x, 0} + m_{s e m i} {(\frac{3 R}{2})}^{2}$ $I_{s e m i, x} = \frac{m_{s e m i} R^{2}}{2} + \frac{9 m_{s e m i} R^{2}}{4} = \frac{11 m_{s e m u} R^{2}}{4}$ $I_{r o d, y} = 0$ $I_{r o d, x} = \frac{m_{r o d} {(3 R)}^{2}}{3} = 3 m_{r o d} R^{2}$ $I_{t o t a l, y} = I_{s e m i, y} + I_{r o d, y} = \frac{m_{s e m i} R^{2}}{2}$ $I_{t o t a l, x} = I_{s e m i, x} + I_{r o d, x} = \frac{11 m_{s e m i} R^{2}}{4} + 3 m_{r o d} R^{2}$ $I_{t o t a l, 0} = I_{t o t a l, y} + I_{t o t a l, x}$ $I_{t o t a l, 0} = \frac{m_{s e m i} R^{2}}{2} + \frac{11 m_{s e m i} R^{2}}{4} + 3 m_{r o d} R^{2}$ $I_{t o t a l, 0} = \frac{13 m_{s e m i} R^{2}}{4} + 3 m_{r o d} R^{2}$ $m_{s e m i} = \frac{π}{3 + π} m_{t o t a l}$ $m_{r o d} = \frac{3}{3 + π} m_{t o t a l}$ $I_{t o t a l, 0} = (\frac{13 π}{4} + 9) \frac{m_{t o t a l} R^{2}}{3 + π} = \frac{(36 + 13 π) m_{t o t a l} R^{2}}{4 (3 + π)}$
	Commented by ajfour last updated on 03/Jan/22

	$I t h o u g h t f i r s t / / a x i s t h e o r e m$ $f o r m o m e n t o f i n e r t i a w d a p p l y$ $f o r s y m m e t r i c a l b o d i e s o n l y,$ $b u t r e a l l y t h a t s t h e r e s t r i c t i o n$ $i n ⊥ a x i s t h e o r e m, s o i h a d$ $c o m m e n t e d s u c h, n y e s s i r, s o r r y$ $f o r t h e m i s t a k e f o r c . m . o f r i n g . .$
	Answered by mr W last updated on 03/Jan/22

	Commented by ajfour last updated on 03/Jan/22

	$A w e z o m e . . \to \infty$
	Commented by mr W last updated on 03/Jan/22

	$\cos φ = \frac{2}{3}$ $e = \frac{2 R}{3 + π}$ $I_{0} = \frac{(36 + 13 π) {m R}^{2}}{4 (3 + π)}$ $r = \sqrt{e^{2} + {(\frac{3 R}{2})}^{2}} = \frac{\sqrt{97 + 54 π + 9 π^{2}} R}{2 (3 + π)}$ $\cos ϕ = \frac{3 R}{2 r} = \frac{3 (3 + π)}{\sqrt{97 + 54 π + 9 π^{2}}}$ $m g r \sin θ = I_{0} (ω \frac{d ω}{d θ})$ $m g r \sin θ = \frac{(36 + 13 π) {m R}^{2}}{4 (3 + π)} (ω \frac{d ω}{d θ})$ $\frac{2 g \sqrt{97 + 54 π + 9 π^{2}}}{(36 + 13 π) R} \sin θ = ω \frac{d ω}{d θ}$ $ω d ω = \frac{2 g \sqrt{97 + 54 π + 9 π^{2}}}{(36 + 13 π) R} \sin θ d θ$ $\frac{ω^{2}}{2} = \frac{2 g \sqrt{97 + 54 π + 9 π^{2}}}{(36 + 13 π) R} [\cos (ϕ) - \cos θ]$ $ω^{2} = \frac{4 g \sqrt{97 + 54 π + 9 π^{2}}}{(36 + 13 π) R} [\frac{3 (3 + π)}{\sqrt{97 + 54 π + 9 π^{2}}} - \cos θ]$ $ω^{2} = \frac{g}{a^{2} R} (b - \cos θ)$ $a = \frac{1}{2} \sqrt{\frac{36 + 13 π}{\sqrt{97 + 54 π + 9 π^{2}}}} \approx 1.009402$ $b = \cos ϕ = \frac{3 (3 + π)}{\sqrt{97 + 54 π + 9 π^{2}}} \approx 0.977236$ $ω = \frac{d θ}{d t} = \frac{\sqrt{b - \cos θ}}{a} \sqrt{\frac{g}{R}}$ $\sqrt{\frac{R}{g}} \times \frac{a d θ}{\sqrt{b - \cos θ}} = d t$ $T = \sqrt{\frac{R}{g}} \int_{ϕ}^{φ + ϕ} \frac{a d θ}{\sqrt{b - \cos θ}} \approx 3.2956 \sqrt{\frac{R}{g}}$
	Commented by mr W last updated on 03/Jan/22

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