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Question Number 129660 by ajfour last updated on 17/Jan/21

Commented by ajfour last updated on 17/Jan/21

$I f a p l a n k i s r e l e a s e d i n c o n t a c t$ $w i t h a s m o o t h c y l i n d e r o f r a d i u s$ $R, a t a n a n g l e α . F i n d t h e a n g l e$ $θ w h e n p l a n k i s a b o u t t o l e a v e$ $c o n t a c t w i t h c y l i n d e r .$
	Answered by mr W last updated on 18/Jan/21

	Commented by mr W last updated on 19/Jan/21
	$assumed that the contact point A is the end of the rod when the contact is about to get lost, i.e. θ≤θ_m =2 tan^(−1) (R/L). let λ=(L/R) y_A =R+R cos ϕ=L sin θ ⇒cos ϕ=λ sin θ−1 ⇒sin ϕ=(√(1−(λ sin θ−1)^2 ))=(√(λsin θ(2−λsin θ))) y_C =(L/2) sin θ=R×(λ/2) sin θ x_C =R sin ϕ+(L/2) cos θ=R[(√(λ sin θ(2−λ sin θ)))+(λ/2) cos θ] ω=(dθ/dt) v_(C,x) =(dx_C /dt)=ω(dx_C /dθ)=((ωλR)/2)[((2cos θ(1−λsin θ))/( (√(λ sin θ(2−λ sin θ)))))−sin θ] let η=((cos θ(1−λsin θ))/( (√(λ sin θ(2−λ sin θ))))) ⇒v_(C,x) =((ωλR)/2)(2η−sin θ) v_(C,y) =(dy_C /dt)=ω(dy_C /dθ)=((ωλR)/2)cos θ v_C ^2 =v_(C,x) ^2 +v_(C,y) ^2 =(((ωλR)/2))^2 {(2η−sin θ)^2 +cos^2 θ} ⇒v_C ^2 =(((ωL)/2))^2 [4η(η−sin θ)+1] (1/2)mv_C ^2 +(1/2)(((mL^2 )/(12)))ω^2 =mg(L/2)(sin α−sin θ) v_C ^2 +((L^2 ω^2 )/(12))=gL(sin α−sin θ) ((ω^2 L^2 )/4)[4η(η−sin θ)+1]+((L^2 ω^2 )/(12))=gL(sin α−sin θ) ω^2 [3η(η−sin θ)+1]=((3g)/L)(sin α−sin θ) ⇒ω=(√((3g)/L))×(√((sin α−sin θ)/(3η(η−sin θ)+1))) v_(C,x) =((ωL)/2)(2η−sin θ) a_(C,x) =(dv_(C,x) /dt)=ω(dv_(C,x) /dθ)=ω(d/dθ){((ωL)/2)(2η−sin θ)} N sin ϕ=ma_(C,x) N=0 ⇒ a_(C,x) =0 ⇒(d/dθ){(√((sin α−sin θ)/(3η(η−sin θ)+1)))(η−((sin θ)/2))}=0 that means (√((sin α−sin θ)/(3η(η−sin θ)+1)))(η−((sin θ)/2))=extremal example: λ=(L/R)=3, α=60° ⇒θ≈32.1111° < 2 tan^(−1) (1/3)≈36.87°✓$
