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Question Number 100928 by ajfour last updated on 29/Jun/20

Commented by ajfour last updated on 29/Jun/20

$F i n d r i n t e r m s o f R .$
Commented by ajfour last updated on 30/Jun/20

$I t s p h y s i c s p l u s m a t h s, S i r .$
	Answered by bramlex last updated on 29/Jun/20

	Commented by bramlex last updated on 29/Jun/20

	$x^{2} + {(R - r)}^{2} = {(R + r)}^{2}$ $x^{2} - 2 r R = 2 r R \to x = 2 \sqrt{r R}$ $(1) {(x - R)}^{2} + {(y - R)}^{2} = R^{2}$ $\to x^{2} + y^{2} - 2 R x - 2 R y + R^{2} = 0$ $(2) (x - {(2 \sqrt{r R} + r)}^{2} + {(y - r)}^{2} = r^{2}$ $\to x^{2} - 2 (r + 2 \sqrt{r R}) x + {(r + 2 \sqrt{r R})}^{2} + y^{2} - 2 r y = 0$ $. . .$
	Answered by mr W last updated on 29/Jun/20

	Commented by mr W last updated on 29/Jun/20

	$a t p o s i t i o n θ :$ $\frac{1}{2} {m u}^{2} = R (1 - \cos θ) m g$ $\Rightarrow u = \sqrt{2 R (1 - \cos θ) g}$ $m g \cos θ - N = \frac{{m u}^{2}}{R}$ $N = 0 w h e n c o n t a c t g e t s l o s t$ $m g \cos θ = \frac{m \times 2 R (1 - \cos θ) g}{R}$ $\Rightarrow \cos θ = \frac{2}{3}$ $\Rightarrow u = \sqrt{\frac{2 R g}{3}}$ $t r a c k o f m a s s m a f t e r p o i n t B :$ $x = R \sin θ + u \cos θ t$ $\Rightarrow x = \frac{\sqrt{5} R}{3} + \frac{2}{3} \sqrt{\frac{2 R g}{3}} t$ $y = R (1 + \cos θ) - u \sin θ t - \frac{1}{2} {g t}^{2}$ $\Rightarrow y = \frac{5 R}{3} - \frac{\sqrt{5}}{3} \sqrt{\frac{2 R g}{3}} t - \frac{1}{2} {g t}^{2}$ $D (\sqrt{{(R + r)}^{2} - {(R - r)}^{2}}, r) \equiv D (2 \sqrt{R r}, r)$ $l e t Φ = {D C}^{2}$ $Φ = {(x - 2 \sqrt{R r})}^{2} + {(y - r)}^{2}$ $Φ = {(\frac{\sqrt{5} R}{3} + \frac{2}{3} \sqrt{\frac{2 R g}{3}} t - 2 \sqrt{R r})}^{2} + {(\frac{5 R}{3} - \frac{\sqrt{5}}{3} \sqrt{\frac{2 R g}{3}} t - \frac{1}{2} {g t}^{2} - r)}^{2}$ $\frac{d Φ}{d t} = 2 (\frac{\sqrt{5} R}{3} + \frac{2}{3} \sqrt{\frac{2 R g}{3}} t - 2 \sqrt{R r}) (\frac{2}{3} \sqrt{\frac{2 R g}{3}}) + 2 (\frac{5 R}{3} - \frac{\sqrt{5}}{3} \sqrt{\frac{2 R g}{3}} t - \frac{1}{2} {g t}^{2} - r) (- \frac{\sqrt{5}}{3} \sqrt{\frac{2 R g}{3}} - g t) = 0$ $\Rightarrow 12 (\sqrt{30} + 4 \sqrt{\frac{g}{R}} t - 6 \sqrt{6} \sqrt{\frac{r}{R}}) = (30 - 2 \sqrt{30} \sqrt{\frac{g}{R}} t - 9 {(\sqrt{\frac{g}{R}} t)}^{2} - 18 \frac{r}{R}) (\sqrt{30} + 9 \sqrt{\frac{g}{R}} t)$ $Φ_{m i n} = r^{2}$ $\Rightarrow {(\frac{\sqrt{5}}{3} + \frac{2 \sqrt{6}}{9} \sqrt{\frac{g}{R}} t - 2 \sqrt{\frac{r}{R}})}^{2} + {(\frac{5}{3} - \frac{\sqrt{30}}{9} \sqrt{\frac{g}{R}} t - \frac{1}{2} {(\sqrt{\frac{g}{R}} t)}^{2} - \frac{r}{R})}^{2} = {(\frac{r}{R})}^{2}$ $l e t δ = \sqrt{\frac{g}{R}} t, λ = \sqrt{\frac{r}{R}}$ $\Rightarrow 12 (\sqrt{30} + 4 δ - 6 \sqrt{6} λ) = (30 - 2 \sqrt{30} δ - 9 δ^{2} - 18 λ^{2}) (\sqrt{30} + 9 δ) . . . (i)$ $\Rightarrow 4 {(3 \sqrt{5} + 2 \sqrt{6} δ - 18 λ)}^{2} + {(30 - 2 \sqrt{30} δ - 9 δ^{2} - 18 λ^{2})}^{2} = 324 λ^{4} . . . (i i)$ $w e g e t f r o m (i) a n d (i i) :$ $λ \approx 0.53716$ $\Rightarrow r = λ^{2} R \approx 0.2885 R$
	Commented by mr W last updated on 29/Jun/20

