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Question Number 130350 by ajfour last updated on 24/Jan/21

Commented by ajfour last updated on 24/Jan/21

$A h o l l o w c y l i n d e r (r a d i u s R,$ $m a s s M) r o l l i n g d o w n a l o n g$ $i n c l i n e, h a s a s m a l l b a l l a t$ $a n a n g u l a r p o s i t i o n ϕ .$ $D e t e r m i n e ϕ i n t e r m s o f$ $α, g, m, M, a n d R .$
	Answered by mr W last updated on 25/Jan/21

	Commented by mr W last updated on 25/Jan/21

	$N \cos ϕ = m g \cos β$ $\Rightarrow N = \frac{\cos β}{\cos ϕ} m g$ $N \sin ϕ + m g \sin β = m a$ $\Rightarrow a = (\tan ϕ \cos β + \sin β) g$ $I_{1} = I_{0} + {M R}^{2} = {M R}^{2} + {M R}^{2} = 2 {M R}^{2}$ $I_{1} α = M g R \sin β - N R \sin ϕ$ $2 {M R}^{2} α = M g R \sin β - \frac{\cos β}{\cos ϕ} m g R \sin ϕ$ $\Rightarrow α = (\sin β - \frac{m}{M} \cos β \tan ϕ) \frac{g}{2 R}$ $a = α R (1 - \cos ϕ) = \frac{g}{2} (\sin β - \frac{m}{M} \cos β \tan ϕ) (1 - \cos ϕ)$ $\frac{g}{2} (\sin β - \frac{m}{M} \cos β \tan ϕ) (1 - \cos ϕ) = (\tan ϕ \cos β + \sin β) g$ $(\sin β - \frac{m}{M} \cos β \tan ϕ) (1 - \cos ϕ) = 2 (\tan ϕ \cos β + \sin β)$ $(\tan β - \frac{m}{M} \tan ϕ) (1 - \cos ϕ) = 2 (\tan ϕ + \tan β)$ $\Rightarrow \frac{m}{M} \sin ϕ = (2 + \frac{m}{M}) \tan ϕ + (1 + \cos ϕ) \tan β$
	Commented by ajfour last updated on 26/Jan/21

	$S i r i t h i n k,$ $a = α R$
	Commented by mr W last updated on 26/Jan/21

	$y o u a r e r i g h t s i r!$
	Commented by ajfour last updated on 26/Jan/21

	$M g R \sin β - N R \sin ϕ = 2 {M R}^{2} α$ $N \cos ϕ = m g \cos β$ $N \sin ϕ + m g \sin β = m a$ $a = α R$ $- - - - - - - - - - - - - - -$ $M g R \sin β - m g R \cos β \tan ϕ$ $= 2 M R (g \sin β + g \cos β \tan ϕ)$ $\tan ϕ = \frac{- M \sin β}{(2 M + m) \cos β}$ $\Rightarrow \tan ϕ = - \frac{\tan β}{(2 + \frac{m}{M})}$
	Commented by ajfour last updated on 26/Jan/21

	$b u t t h e n e i t h e r, ϕ < 0 o r ϕ > \frac{π}{2}$ $I w o n d e r w h i c h o n e ?$ $w h a t i f m \to 0 ?$
	Commented by mr W last updated on 26/Jan/21

	$s i n c e N m u s t b e > 0, s o ϕ < 0 i s$ $c o r r e c t .$
	Commented by ajfour last updated on 26/Jan/21

	$\frac{N}{m} = \frac{g \cos β}{\cos ϕ} = \frac{a - g \sin β}{\sin ϕ}$ $a = \frac{M g \sin β}{(M + \frac{I_{c m}}{R^{2}})} = \frac{g \sin β}{2}$ $\Rightarrow \tan ϕ = - \frac{\tan β}{2}$ $B u t \frac{N}{m} > 0 \Rightarrow \frac{g \cos β}{\cos ϕ} > 0$ $a n d e v e n \frac{N}{m} > 0 \Rightarrow \frac{- g \sin β}{\sin ϕ} > 0$ $\Rightarrow \cos ϕ > 0, \sin ϕ < 0$ $\Rightarrow - \frac{π}{2} < - \tan^{- 1} (\frac{\tan β}{2}) = ϕ < 0$
	Answered by mr W last updated on 26/Jan/21

