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Question Number 23937 by ajfour last updated on 10/Nov/17

Commented by ajfour last updated on 10/Nov/17

$A t h i n r o d p l a c e d p e r p e n d i c u l a r$ $o n a r o u g h h o r i z o n t a l s u r f a c e$ $a n d r e l e a s e d a t r e s t . F i n d a n g l e$ $α a t w h i c h i t s b o t t o m e n d s t a r t s$ $s l i p p i n g . T h e c o e f f i c i e n t o f$ $f r i c t i o n f o r s u r f a c e a n d r o d i s μ .$
	Answered by Physics lover last updated on 10/Nov/17
	$assuming that the rod has turned by an angle θ and further θ + dθ α = ((mg Sin θ (l/2))/((ml^2 )/3)) ⇒ α = ((3g Sin θ)/(2l)) ...i ⇒ ω^(2 ) = ((3g (1−Cos θ))/l) .... ii mg − N = (mω^2 (l/2) Cos θ ) + ((3mgSin^2 θ)/4) ⇒ N = {mg − mω^(2 ) (l/2) Cos θ − ((3mg Sin^2 θ )/4) } ⇒ N = mg{ 1− (((3(1−Cos θ))/2))Cos θ −((3Sin^2 θ)/4)} ⇒ N = ((mg)/4) [4 − 6 Cos θ + 6 Cos^(2 ) θ −3 + 3 Cos^(2 ) θ ] ⇒ N = ((mg)/4)[ 1− 3 Cos θ ]^2 we can have two cases 1. when mg Cos θ > mω^2 (l/2) 2. mω^(2 ) (l/2) > mg Cos θ in first case μN Sin θ will be in opposite direction of mg Cos θ. ⇒mg Cos θ ± μ N Sine θ = mω^2 (l/2) ⇒mg Cos θ ± μ ((mg(1 − 3 Cos θ)^2 )/4)Sin θ = ((3mg(1− Cos θ))/2) ⇒4 Cos θ ± μ∙Sin θ ∙ (1− 3 Cos θ)^2 =6− 6 Cos θ ⇒10 Cos θ ± μ Sin θ(1− 3 Cos θ)^2 = 6 α = θ : it satisfies the above equation$
	$a s s u m i n g t h a t t h e r o d h a s t u r n e d b y$ $a n a n g l e θ a n d f u r t h e r θ + d θ$ $α = \frac{m g S i n θ \frac{l}{2}}{\frac{{m l}^{2}}{3}} \Rightarrow α = \frac{3 g S i n θ}{2 l} . . . i$ $\Rightarrow ω^{2} = \frac{3 g (1 - C o s θ)}{l} . . . . i i$ $m g - N = (m ω^{2} \frac{l}{2} C o s θ) + \frac{3 {m g S i n}^{2} θ}{4}$ $\Rightarrow N = {m g - m ω^{2} \frac{l}{2} C o s θ - \frac{3 m g {S i n}^{2} θ}{4}}$ $\Rightarrow N = m g {1 - (\frac{3 (1 - C o s θ)}{2}) C o s θ - \frac{3 {S i n}^{2} θ}{4}}$ $\Rightarrow N = \frac{m g}{4} [4 - 6 C o s θ + 6 C o s^{2} θ - 3 + 3 C o s^{2} θ]$ $\Rightarrow N = \frac{m g}{4} {[1 - 3 C o s θ]}^{2}$ $w e c a n h a v e t w o c a s e s$ $1 . w h e n m g C o s θ > m ω^{2} \frac{l}{2}$ $2 . m ω^{2} \frac{l}{2} > m g C o s θ$ $i n f i r s t c a s e μ N S i n θ w i l l b e i n$ $o p p o s i t e d i r e c t i o n o f m g C o s θ .$ $\Rightarrow m g C o s θ \pm μ N S i n e θ = m ω^{2} \frac{l}{2}$ $\Rightarrow m g C o s θ \pm μ \frac{m g {(1 - 3 C o s θ)}^{2}}{4} S i n θ = \frac{3 m g (1 - C o s θ)}{2}$ $\Rightarrow 4 C o s θ \pm μ \cdot S i n θ \cdot {(1 - 3 C o s θ)}^{2} = 6 - 6 C o s θ$ $\Rightarrow 10 C o s θ \pm μ S i n θ {(1 - 3 C o s θ)}^{2} = 6$ $α = θ : i t s a t i s f i e s t h e a b o v e e q u a t i o n$
	Commented by ajfour last updated on 10/Nov/17

