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Question Number 72036 by mr W last updated on 23/Oct/19

Commented by mr W last updated on 23/Oct/19

$t h e c u p h a s t h e s h a p e o f h e m i s p h e r e$ $w i t h r a d i u s R . a s t i c k o f m a s s m a n d$ $l e n g t h l (2 R < l < 4 R) i s r e l e a s e d a t$ $b l u e p o s i t i o n . t h e r e i s n o f r i c t i o n$ $b e t w e e n c u p a n d s t i c k .$ $f i n d t h e t i m e t i n t e r m s o f θ .$ $f i n d θ_{m a x} .$
Commented by ajfour last updated on 23/Oct/19

$c u p f i x e d, i g u e s s ?$
Commented by mr W last updated on 23/Oct/19

$y e s s i r .$
	Answered by mr W last updated on 24/Oct/19

	Commented by mr W last updated on 24/Oct/19

	$φ = \frac{θ}{2}$ $ω = \frac{d φ}{d t} = \frac{1}{2} \times \frac{d θ}{d t}$ $α = \frac{d ω}{d t} = ω \frac{d w}{d φ} = 2 ω \frac{d ω}{d θ}$ $O S = R$ $l e t λ = \frac{l}{4 R} < 1$ $e = 2 R \cos θ - \frac{l}{2} \cos φ = 2 R (\cos θ - \frac{l}{4 R} \cos \frac{θ}{2})$ $= 2 R (\cos θ - λ \cos \frac{θ}{2})$ ${S M}^{2} = {(2 R)}^{2} + {(\frac{l}{2})}^{2} - 2 \times 2 R \times \frac{l}{2} \cos φ$ $= 4 R^{2} (1 + λ^{2} - 2 λ \cos \frac{θ}{2})$ $I_{S} = \frac{{m l}^{2}}{12} + m \times {S M}^{2}$ $= 4 {m R}^{2} (1 + \frac{4 λ^{2}}{3} - 2 λ \cos \frac{θ}{2})$ $\frac{1}{2} I_{S} ω^{2} = m g (R \sin θ - \frac{l}{2} \sin \frac{θ}{2})$ $2 R (1 + \frac{4 λ^{2}}{3} - 2 λ \cos \frac{θ}{2}) ω^{2} = g (\sin θ - 2 λ \sin \frac{θ}{2})$ $\Rightarrow ω = \frac{d θ}{2 d t} = \sqrt{\frac{g}{2 R}} \sqrt{\frac{\sin θ - 2 λ \sin \frac{θ}{2}}{1 + \frac{4 λ^{2}}{3} - 2 λ \cos \frac{θ}{2}}}$ $\Rightarrow t = \sqrt{\frac{R}{2 g}} \int_{0}^{θ} \sqrt{\frac{1 + \frac{4 λ^{2}}{3} - 2 λ \cos \frac{θ}{2}}{\sin θ - 2 λ \sin \frac{θ}{2}}} d θ$
	Commented by ajfour last updated on 24/Oct/19

	$α I_{S} = m g e$ $p l e a s e c h e c k s i r, i h a v e l i t t l e$ $d o u b t . . \frac{{d L}_{S}}{d t} = I_{S} α + \frac{{d I}_{S}}{d t} ω = m g e$
	Commented by mr W last updated on 24/Oct/19

	$y o u a r e v e r y r i g h t s i r! i t i s a$ $s y s t e m w i t h v a r i a b l e M o I .$ $b u t$ $\frac{1}{2} I_{S} ω^{2} = m g Δ h$ $s h o u l d b e o k .$
	Commented by ajfour last updated on 24/Oct/19

	$y e s i t o o t h i n k t h e o t h e r o n e o k a y .$
	Commented by mr W last updated on 24/Oct/19