	$a s s u m e d t h a t t h e c o n t a c t p o i n t A i s$ $t h e e n d o f t h e r o d w h e n t h e c o n t a c t$ $i s a b o u t t o g e t l o s t, i . e . θ ⩽ θ_{m} = 2 \tan^{- 1} \frac{R}{L} .$ $l e t λ = \frac{L}{R}$ $y_{A} = R + R \cos φ = L \sin θ$ $\Rightarrow \cos φ = λ \sin θ - 1$ $\Rightarrow \sin φ = \sqrt{1 - {(λ \sin θ - 1)}^{2}} = \sqrt{λ \sin θ (2 - λ \sin θ)}$ $y_{C} = \frac{L}{2} \sin θ = R \times \frac{λ}{2} \sin θ$ $x_{C} = R \sin φ + \frac{L}{2} \cos θ = R [\sqrt{λ \sin θ (2 - λ \sin θ)} + \frac{λ}{2} \cos θ]$ $ω = \frac{d θ}{d t}$ $v_{C, x} = \frac{{d x}_{C}}{d t} = ω \frac{{d x}_{C}}{d θ} = \frac{ω λ R}{2} [\frac{2 \cos θ (1 - λ \sin θ)}{\sqrt{λ \sin θ (2 - λ \sin θ)}} - \sin θ]$ $l e t η = \frac{\cos θ (1 - λ \sin θ)}{\sqrt{λ \sin θ (2 - λ \sin θ)}}$ $\Rightarrow v_{C, x} = \frac{ω λ R}{2} (2 η - \sin θ)$ $v_{C, y} = \frac{{d y}_{C}}{d t} = ω \frac{{d y}_{C}}{d θ} = \frac{ω λ R}{2} \cos θ$ $v_{C}^{2} = v_{C, x}^{2} + v_{C, y}^{2} = {(\frac{ω λ R}{2})}^{2} {{(2 η - \sin θ)}^{2} + \cos^{2} θ}$ $\Rightarrow v_{C}^{2} = {(\frac{ω L}{2})}^{2} [4 η (η - \sin θ) + 1]$ $\frac{1}{2} {m v}_{C}^{2} + \frac{1}{2} (\frac{{m L}^{2}}{12}) ω^{2} = m g \frac{L}{2} (\sin α - \sin θ)$ $v_{C}^{2} + \frac{L^{2} ω^{2}}{12} = g L (\sin α - \sin θ)$ $\frac{ω^{2} L^{2}}{4} [4 η (η - \sin θ) + 1] + \frac{L^{2} ω^{2}}{12} = g L (\sin α - \sin θ)$ $ω^{2} [3 η (η - \sin θ) + 1] = \frac{3 g}{L} (\sin α - \sin θ)$ $\Rightarrow ω = \sqrt{\frac{3 g}{L}} \times \sqrt{\frac{\sin α - \sin θ}{3 η (η - \sin θ) + 1}}$ $v_{C, x} = \frac{ω L}{2} (2 η - \sin θ)$ $a_{C, x} = \frac{{d v}_{C, x}}{d t} = ω \frac{{d v}_{C, x}}{d θ} = ω \frac{d}{d θ} {\frac{ω L}{2} (2 η - \sin θ)}$ $N \sin φ = {m a}_{C, x}$ $N = 0 \Rightarrow a_{C, x} = 0$ $\Rightarrow \frac{d}{d θ} {\sqrt{\frac{\sin α - \sin θ}{3 η (η - \sin θ) + 1}} (η - \frac{\sin θ}{2})} = 0$ $t h a t m e a n s$ $\sqrt{\frac{\sin α - \sin θ}{3 η (η - \sin θ) + 1}} (η - \frac{\sin θ}{2}) = e x t r e m a l$ $e x a m p l e :$ $λ = \frac{L}{R} = 3, α = 60 °$ $\Rightarrow θ \approx 32.1111 ° < 2 \tan^{- 1} \frac{1}{3} \approx 36.87 ° ✓$
	Commented by ajfour last updated on 18/Jan/21