	Commented by ajfour last updated on 29/Jun/20

	${h a v e n}^{'} t b e e n t h r o u g h a l l o f i t s i r, b u t$ $r e a l l y l o o k s s u p e r b s o l u t i o n, i w a n t$ $t o t r y o n m y o w n f o r a l i t t l e w h i l e . . .$
	Commented by ajfour last updated on 29/Jun/20

	$c a n y o u h e l p m e s i r, w i t h t h e e q u a t i o n$ $o f p a r a b o l a w i t h s h o w n a x e s . . ?$
	Commented by mr W last updated on 30/Jun/20

	Commented by ajfour last updated on 02/Jul/20
	$x=Rsin θ+(ucos θ)t y=R(1+cos θ)−(usin θ)t−((gt^2 )/2) As cos θ=(2/3) , sin θ=((√5)/3) , u^2 =((2Rg)/3) y=((5R)/3)−(((u(√5))/3))(((x−((R(√5))/3)))/((((2u)/3))))−(g/2)(((x−((R(√5))/3))^2 )/((((2u)/3))^2 )) y=((5R)/3)−((√5)/2)(x−((R(√5))/3))−((27)/(16R))(x−((R(√5))/3))^2 (dy/dx)=−((√5)/2)−((27)/(8R))(x−((R(√5))/3))=0 ⇒ x_0 =((R(√5))/3)−((8R)/(27))(((√5)/2)) = ((5(√5)R)/(27)) y_0 =((5R)/3)−((√5)/2)(((5(√5)R)/(27))−((9R(√5))/(27))) −((27)/(16R))(((80R^2 )/(27×27))) ⇒ y_0 =((5R)/3)+((10R)/(27))−((5R)/(27))=((50R)/(27)) < 2R Thus eq. of parabola is y=((50R)/(27))−((27)/(16R))(x−((5(√5)R)/(27)))^2 (dy/dx)=−((27)/(8R))(x−((5(√5)R)/(27))) say this parabola touches the smaller circle at P (h,k) and let center of smaller circle is C(2(√(Rr)), r); then h=2(√(Rr))+rcos 𝛃 k=r+rsin 𝛃 cot β=((27)/(8R))(h−((5(√5)R)/(27)))=((h−2(√(Rr)))/(k−r)) ⇒ ((27)/8)z=(((z+((5(√5))/(27)))−2(√s))/(((50)/(27))−((27)/(16))z^2 −s)) ......(i) k=((50R)/(27))−((27)/(16R))(h−((5(√5)R)/(27)))^2 = r+(r/((1+[((27)/(8R))(h−((5(√5)R)/(27)))]^2 ))^(1/) ) ⇒ ((50)/(27))−((27)/(16))z^2 =s+(s/((1+(((27)/(16))z)^2 ))^(1/) ) .....(ii) ⇒ s=(((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) ))) ; Now ⇒ ((27)/8)z=(((z+((5(√5))/(27)))−2{(((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) )))}^(1/2) )/(((50)/(27))−((27)/(16))z^2 −((((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) )))))) ......(A) z is obtained from above eq. Then s=(r/R) = (((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) ))) ■$
