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Question Number 137986 by ajfour last updated on 08/Apr/21

Commented by ajfour last updated on 08/Apr/21

$F i n d m a x i m u m s p e e d g a i n e d$ $b y c y l i n d e r i f m o t i o n i s$ $i n i t i a t e d . B o t t o m c o r n e r i s$ $h i n g e d t o g r o u n d .$
	Answered by mr W last updated on 08/Apr/21

	Commented by mr W last updated on 11/Apr/21
	$let ω=−(dθ/dt) sin φ=((r−(√2)r sin θ)/r)=1−(√2) sin θ s=(√2)r cos θ+r cos φ =(√2)r cos θ+r(√(1−(1−(√2) sin θ)^2 )) =(√2)r cos θ+r(√(2 sin θ((√2)−sin θ))) (ds/dθ)=−(√2)r sin θ+r((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ))))) v=(ds/dt)=−ω(ds/dθ)=ωr[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))] I_A =((m((√2)r)^2 )/6)+mr^2 =((4mr^2 )/3) Iα=mgrsin ((π/4)−θ)+N sin φ s−N cos φ r ((4r)/(3g))α−sin ((π/2)−θ)=(N/(mg))(1−(√2) sin θ)((√2) cos θ) ⇒(N/(mg))=((((4r)/(3g))α−sin ((π/4)−θ))/( (√2)cos θ (1−(√2) sin θ))) when N=0: α=((3g)/(4r)) sin ((π/4)−θ) (1/2)I_A ω^2 +(1/2)Mv^2 =mgr[1−cos ((π/4)−θ)] ((4mr^2 )/3)ω^2 +Mω^2 r^2 [(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 =2mgr[1−cos ((π/4)−θ)] ω^2 {(4/3)+(M/m)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 }=((2g)/r)[1−cos ((π/4)−θ)] ω=(√(g/r))×(√((1−cos ((π/4)−θ))/((1/3)+((2M)/m)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 ))) with Φ(θ)=(√((1−cos ((π/4)−θ))/((1/3)+((2M)/m)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 ))) ⇒ω=(√(g/r))×Φ(θ) α=(dω/dt)=−ω(dω/dθ)=−(g/r)Φ(θ)Φ′(θ) −(g/r)Φ(θ)Φ′(θ)=((3g)/(4r)) sin ((π/4)−θ) ⇒Φ(θ)Φ′(θ)+(3/4) sin ((π/4)−θ)=0 v=(√(gr))×Φ(θ)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))] examples: (M/m)=1 we get N=0 at θ≈0.5795 (=32.20°) v_(max) ≈0.082(√(gr)) (M/m)=10 we get N=0 at θ≈0.5373 (=30.79°) v_(max) ≈0.056(√(gr))$
	$l e t ω = - \frac{d θ}{d t}$ $\sin ϕ = \frac{r - \sqrt{2} r \sin θ}{r} = 1 - \sqrt{2} \sin θ$ $s = \sqrt{2} r \cos θ + r \cos ϕ$ $= \sqrt{2} r \cos θ + r \sqrt{1 - {(1 - \sqrt{2} \sin θ)}^{2}}$ $= \sqrt{2} r \cos θ + r \sqrt{2 \sin θ (\sqrt{2} - \sin θ)}$ $\frac{d s}{d θ} = - \sqrt{2} r \sin θ + r \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}$ $v = \frac{d s}{d t} = - ω \frac{d s}{d θ} = ω r [\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}]$ $I_{A} = \frac{m {(\sqrt{2} r)}^{2}}{6} + {m r}^{2} = \frac{4 {m r}^{2}}{3}$ $I α = m g r \sin (\frac{π}{4} - θ) + N \sin ϕ s - N \cos ϕ r$ $\frac{4 r}{3 g} α - \sin (\frac{π}{2} - θ) = \frac{N}{m g} (1 - \sqrt{2} \sin θ) (\sqrt{2} \cos θ)$ $\Rightarrow \frac{N}{m g} = \frac{\frac{4 r}{3 g} α - \sin (\frac{π}{4} - θ)}{\sqrt{2} \cos θ (1 - \sqrt{2} \sin θ)}$ $w h e n N = 0 :$ $α = \frac{3 g}{4 r} \sin (\frac{π}{4} - θ)$ $\frac{1}{2} I_{A} ω^{2} + \frac{1}{2} {M v}^{2} = m g r [1 - \cos (\frac{π}{4} - θ)]$ $\frac{4 {m r}^{2}}{3} ω^{2} + M