	Commented by mr W last updated on 26/Jan/21

	$t h e q u e s t i o n i s t h e s a m e a s i f a s m a l l$ $m a s s i s s u s p e n d e d w i t h a s t r i n g o n$ $t h e c e n t e r o f t h e c y l i n d e r .$ $s a y t h e l e n g t h o f t h e s t r i n g i s r .$ $N \cos ϕ = m g \cos β$ $\Rightarrow N = \frac{\cos β}{\cos ϕ} m g$ $m g \sin β - N \sin ϕ = m a$ $\Rightarrow a = (\sin β - \cos β \tan ϕ) g$ $I_{1} = I_{0} + {M R}^{2} = 2 {M R}^{2}$ $I_{1} α = M g R \sin β + N R \sin ϕ$ $\Rightarrow α = (\sin β + \frac{m}{M} \times \cos β \tan ϕ) \frac{g}{2 R}$ $a = α R$ $(\sin β + \frac{m}{M} \times \cos β \tan ϕ) \frac{g}{2} = (\sin β - \cos β \tan ϕ) g$ $(2 + \frac{m}{M}) \tan ϕ = \tan β$ $\Rightarrow \tan ϕ = \frac{\tan β}{2 + \frac{m}{M}}$ $\Rightarrow ϕ = \tan^{- 1} (\frac{\tan β}{2 + \frac{m}{M}})$ $i t i s t o s e e t h a t ϕ < β a n d t h i s i s$ $i n d e p e n d e n t f r o m t h e l e n g t h o f t h e$ $s t r i n g r! i n o u r c a s e r = R .$
	Answered by mr W last updated on 27/Jan/21