	$m g - N = m ω^{2} \frac{l}{2} \cos θ + \frac{m α l}{2} \sin θ$ $\Rightarrow m g - N = \frac{3 m g \cos θ (1 - \cos θ)}{2}$ $+ \frac{3 m g}{4} \sin^{2} θ$ $\Rightarrow N = m g [1 - \frac{3}{2} \cos θ (1 - \cos θ)$ $- \frac{3}{4} \sin^{2} θ]$ $N = \frac{m g}{4} [4 - 6 \cos θ + 6 \cos^{2} θ - 3 \sin^{2} θ]$ $= \frac{m g}{4} (1 - 6 \cos θ + 9 \cos^{2} θ)$ $N = \frac{m g}{4} {(3 \cos θ - 1)}^{2}$ $\Rightarrow N i s z e r o f o r θ = \cos^{- 1} (\frac{1}{3}) .$ $I f f ⩾ μ N f o r θ < \cos^{- 1} (\frac{1}{3}) t h e n$ $s l i p p i n g o c c u r s a t a l e s s e r a n g l e .$ $f = m (\frac{α l}{2} \cos θ - \frac{ω^{2} l}{2} \sin θ)$ $= \frac{3 m g}{4} \sin θ \cos θ - \frac{3 m g}{2} \sin θ (1 - \cos θ)$ $= \frac{3 m g}{4} \sin θ [\cos θ - 2 + 2 \cos θ]$ $= \frac{3 m g}{4} \sin θ (3 \cos θ - 2)$ $f > μ N \Rightarrow$ $\frac{3 m g}{4} \sin θ (3 \cos θ - 2) > \frac{μ m g}{4} {(3 \cos θ - 1)}^{2}$ $o r μ < \frac{3 \sin θ (3 \cos θ - 2)}{{(3 \cos θ - 1)}^{2}}$ $S o i f s l i p p i n g {d o e s n}^{'} t t a k e p l a c e$ $t i l l θ ⩽ \cos^{- 1} (\frac{1}{3}), c o n t a c t i t s e l f$ $b r e a k s a t θ = \cos^{- 1} (\frac{1}{3});$ $a n d f o r s l i p p i n g$ $t o o c c u r a t a l e s s e r a n g l e, θ$ $s h o u l d b e s u c h t h a t$ $\frac{\sin θ (\cos θ - \frac{2}{3})}{{(\cos θ - \frac{1}{3})}^{2}} > μ .$
	Commented by Physics lover last updated on 10/Nov/17

	$w e n e e d t o c o n s i d e r b o t h t h e c a s e s$ $, d o n t w e ? s l i p p i n g m a y o c c u r$ $d u e t o m g S i n θ o r m ω^{2} l / 2$
	Commented by Physics lover last updated on 10/Nov/17

	$I s m y a n s w e r c o r r e c t ?$
	Commented by ajfour last updated on 10/Nov/17

	$y o u m i s s e d a \sin θ i n t h e l a s t t e r m$ $o f e x p r e s s i o n o f N (i t w i l l h a v e$ $\sin^{2} θ) . .$
	Commented by Physics lover last updated on 10/Nov/17

	$y e s s i r i t s {S i n}^{2} θ$
	Commented by Physics lover last updated on 10/Nov/17

	$C h e c k i t n o w, p l z .$
	Commented by Physics lover last updated on 10/Nov/17

	$N = \frac{m g}{4} (1 - 6 C o s θ + 9 {C o s}^{2} θ)$
	Commented by mrW1 last updated on 10/Nov/17

	$I g o t :$ $μ = 0.4 : n e v e r s l i p p i n g$ $μ = 0.3 : s l i p p i n g a t α = 24.2 °$ $μ = 0.2 : s l i p p i n g a t α = 15.5 °$ $μ = 0.1 : s l i p p i n g a t α = 7.6 °$
	Commented by Physics lover last updated on 10/Nov/17

	$a n d t h e + v e a n d - v e s i g n s a l s o$ $m a k e s e n s e$
	Commented by ajfour last updated on 10/Nov/17

	$y e s M r W 1 s i r, m y c o r r e c t e d r e s u l t s$ $p r e d i c t t h e s a m e p e c i f i c v a l u e s .$ $T h a n k y o u s o m u c h, S i r .$ $T h a n k y o u P h y s i c s l o v e r$ $f o r y o u r c o o p e r a t i o n .$
	Commented by Physics lover last updated on 10/Nov/17

	$y o u a r e c o r r e c t M r a j f o u r .$ $w e c a n n o t t a k e t h e c a s e θ > {c o s}^{- 1} (1 / 3)$ $E x c e l l e n t j o b!!!!$
	Commented by Physics lover last updated on 10/Nov/17

	$b u t i t w i l l a l w a y s s l i p w h a t e v e r$ $μ m a y b e .$
	Commented by mrW1 last updated on 10/Nov/17

	$y o u a r e r i g h t s i r . s l i p p i n g o c c u r s f o r$ $a n y v a l u e o f μ .$ $f o r 0 ⩽ μ ⩽ 0.37 :$ $s l i p p i n g a t 0 ° ⩽ α ⩽ 35.1 °$ $f o r μ > 0.37 :$ $s l i p p i n g a t 50.2 ° ⩽ α ⩽ 70.5 °$
	Commented by mrW1 last updated on 10/Nov/17

