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Question Number 187640 by ajfour last updated on 19/Feb/23

Commented by ajfour last updated on 19/Feb/23

$I f t h e h e m i s p h e r e r a d i u s R . T h e$ $b a l l r a d i u s b a n d m a s s m . F i n d$ $T e n s i o n i n s t r i n g i n t e r m s o f m g,$ $θ, b, R .$
	Answered by a.lgnaoui last updated on 19/Feb/23
	$tan θ=((ma_c )/(mg))=(v^2 /(Rg)) v=R(dθ/dt) T { ((Tx=Tcos θ=−mg)),((Ty=Tsin θ=ma_c )) :} T=m(√(g^2 +R((dθ/dt) )^2 )) (suite) η=(m/V)=(m/(4/3πb^3 )) T=(4/3)πb^3 η(√(g^2 +R((dθ/dt))^2 )) η=masse volumique$
	$\tan θ = \frac{{m a}_{c}}{m g} = \frac{v^{2}}{R g} v = R \frac{d θ}{d t}$ $T {\begin{cases} T x = T \cos θ = - m g \\ T y = T \sin θ = {m a}_{c} \end{cases}$ $T = m \sqrt{g^{2} + R {(\frac{d θ}{d t})}^{2}}$ $(s u i t e)$ $η = \frac{m}{V} = \frac{m}{4 / 3 π b^{3}}$ $T = \frac{4}{3} π b^{3} η \sqrt{g^{2} + R {(\frac{d θ}{d t})}^{2}}$ $η = m a s s e v o l u m i q u e$
	Answered by mr W last updated on 20/Feb/23

	Commented by mr W last updated on 20/Feb/23

	$\cos ϕ = \frac{R}{R + b}$ $T \cos ϕ = m g \cos θ$ $\Rightarrow T = \frac{m g \cos θ}{\cos ϕ} = (1 + \frac{b}{R}) m g \cos θ ✓$
	Commented by ajfour last updated on 20/Feb/23

	$v e r y f i n e, t h a n k s s i r!$ $W h a t i f s t r i n g b e a t t a c h e d t o$ $t h e s u r f a c e o f t h e b a l l (a n d c o u l d$ $b e h e a v y a s w e l l) ?$
	Commented by mr W last updated on 20/Feb/23

	$i f s t r i n g i s w e i g h t l e s s, t h e n i t m a k e s$ $n o d i f f e r e n c e w h e t h e r t h e s t r i n g i s$ $a t t a c h e d t o t h e s u r f a c e o r t o t h e$ $c e n t e r o f t h e b a l l, t h e t e n s i o n o f$ $s t r i n g p a s s e s a l w a y s t h e c e n t e r o f$ $t h e b a l l w h e r e N a n d m g i n t e r s e c t .$
	Commented by ajfour last updated on 20/Feb/23

	$y e s s i r w i t h o u t f r i c t i o n a n d l i g h t$ $s t r i n g t h e s o l u t i o n i s c o m p l e t e .$
	Commented by mr W last updated on 20/Feb/23

	$i f t h e s t r i n g h a s u n i f o r m m a s s ρ,$ $t h e n w e h a v e f o l l o w i n g s i t u a t i o n :$ $(B C = h a n g i n g s t r i n g)$
	Commented by mr W last updated on 20/Feb/23

	Commented by mr W last updated on 20/Feb/23

	Commented by mr W last updated on 20/Feb/23

	$T \sin ϕ = m g \cos θ$ $\Rightarrow T = \frac{m g \cos θ}{\sin ϕ}$ $T_{0} = T \cos (ϕ - θ)$ $a = \frac{T_{0}}{ρ g} = \frac{m g \cos θ \cos (ϕ - θ)}{ρ g \sin ϕ}$ $= \frac{m}{2 ρ} (\frac{1 + \cos 2 θ}{\tan ϕ} + \sin 2 θ)$ $i n x y - s y s t e m w i t h o r i g i n a t c e n t e r$ $o f h e m i s p h e r e :$ $x_{B} = R \cos (φ + θ)$ $y_{B} = R \sin (φ + θ)$ $x_{C} = (R + b) \cos θ - b \cos (ϕ - θ)$ $y_{C} = (R + b) \sin θ + b \sin (ϕ - θ)$ $\Rightarrow x_{C} - x_{B} = (R + b) \cos θ - b \cos (ϕ - θ) - R \cos (φ + θ)$ $\Rightarrow y_{B} - y_{C} = R \sin (φ + θ) - (R + b) \sin θ - b \sin (ϕ - θ)$ $h a n g i n g s t r i n g B C i s a s e g m e n t o f$ $c a t e n a r y w i t h f o l l o w i n g e q u a t i o n$ $i n x y - s y s t e m a s s h o w n :$ $y = a \cosh \frac{x}{a}$ $a t p o i n t C :$ $\tan (ϕ - θ) = \sinh \frac{x_{C}}{a}$ $\frac{x_{C}}{a} = \sinh^{- 1} (\tan (ϕ - θ)$ $y_{C} = a \cosh \frac{x_{C}}{a}$ $\Rightarrow y_{C} = a \sqrt{1 + \tan^{2} (ϕ - θ)}$ $a t p o i n t B :$ $\frac{1}{\tan (φ + θ)} = \sinh \frac{x_{B}}{a}$ $\Rightarrow \frac{x_{B}}{a} = \sinh^{- 1} \frac{1}{\tan (φ + θ)}$ $y_{B} = a \cosh \frac{x_{B}}{a}$ $\Rightarrow y_{B} = a \sqrt{1 + \frac{1}{\tan^{2} (φ + θ)}}$ $x_{B} - x_{C} = a [\sinh^{- 1} \frac{1}{\tan (φ + θ)} - \sinh^{- 1} (\tan (ϕ - θ))]$ $\Rightarrow (1 + \frac{b}{R}) \cos θ - \frac{b}{R} \cos (ϕ - θ) - \cos (φ + θ) = \frac{m}{2 R ρ} (\frac{1 + \cos 2 θ}{\tan ϕ} + \sin 2 θ) [\sinh^{- 1} \frac{1}{\tan (φ + θ)} - \sinh^{- 1} (\tan (ϕ - θ))] . . . (i)$ $y_{B} - y_{C} = a [\sqrt{1 + \frac{1}{\tan^{2} (φ + θ)}} - \sqrt{1 + \tan^{2} (ϕ - θ)}]$ $\Rightarrow \sin (φ + θ) - (1 + \frac{b}{R}) \sin θ - \frac{b}{R} \sin (ϕ - θ) = \frac{m}{2 R ρ} (\frac{1 + \cos 2 θ}{\tan ϕ} + \sin 2 θ) [\sqrt{1 + \frac{1}{\tan^{2} (φ + θ)}} - \sqrt{1 + \tan^{2} (ϕ - θ)}] . . . (i i)$ $w e c a n s o l v e (i) a n d (i i) f o r φ a n d ϕ .$
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