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Question Number 68451 by mr W last updated on 10/Sep/19

Commented by mr W last updated on 11/Sep/19

$i f M = 0, t h e b l o c k m j u s t h a s f r e e$ $f a l l, t h e r e f o r e T = \sqrt{2 g h} = \sqrt{2 g R} .$
Commented by ajfour last updated on 11/Sep/19

$s o r r y s i r m y m i s t a k e f i r s t,$ $i t m a t c h e d b u t i n t h e e n d i$ $t y p e d i n \sqrt{2 g R} i t i s t i m e n o t$ $v e l o c i t y i t i s \sqrt{\frac{2 R}{g}} (h a . . h a)$ $B y t h e w a y p l e a s e c h e c k, i g o t$ $s o m e a n s w e r f o r f i x e d w e d g e t o o .$
Commented by mr W last updated on 10/Sep/19

$t h e b l o c k m i s r e l e a s e d a t t h e t o p o f$ $t h e w e d g e . t h e r e i s n o f r i c t i o n .$ $f i n d t h e v e l o c i t y o f b o t h o b j e c t s w h e n$ $t h e s m a l l b l o c k r e a c h e s t h e b o t t o m$ $o f t h e w e d g e .$
Commented by ajfour last updated on 11/Sep/19

Commented by ajfour last updated on 11/Sep/19

$L e t s t r y a n d f i n d t i m e i t t a k e s$ $t o s l i p d o w n .$ $N + m A \cos θ - m g \sin θ = \frac{{m v}_{r}^{2}}{R}$ $m A \sin θ + m g \cos θ = \frac{{m v}_{r} {d v}_{r}}{R d θ}$ $N \cos θ = M A$ $\Rightarrow i f w e l e t M = μ m$ $μ A \sec θ + A \cos θ - g \sin θ = \frac{v_{r}^{2}}{R} . . (i)$ $A \sin θ + g \cos θ = \frac{v_{r} {d v}_{r}}{R d θ}$ $\Rightarrow A = (\frac{v_{r} {d v}_{r}}{R d θ} - g \cos θ) cosec θ . . . (i i)$ $\Rightarrow$ $(μ \sec θ + \cos θ) (\frac{v_{r} {d v}_{r}}{R d θ} - g \cos θ) cosec θ$ $- g \sin θ = \frac{v_{r}^{2}}{R}$ $l e t v_{r}^{2} / R = 2 s \Rightarrow v_{r} {d v}_{r} / R = d s$ $N o w$ $(μ \sec θ + \cos θ) (cosec θ) \frac{d s}{d θ} - 2 s$ $= g \sin θ + g \cos θ cosec θ (μ \sec θ + \cos θ)$ $o r$ $\frac{d s}{d θ} - \frac{2 s (\sin θ)}{(μ \sec θ + \cos θ)}$ $= g (\frac{\sin^{2} θ}{μ \sec θ + \cos θ} + \cos θ)$ $\Rightarrow$ $\frac{d s}{\cos θ d θ} - \frac{2 s (\sin θ)}{(μ + \cos^{2} θ)}$ $= g (1 + \frac{\sin^{2} θ}{μ + \cos^{2} θ})$ $l e t \sin θ = u \Rightarrow \cos θ d θ = d u$ $\Rightarrow$ $\frac{d s}{d u} - \frac{2 s u}{μ + 1 - u^{2}} = \frac{g (μ + 1)}{μ + 1 - u^{2}}$ $l e t μ + 1 = λ \Rightarrow$ $\frac{d s}{d u} + \frac{2 s u}{u^{2} - λ} = \frac{λ g}{λ - u^{2}}$ $\Rightarrow (u^{2} - λ) \frac{d s}{d u} + 2 s u = - λ g$ $\Rightarrow \int_{0}^{s (u^{2} - λ)} d [s (u^{2} - λ)] = - λ g \int_{0}^{u} d u$ $\Rightarrow s (u^{2} - λ) = - λ g u$ $\Rightarrow [\frac{v_{r}^{2}}{2 R} (\sin^{2} θ - λ) = - λ g \sin θ]$ $b u t i f w e c o n t i n u e w i t h s a m e$ $v a r i a b l e u = \sin θ, t h e n$ $\Rightarrow v_{r} = \frac{d (R d θ)}{d t} = \sqrt{2 s R} = \sqrt{2 R (\frac{λ g u}{λ - u^{2}})}$ $\int_{0}^{T} d t = \int_{0}^{1} \frac{R d u}{\sqrt{1 - u^{2}}} \sqrt{\frac{λ - u^{2}}{2 λ g R u}}$ $T = \sqrt{\frac{R}{2 λ g}} \int_{0}^{1} \sqrt{\frac{λ - u^{2}}{u (1 - u^{2})}} d u$ $I f w e d g e i s f a r t o o h e a v i e r, t h e n$ $i t i s a s g o o d a s f i x e d;$ $μ = \frac{M}{m} \to \infty$ $T = \sqrt{\frac{R}{2 g}} \int_{0}^{1} \frac{d u}{\sqrt{u (1 - u^{2})}}$ $= t_{0} \int_{0}^{π / 2} \frac{\cos θ d θ}{\sqrt{\sin θ \cos^{2} θ}}$ $= t_{0} \int_{0}^{π / 2} \frac{d θ}{\sqrt{\sin θ}} l e t \sin θ = t^{2}$ $\Rightarrow 2 t d t = \cos θ d θ$ $\Rightarrow d θ = \frac{2 t d t}{\sqrt{1 - t^{2}}}$ $T = t_{0} \int_{0}^{1} \frac{2 t d t}{t \sqrt{1 - t^{2}}} = 2 t_{0} \int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}}$ $= 2 t_{0} (\frac{π}{2}) = π t_{0}$ $T = π \sqrt{\frac{R}{2 g}}!$ $I f t h e r e i s a v e r y l i g h t w e d g e$ $(j u s t t o c h e c k)$ $μ = \frac{M}{m} \to 0 \Rightarrow λ \to 1$ $T = \sqrt{\frac{R}{2 g}} \int_{0}^{1} \frac{d u}{\sqrt{u}}$ $T = \sqrt{\frac{2 R}{g}} (f a b u l o u s!)$ $b u t i t h i n k o t h e r w i s e s o l v i n g t h e$ $i n t e g r a t i o n i s b e y o n d m y$ $u n d e r s t a n d i n g . .$
Commented by mr W last updated on 11/Sep/19

$t h a n k s s i r f o r s o l v i n g!$ $c a n w e f i n d t h e t i m e i f t h e w e d g e i s$ $f i x e d o n t h e g r o u n d ?$
Commented by mr W last updated on 11/Sep/19

$f a n t a s t i c s i r! c o r r e c t r e s u l t .$
	Answered by ajfour last updated on 10/Sep/19

	$m g R = \frac{1}{2} {m v}^{2} + \frac{1}{2} {M V}^{2}$ $m v = M V$ $2 m g R = [m + M {(\frac{m}{M})}^{2}] v^{2}$ $v = \sqrt{\frac{2 M g R}{M + m}}, V = \sqrt{\frac{2 m g R}{M + m}} .$
	Commented by mr W last updated on 11/Sep/19

	$s i m p l e a n d d i r e c t, g r e a t!$
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