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Question Number 80648 by mr W last updated on 05/Feb/20

Commented by mr W last updated on 05/Feb/20

$t h r e e t h i n - w a l l t u b e s w i t h i d e n t i c a l$ $s i z e a n d m a s s a r e r e l e a s e d f r o m t h e$ $s t a t e a s s h o w n .$ $f i n d t h e t i m e w h i c h t h e u p p e r t u b e$ $t a k e s t o h i t t h e g r o u n d .$ $t h e r e i s n o f r i c t i o n b e t w e e n t h e t u b e s,$ $b u t t h e g r o u n d i s r o u g h s u c h t h a t t h e$ $t u b e s {c a n}^{'} t s l i p o v e r i t .$
Commented by ajfour last updated on 05/Feb/20

Commented by ajfour last updated on 05/Feb/20
$V=−(dy/dt) , A=(dV/dt) y=2rsin θ ⇒ −V=2rcos θ((dθ/dt)) friction on right tube is f^← fr=(mr^2 )(a/r) ⇒ f=ma Ncos θ−f=m(dv/dt) = ma ⇒ N=((2ma)/(cos θ)) mg−2Nsin θ=((mVdV)/dy) = mA ⇒ mg−((4masin θ)/(cos θ))=mA ⇒ A=g−4atan θ y=2rsin θ , x=2rcos θ Vsin θ=vcos θ .......(I) V=((vcos θ)/(sin θ))=−2rcos θ((dθ/dt)) ⇒ v=−2rsin θ((dθ/dt)) .....(II) differentiating (I) ⇒ V=vcot θ ⇒ Asin θ+Vcos θ(dθ/dt) = acos θ−vsin θ(dθ/dt) ⇒ gsin θ−4asin θtan θ +vcos θcot θ((dθ/dt)) = acos θ−vsin θ((dθ/dt)) ⇒ gsin θ−4(((sin^2 θ)/(cos θ)))(dv/dt) +((vcos^2 θ)/(sin θ))((dθ/dt))=(dv/dt)cos θ−vsin θ(dθ/dt) using (II) mgr(√3)−2mgrsin θ = 2mv^2 +((mV^( 2) )/2) gr(√3)−2grsin θ=2v^2 +((v^2 cos^2 θ)/(2sin^2 θ)) v^2 =((2gr((√3)−2sin θ)sin^2 θ)/(3+sin^2 θ)) ⇒ 2v(dv/dt)=2gr{(((3+sin^2 θ)(2(√3)sin θcos θ−6sin^2 θcos θ)−2sin θcos θ((√3)−2sin θ)sin^2 θ)/((1+sin^2 θ)^2 ))} a=(dv/dt)= 0 ⇒ N=0 ⇒ (3+sin^2 θ)(2(√3)cos θ−6sin θcos θ)=2cos θ((√3)−2sin θ)sin^2 θ θ ≈ 33.92046° .....$
$V = - \frac{d y}{d t}, A = \frac{d V}{d t}$ $y = 2 r \sin θ \Rightarrow - V = 2 r \cos θ (\frac{d θ}{d t})$ $f r i c t i o n o n r i g h t t u b e i s \overset{\leftarrow}{f}$ $f r = ({m r}^{2}) \frac{a}{r} \Rightarrow f = m a$ $N \cos θ - f = m \frac{d v}{d t} = m a$ $\Rightarrow N = \frac{2 m a}{\cos θ}$ $m g - 2 N \sin θ = \frac{m V d V}{d y} = m A$ $\Rightarrow m g - \frac{4 m a \sin θ}{\cos θ} = m A$ $\Rightarrow A = g - 4 a \tan θ$ $y = 2 r \sin θ, x = 2 r \cos θ$ $V \sin θ = v \cos θ . . . . . . . (I)$ $V = \frac{v \cos θ}{\sin θ} = - 2 r \cos θ (\frac{d θ}{d t})$ $\Rightarrow v = - 2 r \sin θ (\frac{d θ}{d t}) . . . . . (I I)$ $d i f f e r e n t i a t i n g (I)$ $\Rightarrow V = v \cot θ$ $\Rightarrow A \sin θ + V \cos θ \frac{d θ}{d t} =$ $a \cos θ - v \sin θ \frac{d θ}{d t}$ $\Rightarrow g \sin θ - 4 a \sin θ \tan θ$ $+ v \cos θ \cot θ (\frac{d θ}{d t}) =$ $a \cos θ - v \sin θ (\frac{d θ}{d t})$ $\Rightarrow g \sin θ - 4 (\frac{\sin^{2} θ}{\cos θ}) \frac{d v}{d t}$ $+ \frac{v \cos^{2} θ}{\sin θ} (\frac{d θ}{d t}) = \frac{d v}{d t} \cos θ - v \sin θ \frac{d θ}{d t}$ $u s i n g (I I)$ $m g r \sqrt{3} - 2 m g r \sin θ$ $= 2 {m v}^{2} + \frac{{m V}^{2}}{2}$ $g r \sqrt{3} - 2 g r \sin θ = 2 v^{2} + \frac{v^{2} \cos^{2} θ}{2 \sin^{2} θ}$ $v^{2} = \frac{2 g r (\sqrt{3} - 2 \sin θ) \sin^{2} θ}{3 + \sin^{2} θ}$ $\Rightarrow 2 v \frac{d v}{d t} = 2 g r {\frac{(3 + \sin^{2} θ) (2 \sqrt{3} \sin θ \cos θ - 6 \sin^{2} θ \cos θ) - 2 \sin θ \cos θ (\sqrt{3} - 2 \sin θ) \sin^{2} θ}{{(1 + \sin^{2} θ)}^{2}}}$ $a = \frac{d v}{d t} = 0 \Rightarrow N = 0$ $\Rightarrow$ $(3 + \sin^{2} θ) (2 \sqrt{3} \cos θ - 6 \sin θ \cos θ) = 2 \cos θ (\sqrt{3} - 2 \sin θ) \sin^{2} θ$ $θ \approx 33.92046 °$ $. . . . .$
Commented by mr W last updated on 05/Feb/20

$s u p e r!$
Commented by mr W last updated on 06/Feb/20

$f i n a l e q u a t i o n c a n b e s i m p l i f i e d t o :$ $\sin^{3} θ + 9 \sin θ - 3 \sqrt{3} = 0$ $\sin θ = \sqrt[3]{\frac{3 \sqrt{3} (\sqrt{5} + 1)}{2}} - \sqrt[3]{\frac{3 \sqrt{3} (\sqrt{5} - 1)}{2}}$ $\Rightarrow θ = 33.920456 °$
Commented by ajfour last updated on 06/Feb/20

Commented by ajfour last updated on 06/Feb/20

$l o s s i n p o t e n t i a l e n e r g y i s$ $m g (r \sqrt{3} - 2 r \sin θ)$ ${s h o u l d n}^{'} t i t b e s o, S i r ?$
Commented by mr W last updated on 06/Feb/20

$c e r t a i n l y s o . i h a d a s h o r t c i r c u i t .$ $s o r r y!$
Commented by zainal tanjung last updated on 06/Feb/20


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