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Question Number 176660 by mr W last updated on 24/Sep/22

Commented by mr W last updated on 25/Sep/22

$f i n d t h e t i m e t h e r o d t a k e s t o r e a c h$ $t h e g r o u n d .$ $t h e c o e f f i c i e n t o f f r i c t i o n b e t w e e n$ $r o d a n d w a l l i s μ w h i c h i s s m a l l e n o u g h$ $s o t h a t t h e r o d c a n s l i p o n t h e w a l l .$ $t h e f r i c t i o n b e t w e e n r o d a n d g r o u n d$ $i s l a r g e e n o u g h s o t h a t t h e r o d {d o e s n}^{'} t$ $s l i p o n t h e g r o u n d .$
	Answered by mr W last updated on 25/Sep/22

	Commented by mr W last updated on 26/Sep/22
	$L=(√(b^2 +h^2 )) I=((ML^2 )/3)=((M(b^2 +h^2 ))/3) ω=(dθ/dt) α=(d^2 θ/dt^2 )=(dω/dt)=ω(dω/dθ) b tan ϕ=h sin θ ⇒ϕ=tan^(−1) (((h sin θ)/b)) (dϕ/dt)=(((h cos θ)/b)/(1+(((h sin θ)/b))^2 ))×ω (d^2 ϕ/dt^2 )=(((h cos θ)/b)/(1+(((h sin θ)/b))^2 ))×ω×(dω/dθ)+ω^2 ×[−(((((h cos θ)/b))^2 )/((1+(((h sin θ)/b))^2 )^2 ))−(((h sin θ)/b)/(1+(((h sin θ)/b))^2 ))] (d^2 ϕ/dt^2 )=(((h cos θ)/b)/(1+(((h sin θ)/b))^2 ))×ω×(dω/dθ)−(ω^2 /(1+(((h sin θ)/b))^2 ))×[(((((h cos θ)/b))^2 )/(1+(((h sin θ)/b))^2 ))+((h sin θ)/b)] I(d^2 ϕ/dt^2 )=N h sin θ N=((M(b^2 +h^2 ))/(3h sin θ (1+(((h sin θ)/b))^2 ))){((h cos θ)/b)×ω×(dω/dθ)−ω^2 [(((((h cos θ)/b))^2 )/(1+(((h sin θ)/b))^2 ))+((h sin θ)/b)]} I(d^2 θ/dt^2 )=−fh+((Mgh sin θ)/2) I(d^2 θ/dt^2 )=−μNh+((Mgh sin θ)/2) ω(dω/dθ)=−(μ/(sin θ (1+(((h sin θ)/b))^2 ))){((h cos θ)/b)×ω×(dω/dθ)−ω^2 [(((((h cos θ)/b))^2 )/(1+(((h sin θ)/b))^2 ))+((h sin θ)/b)]}+((3gh sin θ)/(2(b^2 +h^2 ))) let λ=(h/b) ω(dω/dθ)=−(μ/(sin θ (1+λ^2 sin^2 θ))){λ cos θ×ω×(dω/dθ)−ω^2 (((λ^2 cos^2 θ)/(1+λ^2 sin^2 θ))+λ sin θ)}+((3gλ sin θ)/(2b(1+λ^2 ))) (1/2)((1/λ)+((μ cos θ)/(sin θ (1+λ^2 sin^2 θ))))(dω^2 /dθ)−(μ/(sin θ (1+λ^2 sin^2 θ)))(((λcos^2 θ)/(1+λ^2 sin^2 θ))+sin θ)ω^2 −((3g sin θ)/(2b(1+λ^2 )))=0 (dω^2 /dθ)−((2μλ(λcos^2 θ+λ^2 sin^3 θ+sin θ))/((sin θ (1+λ^2 sin^2 θ)+μλ cos