	$x = R \sin θ + (u \cos θ) t$ $y = R (1 + \cos θ) - (u \sin θ) t - \frac{{g t}^{2}}{2}$ $A s \cos θ = \frac{2}{3}, \sin θ = \frac{\sqrt{5}}{3}, u^{2} = \frac{2 R g}{3}$ $y = \frac{5 R}{3} - (\frac{u \sqrt{5}}{3}) \frac{(x - \frac{R \sqrt{5}}{3})}{(\frac{2 u}{3})} - \frac{g}{2} \frac{{(x - \frac{R \sqrt{5}}{3})}^{2}}{{(\frac{2 u}{3})}^{2}}$ $y = \frac{5 R}{3} - \frac{\sqrt{5}}{2} (x - \frac{R \sqrt{5}}{3}) - \frac{27}{16 R} {(x - \frac{R \sqrt{5}}{3})}^{2}$ $\frac{d y}{d x} = - \frac{\sqrt{5}}{2} - \frac{27}{8 R} (x - \frac{R \sqrt{5}}{3}) = 0$ $\Rightarrow x_{0} = \frac{R \sqrt{5}}{3} - \frac{8 R}{27} (\frac{\sqrt{5}}{2}) = \frac{5 \sqrt{5} R}{27}$ $y_{0} = \frac{5 R}{3} - \frac{\sqrt{5}}{2} (\frac{5 \sqrt{5} R}{27} - \frac{9 R \sqrt{5}}{27})$ $- \frac{27}{16 R} (\frac{80 R^{2}}{27 \times 27})$ $\Rightarrow y_{0} = \frac{5 R}{3} + \frac{10 R}{27} - \frac{5 R}{27} = \frac{50 R}{27} < 2 R$ $T h u s e q . o f p a r a b o l a i s$ $y = \frac{50 R}{27} - \frac{27}{16 R} {(x - \frac{5 \sqrt{5} R}{27})}^{2}$ $\frac{d y}{d x} = - \frac{27}{8 R} (x - \frac{5 \sqrt{5} R}{27})$ $s a y t h i s p a r a b o l a t o u c h e s t h e s m a l l e r$ $c i r c l e a t P (h, k) a n d l e t c e n t e r o f$ $s m a l l e r c i r c l e i s C (2 \sqrt{R r}, r); t h e n$ $h = 2 \sqrt{R r} + r \cos β$ $k = r + r \sin β$ $\cot β = \frac{27}{8 R} (h - \frac{5 \sqrt{5} R}{27}) = \frac{h - 2 \sqrt{R r}}{k - r}$ $\Rightarrow \frac{27}{8} z = \frac{(z + \frac{5 \sqrt{5}}{27}) - 2 \sqrt{s}}{\frac{50}{27} - \frac{27}{16} z^{2} - s} . . . . . . (i)$ $k = \frac{50 R}{27} - \frac{27}{16 R} {(h - \frac{5 \sqrt{5} R}{27})}^{2}$ $= r + \frac{r}{\sqrt[]{1 + {[\frac{27}{8 R} (h - \frac{5 \sqrt{5} R}{27})]}^{2}}}$ $\Rightarrow \frac{50}{27} - \frac{27}{16} z^{2} = s + \frac{s}{\sqrt[]{1 + {(\frac{27}{16} z)}^{2}}} . . . . . (i i)$ $\Rightarrow s = \frac{(\frac{50}{27} - \frac{27}{16} z^{2})}{1 + \frac{1}{\sqrt[]{1 + {(\frac{27 z}{16})}^{2}}}}; N o w$ $\Rightarrow \frac{27}{8} z = \frac{(z + \frac{5 \sqrt{5}}{27}) - 2 {\frac{(\frac{50}{27} - \frac{27}{16} z^{2})}{1 + \frac{1}{\sqrt[]{1 + {(\frac{27 z}{16})}^{2}}}}}^{1 / 2}}{\frac{50}{27} - \frac{27}{16} z^{2} - (\frac{(\frac{50}{27} - \frac{27}{16} z^{2})}{1 + \frac{1}{\sqrt[]{1 + {(\frac{27 z}{16})}^{2}}}})} . . . . . . (A)$ $z i s o b t a i n e d f r o m a b o v e e q .$ $T h e n s = \frac{r}{R} = \frac{(\frac{50}{27} - \frac{27}{16} z^{2})}{1 + \frac{1}{\sqrt[]{1 + {(\frac{27 z}{16})}^{2}}}} ◼$
	Commented by ajfour last updated on 02/Jul/20

	Commented by mr W last updated on 02/Jul/20

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