	$T h a n k y o u, S i r, f o r t h e d i v i n e$ $l i g h t, i w a s i n g r e a t n e e d o f i t . .$
	Commented by ajfour last updated on 17/Jan/21

	$T h a n k s S i r, l e t s c o n t i n u e$ $t h e d i s c u s s i o n t o m o r r o w . .$
	Commented by mr W last updated on 18/Jan/21

	Commented by mr W last updated on 18/Jan/21

	$p h a s e I : θ_{m} < θ < α$ $p h a s e I I : θ ⩽ θ_{m}$
	Commented by mr W last updated on 18/Jan/21

	Answered by mr W last updated on 18/Jan/21

	Commented by mr W last updated on 22/Jan/21

	$p h a s e I$ $λ = \frac{L}{R}$ $θ_{m} = 2 \tan^{- 1} \frac{1}{λ}$ $θ_{m} < θ < α$ $x_{C} = \frac{R}{\tan \frac{θ}{2}} - \frac{L}{2} \cos θ = R (\frac{1}{\tan \frac{θ}{2}} - \frac{λ \cos θ}{2})$ $y_{C} = \frac{L}{2} \sin θ = R \frac{λ \sin θ}{2}$ $ω = \frac{d θ}{d t}$ $v_{C,} x = \frac{{d x}_{C}}{d t} = ω \frac{{d x}_{C}}{d θ} = \frac{ω R}{2} (- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ)$ $v_{C, y} = ω \frac{{d y}_{C}}{d θ} = \frac{ω R}{2} λ \cos θ$ $v_{C}^{2} = \frac{ω^{2} R^{2}}{4} [{(- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ)}^{2} + λ^{2} \cos^{2} θ]$ $v_{C}^{2} = \frac{ω^{2} R^{2}}{4} (\frac{1}{\sin^{4} \frac{θ}{2}} - \frac{4 λ}{\tan \frac{θ}{2}} + λ^{2})$ $\frac{1}{2} {m v}_{C}^{2} + \frac{1}{2} (\frac{{m L}^{2}}{12}) ω^{2} = m g \frac{L}{2} (\sin α - \sin θ)$ $\frac{ω^{2} R^{2}}{4} (\frac{1}{\sin^{4} \frac{θ}{2}} - \frac{4 λ}{\tan \frac{θ}{2}} + λ^{2}) + (\frac{L^{2}}{12}) ω^{2} = g L (\sin α - \sin θ)$ $[\frac{3}{\sin^{4} \frac{θ}{2}} - \frac{12 λ}{\tan \frac{θ}{2}} + 4 λ^{2}] ω^{2} = \frac{12 g}{L} λ^{2} (\sin α - \sin θ)$ $\Rightarrow ω = \sqrt{\frac{12 g}{L}} \times \sqrt{\frac{\sin α - \sin θ}{\frac{3}{λ^{2} \sin^{4} \frac{θ}{2}} - \frac{12}{λ \tan \frac{θ}{2}} + 4}}$ $v_{C, x} = \frac{ω R}{2} (- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ)$ $= R \sqrt{\frac{3 g}{L}} \times \sqrt{\frac{\sin α - \sin θ}{\frac{3}{λ^{2} \sin^{4} \frac{θ}{2}} - \frac{12}{λ \tan \frac{θ}{2}} + 4}} (- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ)$ $a_{C, x} = ω \frac{{d v}_{C, x}}{d θ}$ $a_{C, x} = ω R \sqrt{\frac{3 g}{L}} \times \frac{d}{d θ} [\sqrt{\frac{\sin α - \sin θ}{\frac{3}{λ^{2} \sin^{4} \frac{θ}{2}} - \frac{12}{λ \tan \frac{θ}{2}} + 4}} (- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ)]$ $N \sin θ = {m a}_{C, x}$ $N = 0 \Rightarrow a_{C, x} = 0$ $\Rightarrow \frac{d}{d θ} [\sqrt{\frac{\sin α - \sin θ}{\frac{3}{λ^{2} \sin^{4} \frac{θ}{2}} - \frac{12}{λ \tan \frac{θ}{2}} + 4}} (- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ)] = 0$ $i . e .$ $\sqrt{\frac{\sin α - \sin θ}{\frac{3}{λ^{2} \sin^{4} \frac{θ}{2}} - \frac{12}{λ \tan \frac{θ}{2}} + 4}} (- \frac{1}{\sin^{2} \frac{θ}{2}} + λ \sin θ) = e x t e r m a l$ $w i t h θ_{m} < θ < α$ $t h i s h a s n o r m a l l y n o s o l u t i o n, i . e .$ $i n p h a s e I w e {c a n}^{'} t g e t t h e c a s e N = 0 .$
	Commented by mr W last updated on 19/Jan/21