θ)(1+λ^2 sin^2 θ)))ω^2 −((3gλ sin^2 θ (1+λ^2 sin^2 θ))/(b(1+λ^2 )(sin θ (1+λ^2 sin^2 θ)+μλ cos θ)))=0 let Φ=ω^2 (dΦ/dθ)−((2μλ(λcos^2 θ+λ^2 sin^3 θ+sin θ))/([sin θ (1+λ^2 sin^2 θ)+μλ cos θ](1+λ^2 sin^2 θ)))Φ−((3gλ sin^2 θ (1+λ^2 sin^2 θ))/(b(1+λ^2 )[sin θ (1+λ^2 sin^2 θ)+μλ cos θ]))=0 with p(θ)=((2μλ(λcos^2 θ+λ^2 sin^3 θ+sin θ))/([sin θ (1+λ^2 sin^2 θ)+μλ cos θ](1+λ^2 sin^2 θ))) q(θ)=((3gλ sin^2 θ (1+λ^2 sin^2 θ))/(b(1+λ^2 )[sin θ (1+λ^2 sin^2 θ)+μλ cos θ])) ⇒(dΦ/dθ)−p(θ)Φ−q(θ)=0 we get Φ=Φ(θ) ⇒ω=(√(Φ(θ))) ⇒(dθ/dt)=(√(Φ(θ))) ⇒dt=(dθ/( (√(Φ(θ))))) ⇒∫_0 ^T dt=∫_0 ^(π/2) (dθ/( (√(Φ(θ))))) ⇒T=∫_0 ^(π/2) (dθ/( (√(Φ(θ)))))$
	$L = \sqrt{b^{2} + h^{2}}$ $I = \frac{{M L}^{2}}{3} = \frac{M (b^{2} + h^{2})}{3}$ $ω = \frac{d θ}{d t}$ $α = \frac{d^{2} θ}{{d t}^{2}} = \frac{d ω}{d t} = ω \frac{d ω}{d θ}$ $b \tan φ = h \sin θ$ $\Rightarrow φ = \tan^{- 1} (\frac{h \sin θ}{b})$ $\frac{d φ}{d t} = \frac{\frac{h \cos θ}{b}}{1 + {(\frac{h \sin θ}{b})}^{2}} \times ω$ $\frac{d^{2} φ}{{d t}^{2}} = \frac{\frac{h \cos θ}{b}}{1 + {(\frac{h \sin θ}{b})}^{2}} \times ω \times \frac{d ω}{d θ} + ω^{2} \times [- \frac{{(\frac{h \cos θ}{b})}^{2}}{{(1 + {(\frac{h \sin θ}{b})}^{2})}^{2}} - \frac{\frac{h \sin θ}{b}}{1 + {(\frac{h \sin θ}{b})}^{2}}]$ $\frac{d^{2} φ}{{d t}^{2}} = \frac{\frac{h \cos θ}{b}}{1 + {(\frac{h \sin θ}{b})}^{2}} \times ω \times \frac{d ω}{d θ} - \frac{ω^{2}}{1 + {(\frac{h \sin θ}{b})}^{2}} \times [\frac{{(\frac{h \cos θ}{b})}^{2}}{1 + {(\frac{h \sin θ}{b})}^{2}} + \frac{h \sin θ}{b}]$ $I \frac{d^{2} φ}{{d t}^{2}} = N h \sin θ$ $N = \frac{M (b^{2} + h^{2})}{3 h \sin θ (1 + {(\frac{h \sin θ}{b})}^{2})} {\frac{h \cos θ}{b} \times ω \times \frac{d ω}{d θ} - ω^{2} [\frac{{(\frac{h \cos θ}{b})}^{2}}{1 + {(\frac{h \sin θ}{b})}^{2}} + \frac{h \sin θ}{b}]}$ $I \frac{d^{2} θ}{{d t}^{2}} = - f h + \frac{M g h \sin θ}{2}$ $I \frac{d^{2} θ}{{d t}^{2}} = - μ N h + \frac{M g h \sin θ}{2}$ $ω \frac{d ω}{d θ} = - \frac{μ}{\sin θ (1 + {(\frac{h \sin