	Commented by Physics lover last updated on 10/Nov/17

	Commented by Physics lover last updated on 10/Nov/17

	$t h e r e i s s o m e p r o b l e m . A t θ = 0.8411$ $\Rightarrow μ = 0 \Rightarrow i t w i l l s l i p a t t h a t p o i n t .$ $w h i c h i s n o t t r u e .$ $\Rightarrow t h e μ w h i c h w e a r e c a l c u l a i n g$ $i s f o r a g i v e n θ .$ $i m e a n, i f w e w a n t t o s t a r t$ $s l i p p i n g a t 0.841 \Rightarrow μ = 0.1873$ $B^{'} c u z 0.5367 w i l l h a v e t o b e$ $c o v e r e d b e f o r e 0.841 .$ $A n d i t$ $i s 2 f o r π / 2 . b u t$ $a c c o r d i n g t o t h e f o l l o w i n g c a l c u l a t i o n d$ $i t s h o u l d b e 6 .$
	Commented by Physics lover last updated on 10/Nov/17

	$a s s u m g t h a t t h e r o d i s h i n g e d$ $a n d c o v e r s π / 2 a n d h o r i z o n t a l$ $h i n g e r e a c t i o n a s μ N \Rightarrow μ = 6 .$
	Commented by mrW1 last updated on 11/Nov/17

	$I t h i n k y o u r g r a p h i s n o t c o r r e c t .$ $T h e f u n c t i o n s h o u l d b e :$ $\frac{f}{N} = y = \frac{\sin x (\cos x - \frac{2}{3})}{{(\cos x - \frac{1}{3})}^{2}}$
	Commented by mrW1 last updated on 10/Nov/17

	Commented by mrW1 last updated on 11/Nov/17

	$a t x = 0.8411 (= 48.2 °), f = 0, t h i s$ $i s p o s s i b l e . A n d i t {d o e s n}^{'} t m e a n$ $t h a t μ = 0 a n d i t {d o e s n}^{'} t m e a n t h a t$ $t h e r o d s l i p s .$
	Commented by mrW1 last updated on 11/Nov/17

	Commented by mrW1 last updated on 11/Nov/17

	$M y o p i n i o n i s t h a t s l i p p i n g a l w a y s$ $o c c u r a t a n a n g l e l e s s t h a n \cos^{- 1} \frac{1}{3}$ $(\approx 70.5 °) .$ $I f μ ⩽ 0.3706, t h e a n g l e α a t w h i c h$ $t h e r o d b e g i n s t o s l i p f u l f i l l s$ $\frac{\sin α (\cos α - \frac{2}{3})}{{(\cos α - \frac{1}{3})}^{2}} = μ$ $a n d 0 ⩽ α ⩽ 35.1 °$ $I f μ > 0.3706, t h e a n g l e α a t w h i c h$ $t h e r o d b e g i n s t o s l i p f u l f i l l s$ $\frac{\sin α (\cos α - \frac{2}{3})}{{(\cos α - \frac{1}{3})}^{2}} = - μ$ $a n d 50.2 ° < α < 70.5 °$
	Commented by mrW1 last updated on 11/Nov/17

	$T o M r P h y s i c s l o v e r :$ $W i t h μ = 6, t h e r o d c a n n o t r e a c h$ $t h e p o s i t i o n \frac{π}{2}, b e c a u s e a l r e a d y$ $a t 60.5 ° i t b e g i n s t o s l i p .$
	Commented by ajfour last updated on 11/Nov/17

	$t h a n k y o u S i r . T h i s i s f i n a l l y$ $v e r y s a t i s f a c t o r y .$ $b e f o r e θ b e c o m e s \cos^{- 1} (\frac{2}{3})$ $r o d s l i p s i f μ h a p p e n s t o b e$ $l e s s t h a n s o m e c a l c u l a b l e v a l u e$ $(0.37 a s s h o w n i n y o u r g r a p h p o s t)$ $a n d i f i t e s c a p e s s l i p p i n g h e r e a t$ $t h i s a n g l e t h e n a t θ = \cos^{- 1} (\frac{2}{3})$ $N = \frac{m g}{4} w h i l e f r i c t i o n n e e d n o t$ $a c t (i s z e r o) . B u t a s a n g l e g r o w s$ $b e y o n d t h i s f r i c t i o n r e v e r s e s$ $d i r e c t i o n a n d g r o w s . N o r m a l$ $f o r c e m u s t b e d e c r e a s i n g a n d$ $b e c o m e s z e r o a t θ = \cos^{- 1} (\frac{1}{3}) .$ $N o v a l u e o f μ c a n p r e v e n t$ $s l i p p i n g b y t h i s t i m e (f o r t h i s$ $c a s e) s o f o r s o m e g i v e n v a l u e$ $o f μ > 0.37 r o d s l i p s f o r$ $\cos^{- 1} (\frac{2}{3}) < α < \cos^{- 1} (\frac{1}{3});$ $w h i l e i f μ < 0.37 i t s l i p s w a y$ $b e f o r e θ = \cos^{- 1} (\frac{2}{3}) .$ $(T h i s i s m y c o n c l u s i o n S i r,$ $b e c a u s e o f y o u r k i n d e f f o r t s .)$ $T h a n k y o u S i r .$
	Commented by Physics lover last updated on 11/Nov/17

	$h m m m . f i n a l l y .$
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