	$t h a n k s a l o t s i r!$
	Answered by ajfour last updated on 23/Oct/19

	Commented by ajfour last updated on 24/Oct/19
	$AP =2Rcos (θ/2) ; x=l−2Rcos (θ/2) let φ=(θ/2) τ_P =−(Nsin φ)(l−x)+(mgcos φ)((l/2)−x) .....(i) Also along rod Ncos φ−mgsin φ=m(d^2 x/dt^2 )+mω^2 ((l/2)−x) ........(ii) L=I_P ω let ω=(dφ/dt)=((d(θ/2))/dt) I_P =((ml^2 )/(12))+m((l/2)−x)^2 (dL/dt)=2ωm((l/2)−x)(−(dx/dt))+ [((ml^2 )/2)+m((l/2)−x)^2 ](dω/dt) ..(iii) x=l−2Rcos φ ⇒ −(dx/dt)=−ωRsin φ (d^2 x/dt^2 )=ω^2 Rcos φ Now from (ii) Ncos φ−mgsin φ=m(d^2 x/dt^2 )+mω^2 ((l/2)−x) ⇒ N=mgtan φ+mω^2 R+mω^2 ((l/2)−x)sec φ using this and (i), (iii) in (dL/dt)=τ_P ⇒ 2ωm((l/2)−x)(−(dx/dt))+[((ml^2 )/(12))+m((l/2)−x)^2 ](dω/dt) =−(Nsin φ)(l−x)+(mgcos φ)((l/2)−x) ⇒ 2ω(2Rcos φ−(l/2))(−ωRsin φ) +[(l^2 /(12))+(2Rcos φ)^2 ]((ωdω)/dφ) = −{gtan φ+ω^2 R+ω^2 (2Rcos φ−(l/2))sec φ}sin φ(2Rcos φ) +gcos φ(2Rcos φ−(l/2)) ⇒ ((l^2 /(12))+4R^2 cos^2 φ)((ωdω)/dφ) +2ω^2 R^2 sin φcos φ = 2gR(cos^2 φ−sin^2 φ)−((glcos φ)/2) let ω^2 =z ⇒ (dz/dφ)+(((4R^2 sin φcos φ)/((l^2 /(12))+4R^2 cos^2 φ)))z = ((4gR(cos^2 φ−sin^2 φ)−glcos φ)/(l^2 /12+4R^2 cos^2 φ)) e^(∫Pdx) =e^(∫((4R^2 sin φcos φdφ)/(l^2 /12+4R^2 cos^2 φ))) =−(1/(√(l^2 /12+4R^2 cos^2 φ))) ⇒ −(z/(√(l^2 /12+4R^2 cos^2 φ))) =∫(([4gR(cos^2 φ−sin^2 φ)−glcos φ]dφ)/((l^2 /12+4R^2 cos^2 φ)^(3/2) )) .....$
	$A P = 2 R \cos \frac{θ}{2}; x = l - 2 R \cos \frac{θ}{2}$ $l e t ϕ = \frac{θ}{2}$ $τ_{P} = - (N \sin ϕ) (l - x) + (m g \cos ϕ) (\frac{l}{2} - x)$ $. . . . . (i)$ $A l s o a l o n g r o d$ $N \cos ϕ - m g \sin ϕ = m \frac{d^{2} x}{{d t}^{2}} + m ω^{2} (\frac{l}{2} - x)$ $. . . . . . . . (i i)$ $L = I_{P} ω l e t ω = \frac{d ϕ}{d t} = \frac{d (θ / 2)}{d t}$ $I_{P} = \frac{{m l}^{2}}{12} + m {(\frac{l}{2} - x)}^{2}$ $\frac{d L}{d t} = 2 ω m (\frac{l}{2} - x) (- \frac{d x}{d t}) +$ $[\frac{{m l}^{2}}{2} + m {(\frac{l}{2} - x)}^{2}] \frac{d ω}{d t} . . (i i i)$ $x = l - 2 R \cos ϕ$ $\Rightarrow - \frac{d x}{d t} = - ω R \sin ϕ$ $\frac{d^{2} x}{{d t}^{2}} = ω^{2} R \cos ϕ$ $N o w f r o m (i i)$ $N \cos ϕ - m g \sin ϕ = m \frac{d^{2} x}{{d t}^{2}} + m ω^{2} (\frac{l}{2} - x)$ $\Rightarrow N = m g \tan ϕ + m ω^{2} R + m ω^{2} (\frac{l}{2} - x) \sec ϕ$ $u s i n g t h i s a n d (i), (i i i) i n$ $\frac{d L}{d t} = τ_{P} \Rightarrow$ $2 ω m (\frac{l}{2} - x) (- \frac{d x}{d t}) + [\frac{{m l}^{2}}{12} + m {(\frac{l}{2} - x)}^{2}] \frac{d ω}{d t}$ $= - (N \sin ϕ) (l - x) + (m g \cos ϕ) (\frac{l}{2} - x)$ $\Rightarrow$ $2 ω (2 R \cos ϕ - \frac{l}{2}) (- ω R \sin ϕ)$ $+ [\frac{l^{2}}{12} + {(2 R \cos ϕ)}^{2}] \frac{ω d ω}{d ϕ} =$ $- {g \tan ϕ + ω^{2} R + ω^{2} (2 R \cos ϕ - \frac{l}{2}) \sec ϕ} \sin ϕ (2 R \cos ϕ)$ $+ g \cos ϕ (2 R \cos ϕ - \frac{l}{2})$ $\Rightarrow$ $(\frac{l^{2}}{12} + 4 R^{2} \cos^{2} ϕ) \frac{ω d ω}{d ϕ}$ $+ 2 ω^{2} R^{2} \sin ϕ \cos ϕ =$ $2 g R (\cos^{2} ϕ - \sin^{2} ϕ) - \frac{g l \cos ϕ}{2}$ $l e t ω^{2} = z \Rightarrow$ $\frac{d z}{d ϕ} + (\frac{4 R^{2} \sin ϕ \cos ϕ}{\frac{l^{2}}{12} + 4 R^{2} \cos^{2} ϕ}) z$ $= \frac{4 g R (\cos^{2} ϕ - \sin^{2} ϕ) - g l \cos ϕ}{l^{2} / 12 + 4 R^{2} \cos^{2} ϕ}$ $e^{\int P d x} = e^{\int \frac{4 R^{2} \sin ϕ \cos ϕ d ϕ}{l^{2} / 12 + 4 R^{2} \cos^{2} ϕ}}$ $= - \frac{1}{\sqrt{l^{2} / 12 + 4 R^{2} \cos^{2} ϕ}}$ $\Rightarrow - \frac{z}{\sqrt{l^{2} / 12 + 4 R^{2} \cos^{2} ϕ}}$ $= \int \frac{[4 g R (\cos^{2} ϕ - \sin^{2} ϕ) - g l \cos ϕ] d ϕ}{{(l^{2} / 12 + 4 R^{2} \cos^{2} ϕ)}^{3 / 2}}$ $. . . . .$
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