ω^{2} r^{2} {[\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}]}^{2} = 2 m g r [1 - \cos (\frac{π}{4} - θ)]$ $ω^{2} {\frac{4}{3} + \frac{M}{m} {[\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}]}^{2}} = \frac{2 g}{r} [1 - \cos (\frac{π}{4} - θ)]$ $ω = \sqrt{\frac{g}{r}} \times \sqrt{\frac{1 - \cos (\frac{π}{4} - θ)}{\frac{1}{3} + \frac{2 M}{m} {[\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}]}^{2}}}$ $w i t h Φ (θ) = \sqrt{\frac{1 - \cos (\frac{π}{4} - θ)}{\frac{1}{3} + \frac{2 M}{m} {[\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}]}^{2}}}$ $\Rightarrow ω = \sqrt{\frac{g}{r}} \times Φ (θ)$ $α = \frac{d ω}{d t} = - ω \frac{d ω}{d θ} = - \frac{g}{r} Φ (θ) Φ^{'} (θ)$ $- \frac{g}{r} Φ (θ) Φ^{'} (θ) = \frac{3 g}{4 r} \sin (\frac{π}{4} - θ)$ $\Rightarrow Φ (θ) Φ^{'} (θ) + \frac{3}{4} \sin (\frac{π}{4} - θ) = 0$ $v = \sqrt{g r} \times Φ (θ) [\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}}]$ $e x a m p l e s :$ $\frac{M}{m} = 1$ $w e g e t N = 0 a t θ \approx 0.5795 (= 32.20 °)$ $v_{m a x} \approx 0.082 \sqrt{g r}$ $\frac{M}{m} = 10$ $w e g e t N = 0 a t θ \approx 0.5373 (= 30.79 °)$ $v_{m a x} \approx 0.056 \sqrt{g r}$
	Commented by ajfour last updated on 12/Apr/21
	$Ncos φ=MA mgrsin (π/4−θ)−Nr(√2)sin (θ−φ) =(((4mr^2 )/3))α when contact breaks N=0, A=0, 3gsin (π/4−θ)=4αr αsin (θ−φ)=ω^2 cos (θ−φ) 3gsin (π/4−θ)sin (θ−φ)=4ω^2 rcos (θ−φ) .(i) ωr(√2)sin (θ−φ)=Vcos φ ...(ii) (1/2)(((4mr^2 )/3))ω^2 +(1/2)MV^2 =mgr[1−cos (π/4−θ)] ....(iii) 3g(cos θ−sin θ)sin (θ−φ)=4(√2)ω^2 rcos (θ−φ) ⇒ ((2/3)+((Msin^2 (θ−φ))/(mcos^2 φ)))[((3(cos θ−sin θ)sin (θ−φ))/(4cos (θ−φ)))] = (√2)−cos θ−sin θ with (√2)sin θ=1−sin φ θ is obtained from above eq. Now V=((ωr(√2)sin (θ−φ))/(cos φ)) , hence V^( 2) ={((3(cos θ−sin θ)sin^3 (θ−φ))/( 2(√2)cos^2 φcos (θ−φ)))}gr$
	$N \cos ϕ = M A$ $m g r \sin (π / 4 - θ) - N r \sqrt{2} \sin (θ - ϕ)$ $= (\frac{4 {m r}^{2}}{3}) α$ $w h e n c o n t a c t b r e a k s$ $N = 0, A = 0,$ $3 g \sin (π / 4 - θ) = 4 α r$ $α \sin (θ - ϕ) = ω^{2} \cos (θ - ϕ)$ $3 g \sin (π / 4 - θ) \sin (θ - ϕ) = 4 ω^{2} r \cos (θ - ϕ) . (i)$ $ω r \sqrt{2} \sin (θ - ϕ) = V \cos ϕ . . . (i i)$ $\frac{1}{2} (\frac{4 {m r}^{2}}{3}) ω^{2} + \frac{1}{2} {M V}^{2}$ $= m g r [1 - \cos (π / 4 - θ)] . . . . (i i i)$ $3 g (\cos θ - \sin θ) \sin (θ - ϕ) = 4 \sqrt{2} ω^{2} r \cos (θ - ϕ)$ $\Rightarrow$ $(\frac{2}{3} + \frac{M \sin^{2} (θ - ϕ)}{m \cos^{2} ϕ}) [\frac{3 (\cos θ - \sin θ) \sin (θ - ϕ)}{4 \cos (θ - ϕ)}]$ $= \sqrt{2} - \cos θ - \sin θ$ $w i t h$ $\sqrt{2} \sin θ = 1 - \sin ϕ$ $θ i s o b t a i n e d f r o m a b o v e e q .$ $N o w V = \frac{ω r \sqrt{2} \sin (θ - ϕ)}{\cos ϕ}, h e n c e$ $V^{2} = {\frac{3 (\cos θ - \sin θ) \sin^{3} (θ - ϕ)}{2 \sqrt{2} \cos^{2} ϕ \cos (θ - ϕ)}} g r$
	Commented by mr W last updated on 12/Apr/21

