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Question Number 70831 by ajfour last updated on 08/Oct/19

Commented by ajfour last updated on 08/Oct/19

$S t i c k o f m a s s m, l e n g t h b i s$ $r e l e a s e d a s s h o w n . A s i t s l i d e s$ $d o w n a n d b r e a k s c o n t a c t w i t h$ $h e m i s p h e r e, t h e t o p e n d i s a t$ $a n g u l a r p o s i t i o n β . A s s u m e$ $n o f r i c t i o n, f i n d β .$
	Answered by ajfour last updated on 08/Oct/19

	Answered by mr W last updated on 08/Oct/19

	Commented by ajfour last updated on 10/Oct/19

	$s o f a r, s o g o o d, h o w d o w e$ $u t i l i s e ω i n d e t e r m i n i n g β, S i r ?$ $(y o u l i k e t h e q u e s t i o n, s i r ?)$
	Commented by mr W last updated on 09/Oct/19
	$OA=R[cos β+(√((b^2 /R^2 )−sin^2 β))] AC=OA tan β=R[sin β+tan β(√((b^2 /R^2 )−sin^2 β))] BC=((OA)/(cos β))−R=(R/(cos β))(√((b^2 /R^2 )−sin^2 β)) CD^2 =((AC^2 +BC^2 )/2)−(b^2 /4) CD^2 =(R^2 /2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−(b^2 /4) I_C =I_0 +mCD^2 I_C =((mb^2 )/(12))+((mR^2 )/2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−((mb^2 )/4) I_C =((mR^2 )/2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−((mb^2 )/6) Δh=(R/2)((b/R) cos α−sin β) (1/2)I_C ω^2 =mgΔh ((R/2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−(b^2 /(6R^2 )))ω^2 =g((b/R) cos α−sin β) let λ=(b/R) ((1/2){(1/(cos^2 β))(λ^2 −sin^2 β)+sin^2 β[1+(1/(cos β))(√(λ^2 −sin^2 β))]^2 }−(λ^2 /6))ω^2 =(g/R)(λ cos α−sin β) ⇒ω=(√(g/R))(√((6(λ cos α−sin β))/(3{(1/(cos^2 β))(λ^2 −sin^2 β)+sin^2 β[1+(1/(cos β))(√(λ^2 −sin^2 β))]^2 }−λ^2 ))) ....$
	$O A = R [\cos β + \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}]$ $A C = O A \tan β = R [\sin β + \tan β \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}]$ $B C = \frac{O A}{\cos β} - R = \frac{R}{\cos β} \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}$ ${C D}^{2} = \frac{{A C}^{2} + {B C}^{2}}{2} - \frac{b^{2}}{4}$ ${C D}^{2} = \frac{R^{2}}{2} {\frac{1}{\cos^{2} β} (\frac{b^{2}}{R^{2}} - \sin^{2} β) + \sin^{2} β {[1 + \frac{1}{\cos β} \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}]}^{2}} - \frac{b^{2}}{4}$ $I_{C} = I_{0} + {m C D}^{2}$ $I_{C} = \frac{{m b}^{2}}{12} + \frac{{m R}^{2}}{2} {\frac{1}{\cos^{2} β} (\frac{b^{2}}{R^{2}} - \sin^{2} β) + \sin^{2} β {[1 + \frac{1}{\cos β} \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}]}^{2}} - \frac{{m b}^{2}}{4}$ $I_{C} = \frac{{m R}^{2}}{2} {\frac{1}{\cos^{2} β} (\frac{b^{2}}{R^{2}} - \sin^{2} β) + \sin^{2} β {[1 + \frac{1}{\cos β} \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}]}^{2}} - \frac{{m b}^{2}}{6}$ $Δ h = \frac{R}{2} (\frac{b}{R} \cos α - \sin β)$ $\frac{1}{2} I_{C} ω^{2} = m g Δ h$ $(\frac{R}{2} {\frac{1}{\cos^{2} β} (\frac{b^{2}}{R^{2}} - \sin^{2} β) + \sin^{2} β {[1 + \frac{1}{\cos β} \sqrt{\frac{b^{2}}{R^{2}} - \sin^{2} β}]}^{2}} - \frac{b^{2}}{6 R^{2}}) ω^{2} = g (\frac{b}{R} \cos α - \sin β)$ $l e t λ = \frac{b}{R}$ $(\frac{1}{2} {\frac{1}{\cos^{2} β} (λ^{2} - \sin^{2} β) + \sin^{2} β {[1 + \frac{1}{\cos β} \sqrt{λ^{2} - \sin^{2} β}]}^{2}} - \frac{λ^{2}}{6}) ω^{2} = \frac{g}{R} (λ \cos α - \sin β)$ $\Rightarrow ω = \sqrt{\frac{g}{R}} \sqrt{\frac{6 (λ \cos α - \sin β)}{3 {\frac{1}{\cos^{2} β} (λ^{2} - \sin^{2} β) + \sin^{2} β {[1 + \frac{1}{\cos β} \sqrt{λ^{2} - \sin^{2} β}]}^{2}} - λ^{2}}}$ $. . . .$
	Commented by mr W last updated on 10/Oct/19

	$n i c e q u e s t i o n s i r! b u t n o t e a s y .$
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