	Answered by mr W last updated on 19/Jan/21

	Commented by mr W last updated on 20/Jan/21
	$Phase II − an other way R(1+cos ϕ)=L sin θ cos ϕ=((L sin θ)/R)−1=λ sin θ−1 tan ϕ=((√(λ sin θ (2−λ sin θ)))/(λ sin θ−1)) SB=R+((OB)/(tan ϕ))=R+((L cos θ+R sin ϕ)/(tan ϕ)) SB=(1+cos ϕ+((λ cos θ)/(tan ϕ)))R ⇒SB=L(sin θ+((cos θ)/(tan ϕ))) SC^2 =((L/2))^2 +L^2 (sin θ+((cos θ)/(tan ϕ)))^2 −2×(L/2)×L(sin θ+((cos θ)/(tan ϕ)))×cos ((π/2)−θ) SC^2 =L^2 [(1/4)+(sin θ+((cos θ)/(tan ϕ)))^2 −sin θ (sin θ+((cos θ)/(tan ϕ)))] ⇒SC^2 =[(1/4)+((cos θ)/(tan ϕ))(sin θ+((cos θ)/(tan ϕ)))]L^2 I_S =I+m×SC^2 =((mL^2 )/(12))+[(1/4)+((cos θ)/(tan ϕ))(sin θ+((cos θ)/(tan ϕ)))]mL^2 ⇒I_S =[(1/3)+((cos θ)/(tan ϕ))(sin θ+((cos θ)/(tan ϕ)))]mL^2 ω=(dθ/dt) (1/2)I_S ω^2 =mg(L/2)(sin α−sin θ) [(1/3)+((cos θ)/(tan ϕ))(sin θ+((cos θ)/(tan ϕ)))]mL^2 ω^2 =mgL(sin α−sin θ) ⇒ω^2 =((3g)/L)×((sin α−sin θ)/(1+((3 cos θ)/(tan ϕ))(sin θ+((cos θ)/(tan ϕ))))) let ξ=((cos θ)/(tan ϕ))=((cos θ (λ sin θ−1))/( (√(λ sin θ (2−λ sin θ))))) ⇒ω=(√((3g)/L))×(√((sin α−sin θ)/(1+3ξ(sin θ+ξ)))) v_(C,x) =ω(SB−(L/2)sin θ) ⇒v_(C,x) =ω(((cos θ)/(tan ϕ))+((sin θ)/2))L=ω(ξ+((sin θ)/2))L a_(C,x) =ω(dv_(C,x) /dθ)=ω(√(3gL))×(d/dθ){(ξ+((sin θ)/2))(√((sin α−sin θ)/(1+3ξ(sin θ+ξ))))} N sin ϕ=ma_(C,x) =mω(√(3gL))×(d/dθ){(ξ+((sin θ)/2))(√((sin α−sin θ)/(1+3ξ(sin θ+ξ))))} (N/(mg))=(3/(sin ϕ))(√((sin α−sin θ)/(1+3ξ(sin θ+ξ))))×(d/dθ){(ξ+((sin θ)/2))(√((sin α−sin θ)/(1+3ξ(sin θ+ξ))))} ⇒(N/(mg))=3(√((sin α−sin θ)/(λ sin θ (2−λ sin θ)[1+3ξ(sin θ+ξ)])))×(d/dθ){(ξ+((sin θ)/2))(√((sin α−sin θ)/(1+3ξ(sin θ+ξ))))} for N=0 ⇒ a_(C,x) =0 ⇒(d/dθ){(ξ+((sin θ)/2))(√((sin α−sin θ)/(1+3ξ(sin θ+ξ))))}=0$
	$P h a s e I I - a n o t h e r w a y$ $R (1 + \cos φ) = L \sin θ$ $\cos φ = \frac{L \sin θ}{R} - 1 = λ \sin θ - 1$ $\tan φ = \frac{\sqrt{λ \sin θ (2 - λ \sin θ)}}{λ \sin θ - 1}$ $S B = R + \frac{O B}{\tan φ} = R + \frac{L \cos θ + R \sin φ}{\tan φ}$ $S B = (1 + \cos φ + \frac{λ \cos θ}{\tan φ}) R$ $\Rightarrow S B = L (\sin θ + \frac{\cos θ}{\tan φ})$ ${S C}^{2} = {(\frac{L}{2})}^{2} + L^{2} {(\sin θ + \frac{\cos θ}{\tan φ})}^{2} - 2 \times \frac{L}{2} \times L (\sin θ + \frac{\cos θ}{\tan φ}) \times \cos (\frac{π}{2} - θ)$ ${S C}^{2} = L^{2} [\frac{1}{4} + {(\sin θ + \frac{\cos θ}{\tan