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Question Number 212901 by ajfour last updated on 26/Oct/24

Commented by ajfour last updated on 26/Oct/24

$F i n d x a n d u_{m i n} .$
Commented by ajfour last updated on 26/Oct/24
just published this on Magnetic Field (my video lecture on youtube) https://youtu.be/qx0TX9LSlzQ?si=S-eutbZgXcerKPSe
	Answered by mr W last updated on 27/Oct/24

	Commented by ajfour last updated on 27/Oct/24

	$l e t p r o j e c t i l e l a u n c h e d f r o m c y l i n d e r$ $t a n g e n t i a l l y .$ $h - R (1 + \sin φ) = R (1 + \cos φ) \cot φ$ $- \frac{{g R}^{2} {(1 + \cos φ)}^{2}}{2 [u^{2} - 2 g R (1 + \sin φ)] \underset{―}{\sin^{2} φ}}$ $F o r R = 2, h = 6 u = u_{m i n} w h e n$ $φ \approx 0.4401 r a d \approx 25.2 °$ $f r o m a b o v e a p p l y i n g d i f f e r e n t i a l$ $c a l c u l u s w e f i n d u_{m i n} f o r φ = φ_{0}$ $r e w i n d i n g w e s e e b a l l d r o p s t o$ $g r o u n d w h e n r e l e a s e d t a n g e n t i a l l y$ $f r o m c y l i n d e r, s o i f$ $v_{0} = \sqrt{u_{m i n}^{2} - 2 g R (1 + \sin φ_{0})}$ $N o w L = R (1 + \cos φ_{0}) + s$ $- s = (v_{0} \sin φ) t$ $- R (1 + \sin φ_{0}) = (v_{0} \sin φ_{0}) t - \frac{{g t}^{2}}{2}$ $★$
	Commented by mr W last updated on 27/Oct/24
	$x=u cos θ t y=u sin θ t−((gt^2 )/2) y=tan θ x−(((1+tan^2 θ)gx^2 )/(2u^2 )) let m=tan θ y=mx−(((1+m^2 )gx^2 )/(2u^2 )) (dy/dx)=m−(((1+m^2 )gx)/u^2 ) at top of wall: h=mL−(((1+m^2 )gL^2 )/(2u^2 )) ⇒(h/R)=m((L/R))−(((1+m^2 )gR((L/R))^2 )/(2u^2 )) ...(i) at touching point on circle: x_P =L−R−R cos ϕ y_P =R(1+sin ϕ) R(1+sin ϕ)=m(L−R−R cos ϕ)−(((1+m^2 )g(L−R−R cos ϕ)^2 )/(2u^2 )) ⇒(1+sin ϕ)=m((L/R)−1−cos ϕ)−(((1+m^2 )gR((L/R)−1−cos ϕ)^2 )/(2u^2 )) ...(ii) (1/(tan ϕ))=m−(((1+m^2 )g(L−R−R cos ϕ))/u^2 ) ⇒(1/(tan ϕ))=m−(((1+m^2 )gR((L/R)−1−cos ϕ))/u^2 ) ...(iii) let ξ=(L/R), Φ=((gR)/(2u^2 )), μ=(h/R) μ=mξ−(1+m^2 )ξ^2 Φ ...(i) 1+sin ϕ=m(ξ−1−cos ϕ)−(1+m^2 )(ξ−1−cos ϕ)^2 Φ ...(ii) (1/(tan ϕ))=m−2(1+m^2 )(ξ−1−cos ϕ)Φ ...(iii) from (ii) and (iii): 2(1+sin ϕ)=(ξ−1−cos ϕ)(m+(1/(tan ϕ))) ⇒m=((2(1+sin ϕ))/(ξ−1−cos ϕ))−(1/(tan ϕ)) ...(iv) from (i) and (iii): (ξ^2 /(tan ϕ))=mξ^2 −2(ξ−1−cos ϕ)(mξ−μ) ...(v) (iv) into (v): ((1+sin ϕ)/(ξ−1−cos ϕ))−(1/(tan ϕ))−(1−((1+cos ϕ)/ξ))[((2(1+sin ϕ))/(ξ−1−cos ϕ))−(1/(tan ϕ))−(μ/ξ)]=0 ...(II) 2(1+sin ϕ)ξ^3 −[(1+sin ϕ)(3+2 cos ϕ)−(((1+cos ϕ−μ tan ϕ))/(tan ϕ))]ξ^2 −(((1+cos ϕ−2μ tan ϕ)(1+cos ϕ))/(tan ϕ))ξ−μ(1+cos ϕ)^2 =0 ...(II′) it′s clear that ξ>2. (iv) into (i): Φ=(([((2(1+sin ϕ))/(ξ−1−cos ϕ))−(1/(tan ϕ))]ξ−μ)/({1+[((2(1+sin ϕ))/(ξ−1−cos ϕ))−(1/(tan ϕ))]^2 }ξ^2 )) ...(I) now it is to find the maximum of Φ(ξ, ϕ), which means the minimum of u, under the condition (II).$