	$t h a n k s f o r r e v i e w i n g s i r!$
	Commented by ajfour last updated on 12/Apr/21

	Commented by mr W last updated on 12/Apr/21

	Commented by mr W last updated on 12/Apr/21

	$c a s e 2 :$ $v_{x, C_{1}} = v_{x, C_{2}} = V$ $c a s e 1 :$ $v_{x, C_{1}} \neq v_{x, C_{2}} = V$
	Commented by mr W last updated on 12/Apr/21

	$t h e c o r n e r C_{1} {d o e s n}^{'} t h a v e t h e s a m e$ $v_{x} a s t h e p o i n t C_{2} o n t h e c y l i n d e r!$ $b o t h p o i n t s o n t h e c o n t a c t m u s t h a v e$ $t h e s a m e v e l o c i t y i n x^{'} - d i r e c t i o n,$ $n o t i n x - d i r e c t i o n, i t h i n k .$
	Commented by mr W last updated on 12/Apr/21

	Commented by mr W last updated on 12/Apr/21

	$V_{C 2} \cos ϕ = V_{C 1} \sin (θ - ϕ)$ $\Rightarrow V \cos ϕ = ω \sqrt{2} r \sin (θ - ϕ)$ $\Rightarrow V = \frac{ω r \sqrt{2} \sin (θ - ϕ)}{\cos ϕ}$ $\sin ϕ = 1 - \sqrt{2} \sin θ$ $\cos ϕ = \sqrt{2 \sin θ (\sqrt{2} - \sin θ)}$ $\sin (θ - ϕ) = \sin θ \cos ϕ - \cos θ \sin ϕ$ $V = ω r \sqrt{2} (\sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{2 \sin θ (\sqrt{2} - \sin θ)}})$ $V = ω r (\sqrt{2} \sin θ - \frac{\cos θ (1 - \sqrt{2} \sin θ)}{\sqrt{\sin θ (\sqrt{2} - \sin θ)}})$ $t h i s i s t h e s a m e a s i h a d a b o v e .$
	Commented by ajfour last updated on 12/Apr/21

	$Y e s s i r, y o u a r e r i g h t; I^{'} v e e d i t e d .$
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