φ})}^{2} - \sin θ (\sin θ + \frac{\cos θ}{\tan φ})]$ $\Rightarrow {S C}^{2} = [\frac{1}{4} + \frac{\cos θ}{\tan φ} (\sin θ + \frac{\cos θ}{\tan φ})] L^{2}$ $I_{S} = I + m \times {S C}^{2} = \frac{{m L}^{2}}{12} + [\frac{1}{4} + \frac{\cos θ}{\tan φ} (\sin θ + \frac{\cos θ}{\tan φ})] {m L}^{2}$ $\Rightarrow I_{S} = [\frac{1}{3} + \frac{\cos θ}{\tan φ} (\sin θ + \frac{\cos θ}{\tan φ})] {m L}^{2}$ $ω = \frac{d θ}{d t}$ $\frac{1}{2} I_{S} ω^{2} = m g \frac{L}{2} (\sin α - \sin θ)$ $[\frac{1}{3} + \frac{\cos θ}{\tan φ} (\sin θ + \frac{\cos θ}{\tan φ})] {m L}^{2} ω^{2} = m g L (\sin α - \sin θ)$ $\Rightarrow ω^{2} = \frac{3 g}{L} \times \frac{\sin α - \sin θ}{1 + \frac{3 \cos θ}{\tan φ} (\sin θ + \frac{\cos θ}{\tan φ})}$ $l e t ξ = \frac{\cos θ}{\tan φ} = \frac{\cos θ (λ \sin θ - 1)}{\sqrt{λ \sin θ (2 - λ \sin θ)}}$ $\Rightarrow ω = \sqrt{\frac{3 g}{L}} \times \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}}$ $v_{C, x} = ω (S B - \frac{L}{2} \sin θ)$ $\Rightarrow v_{C, x} = ω (\frac{\cos θ}{\tan φ} + \frac{\sin θ}{2}) L = ω (ξ + \frac{\sin θ}{2}) L$ $a_{C, x} = ω \frac{{d v}_{C, x}}{d θ} = ω \sqrt{3 g L} \times \frac{d}{d θ} {(ξ + \frac{\sin θ}{2}) \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}}}$ $N \sin φ = {m a}_{C, x} = m ω \sqrt{3 g L} \times \frac{d}{d θ} {(ξ + \frac{\sin θ}{2}) \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}}}$ $\frac{N}{m g} = \frac{3}{\sin φ} \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}} \times \frac{d}{d θ} {(ξ + \frac{\sin θ}{2}) \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}}}$ $\Rightarrow \frac{N}{m g} = 3 \sqrt{\frac{\sin α - \sin θ}{λ \sin θ (2 - λ \sin θ) [1 + 3 ξ (\sin θ + ξ)]}} \times \frac{d}{d θ} {(ξ + \frac{\sin θ}{2}) \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}}}$ $f o r N = 0 \Rightarrow a_{C, x} = 0$ $\Rightarrow \frac{d}{d θ} {(ξ + \frac{\sin θ}{2}) \sqrt{\frac{\sin α - \sin θ}{1 + 3 ξ (\sin θ + ξ)}}} = 0$
	Commented by ajfour last updated on 23/Jan/21

	$R e v i e w i n g t h e s e s o l u t i o n s, S i r .$ $T h a n k y o u p l e n t i f u l l y!$
	Commented by mr W last updated on 21/Jan/21

	$b e s t w i s h f o r y o u r h e a l t h s i r!$
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