	Commented by mr W last updated on 28/Jan/21
	$friction coefficient between ball and cylinder =μ N cos φ+μN sin φ=mg cos β ⇒N=((cos β)/(cos φ+μ sin φ))mg ma=N sin φ−μN cos φ+mg sin β ⇒a=[((cos β (tan φ−μ))/(1+μ tan φ))+sin β]g 2MR^2 α=MgR sin β−NR sin φ−μNR(1−cos φ) 2Rα={sin β−(m/M)×((cos β(μ+sin φ−μ cos φ))/(cos φ+μ sin φ))}g 2a={sin β−(m/M)×((cos β(μ+sin φ−μ cos φ))/(cos φ+μ sin φ))}g ((2(tan φ−μ))/(1+μ tan φ))+2tan β=tan β−(m/M)×((μ+sin φ−μ cos φ)/(cos φ+μ sin φ)) ⇒((((mμ)/(Mcos φ))+(2+(m/M))(tan φ−μ))/(1+μ tan φ))=−tan β let λ=(m/M), η=tan β ((λμ+(2+λ)(sin φ−μ cos φ))/(cos φ+μ sin φ))=−η (2+λ+μη)sin φ−(2μ+λμ−η) cos φ=−λμ let δ=tan^(−1) ((2μ+λμ−η)/(2+λ+μη)) sin (φ−δ)=−((λμ)/( (√((2+λ+μη)^2 +(2μ+λμ−η)^2 )))) φ=tan^(−1) ((2μ+λμ−η)/(2+λ+μη))−sin^(−1) ((λμ)/( (√((2+λ+μη)^2 +(2μ+λμ−η)^2 )))) ⇒φ=tan^(−1) (((2+(m/M))μ−tan β)/(2+(m/M)+μ tan β))−sin^(−1) (((μm)/M)/( (√((1+μ^2 ) [tan^2 β+(2+(m/M))^2 ])))) examples: β=30°, (m/M)=0.2, μ=1 ⇒φ=26.7306° β=30°, (m/M)=0.2, μ=0.5 ⇒φ=9.6067° β=30°, (m/M)=0.2, μ=((√3)/6)=0.289 ⇒φ=0 β=30°, (m/M)=0.2, μ=0 ⇒φ=−14.7047°$
	$f r i c t i o n c o e f f i c i e n t b e t w e e n b a l l$ $a n d c y l i n d e r = μ$ $N \cos ϕ + μ N \sin ϕ = m g \cos β$ $\Rightarrow N = \frac{\cos β}{\cos ϕ + μ \sin ϕ} m g$ $m a = N \sin ϕ - μ N \cos ϕ + m g \sin β$ $\Rightarrow a = [\frac{\cos β (\tan ϕ - μ)}{1 + μ \tan ϕ} + \sin β] g$ $2 {M R}^{2} α = M g R \sin β - N R \sin ϕ - μ N R (1 - \cos ϕ)$ $2 R α = {\sin β - \frac{m}{M} \times \frac{\cos β (μ + \sin ϕ - μ \cos ϕ)}{\cos ϕ + μ \sin ϕ}} g$ $2 a = {\sin β - \frac{m}{M} \times \frac{\cos β (μ + \sin ϕ - μ \cos ϕ)}{\cos ϕ + μ \sin ϕ}} g$ $\frac{2 (\tan ϕ - μ)}{1 + μ \tan ϕ} + 2 \tan β = \tan β - \frac{m}{M} \times \frac{μ + \sin ϕ - μ \cos ϕ}{\cos ϕ + μ \sin ϕ}$ $\Rightarrow \frac{\frac{m μ}{M \cos ϕ} + (2 + \frac{m}{M}) (\tan ϕ - μ)}{1 + μ \tan ϕ} = - \tan β$ $l e t λ = \frac{m}{M}, η = \tan β$ $\frac{λ μ + (2 + λ) (\sin ϕ - μ \cos ϕ)}{\cos ϕ + μ \sin ϕ} = - η$ $(2 + λ + μ η) \sin ϕ - (2 μ + λ μ - η) \cos ϕ = - λ μ$ $l e t δ = \tan^{- 1} \frac{2 μ + λ μ - η}{2 + λ + μ η}$ $\sin (ϕ - δ) = - \frac{λ μ}{\sqrt{{(2 + λ + μ η)}^{2} + {(2 μ + λ μ - η)}^{2}}}$ $ϕ = \tan^{- 1} \frac{2 μ + λ μ - η}{2 + λ + μ η} - \sin^{- 1} \frac{λ μ}{\sqrt{{(2 + λ + μ η)}^{2} + {(2 μ + λ μ - η)}^{2}}}$ $\Rightarrow ϕ = \tan^{- 1} \frac{(2 + \frac{m}{M}) μ - \tan β}{2 + \frac{m}{M} + μ \tan β} - \sin^{- 1} \frac{\frac{μ m}{M}}{\sqrt{(1 + μ^{2}) [\tan^{2} β + {(2 + \frac{m}{M})}^{2}]}}$ $e x a m p l e s :$ $β = 30 °, \frac{m}{M} = 0.2, μ = 1 \Rightarrow ϕ = 26.7306 °$ $β = 30 °, \frac{m}{M} = 0.2, μ = 0.5 \Rightarrow ϕ = 9.6067 °$ $β = 30 °, \frac{m}{M} = 0.2, μ = \frac{\sqrt{3}}{6} = 0.289 \Rightarrow ϕ = 0$ $β = 30 °, \frac{m}{M} = 0.2, μ = 0 \Rightarrow ϕ = - 14.7047 °$
	Commented by ajfour last updated on 28/Jan/21

	$G r e a t g e n e r a l i s a t i o n, S i r .$ $T h a n k s f o r t h e a n a l y s i s .$
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