	$x = u \cos θ t$ $y = u \sin θ t - \frac{{g t}^{2}}{2}$ $y = \tan θ x - \frac{(1 + \tan^{2} θ) {g x}^{2}}{2 u^{2}}$ $l e t m = \tan θ$ $y = m x - \frac{(1 + m^{2}) {g x}^{2}}{2 u^{2}}$ $\frac{d y}{d x} = m - \frac{(1 + m^{2}) g x}{u^{2}}$ $a t t o p o f w a l l :$ $h = m L - \frac{(1 + m^{2}) {g L}^{2}}{2 u^{2}}$ $\Rightarrow \frac{h}{R} = m (\frac{L}{R}) - \frac{(1 + m^{2}) g R {(\frac{L}{R})}^{2}}{2 u^{2}} . . . (i)$ $a t t o u c h i n g p o i n t o n c i r c l e :$ $x_{P} = L - R - R \cos φ$ $y_{P} = R (1 + \sin φ)$ $R (1 + \sin φ) = m (L - R - R \cos φ) - \frac{(1 + m^{2}) g {(L - R - R \cos φ)}^{2}}{2 u^{2}}$ $\Rightarrow (1 + \sin φ) = m (\frac{L}{R} - 1 - \cos φ) - \frac{(1 + m^{2}) g R {(\frac{L}{R} - 1 - \cos φ)}^{2}}{2 u^{2}} . . . (i i)$ $\frac{1}{\tan φ} = m - \frac{(1 + m^{2}) g (L - R - R \cos φ)}{u^{2}}$ $\Rightarrow \frac{1}{\tan φ} = m - \frac{(1 + m^{2}) g R (\frac{L}{R} - 1 - \cos φ)}{u^{2}} . . . (i i i)$ $l e t ξ = \frac{L}{R}, Φ = \frac{g R}{2 u^{2}}, μ = \frac{h}{R}$ $μ = m ξ - (1 + m^{2}) ξ^{2} Φ . . . (i)$ $1 + \sin φ = m (ξ - 1 - \cos φ) - (1 + m^{2}) {(ξ - 1 - \cos φ)}^{2} Φ . . . (i i)$ $\frac{1}{\tan φ} = m - 2 (1 + m^{2}) (ξ - 1 - \cos φ) Φ . . . (i i i)$ $f r o m (i i) a n d (i i i) :$ $2 (1 + \sin φ) = (ξ - 1 - \cos φ) (m + \frac{1}{\tan φ})$ $\Rightarrow m = \frac{2 (1 + \sin φ)}{ξ - 1 - \cos φ} - \frac{1}{\tan φ} . . . (i v)$ $f r o m (i) a n d (i i i) :$ $\frac{ξ^{2}}{\tan φ} = m ξ^{2} - 2 (ξ - 1 - \cos φ) (m ξ - μ) . . . (v)$ $(i v) i n t o (v) :$ $\frac{1 + \sin φ}{ξ - 1 - \cos φ} - \frac{1}{\tan φ} - (1 - \frac{1 + \cos φ}{ξ}) [\frac{2 (1 + \sin φ)}{ξ - 1 - \cos φ} - \frac{1}{\tan φ} - \frac{μ}{ξ}] = 0 . . . (I I)$ $2 (1 + \sin φ) ξ^{3} - [(1 + \sin φ) (3 + 2 \cos φ) - \frac{(1 + \cos φ - μ \tan φ)}{\tan φ}] ξ^{2} - \frac{(1 + \cos φ - 2 μ \tan φ) (1 + \cos φ)}{\tan φ} ξ - μ {(1 + \cos φ)}^{2} = 0 . . . ({I I}^{'})$ ${i t}^{'} s c l e a r t h a t ξ > 2 .$ $(i v) i n t o (i) :$ $Φ = \frac{[\frac{2 (1 + \sin φ)}{ξ - 1 - \cos φ} - \frac{1}{\tan φ}] ξ - μ}{{1 + {[\frac{2 (1 + \sin φ)}{ξ - 1 - \cos φ} - \frac{1}{\tan φ}]}^{2}} ξ^{2}} . . . (I)$ $n o w i t i s t o f i n d t h e m a x i m u m o f$ $Φ (ξ, φ), w h i c h m e a n s t h e m i n i m u m$ $o f u, u n d e r t h e c o n d i t i o n (I I) .$
	Commented by mr W last updated on 27/Oct/24

	Commented by mr W last updated on 27/Oct/24

	Commented by mr W last updated on 27/Oct/24

	Commented by mr W last updated on 27/Oct/24

	Commented by ajfour last updated on 27/Oct/24

	Commented by ajfour last updated on 27/Oct/24

	$\frac{u^{2}}{2 g R} = 1 + \sin φ$ $+ \frac{{(1 + \cos φ)}^{2} / (4 \sin^{2} φ)}{(1 + \cos φ) \cot φ + \sin φ + 1 - h / R}$
	Commented by mr W last updated on 27/Oct/24

	$y e s, {t h a t}^{'} s r i g h t!$ $g r e a t a p p r o a c h!$
	Commented by mr W last updated on 27/Oct/24

	Commented by ajfour last updated on 27/Oct/24
	Thank you in everyrespect, sir.
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