θ}{b})}^{2})} {\frac{h \cos θ}{b} \times ω \times \frac{d ω}{d θ} - ω^{2} [\frac{{(\frac{h \cos θ}{b})}^{2}}{1 + {(\frac{h \sin θ}{b})}^{2}} + \frac{h \sin θ}{b}]} + \frac{3 g h \sin θ}{2 (b^{2} + h^{2})}$ $l e t λ = \frac{h}{b}$ $ω \frac{d ω}{d θ} = - \frac{μ}{\sin θ (1 + λ^{2} \sin^{2} θ)} {λ \cos θ \times ω \times \frac{d ω}{d θ} - ω^{2} (\frac{λ^{2} \cos^{2} θ}{1 + λ^{2} \sin^{2} θ} + λ \sin θ)} + \frac{3 g λ \sin θ}{2 b (1 + λ^{2})}$ $\frac{1}{2} (\frac{1}{λ} + \frac{μ \cos θ}{\sin θ (1 + λ^{2} \sin^{2} θ)}) \frac{d ω^{2}}{d θ} - \frac{μ}{\sin θ (1 + λ^{2} \sin^{2} θ)} (\frac{λ \cos^{2} θ}{1 + λ^{2} \sin^{2} θ} + \sin θ) ω^{2} - \frac{3 g \sin θ}{2 b (1 + λ^{2})} = 0$ $\frac{d ω^{2}}{d θ} - \frac{2 μ λ (λ \cos^{2} θ + λ^{2} \sin^{3} θ + \sin θ)}{(\sin θ (1 + λ^{2} \sin^{2} θ) + μ λ \cos θ) (1 + λ^{2} \sin^{2} θ)} ω^{2} - \frac{3 g λ \sin^{2} θ (1 + λ^{2} \sin^{2} θ)}{b (1 + λ^{2}) (\sin θ (1 + λ^{2} \sin^{2} θ) + μ λ \cos θ)} = 0$ $l e t Φ = ω^{2}$ $\frac{d Φ}{d θ} - \frac{2 μ λ (λ \cos^{2} θ + λ^{2} \sin^{3} θ + \sin θ)}{[\sin θ (1 + λ^{2} \sin^{2} θ) + μ λ \cos θ] (1 + λ^{2} \sin^{2} θ)} Φ - \frac{3 g λ \sin^{2} θ (1 + λ^{2} \sin^{2} θ)}{b (1 + λ^{2}) [\sin θ (1 + λ^{2} \sin^{2} θ) + μ λ \cos θ]} = 0$ $w i t h p (θ) = \frac{2 μ λ (λ \cos^{2} θ + λ^{2} \sin^{3} θ + \sin θ)}{[\sin θ (1 + λ^{2} \sin^{2} θ) + μ λ \cos θ] (1 + λ^{2} \sin^{2} θ)}$ $q (θ) = \frac{3 g λ \sin^{2} θ (1 + λ^{2} \sin^{2} θ)}{b (1 + λ^{2}) [\sin θ (1 + λ^{2} \sin^{2} θ) + μ λ \cos θ]}$ $\Rightarrow \frac{d Φ}{d θ} - p (θ) Φ - q (θ) = 0$ $w e g e t Φ = Φ (θ)$ $\Rightarrow ω = \sqrt{Φ (θ)}$ $\Rightarrow \frac{d θ}{d t} = \sqrt{Φ (θ)}$ $\Rightarrow d t = \frac{d θ}{\sqrt{Φ (θ)}}$ $\Rightarrow \int_{0}^{T} d t = \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{Φ (θ)}}$ $\Rightarrow T = \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{Φ (θ)}}$
	Commented by Tawa11 last updated on 25/Sep/22

	$Great sir$
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