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Question Number 161783 by ajfour last updated on 22/Dec/21

Commented by ajfour last updated on 22/Dec/21

$C y l i n d e r i s f r e e t o p u r e l y r o l l, f i n d$ $r a d i u s, i f p o i n t m a s s b a l l i s t o b e$ $r e c e i v e d s u c c e s s f u l l y .$
Commented by ajfour last updated on 22/Dec/21
Left d job, shall hunt online..
Commented by mr W last updated on 22/Dec/21

$w e l c o m e b a c k s i r!$
Commented by ajfour last updated on 22/Dec/21

$t h a n k y o u s i r, w o n t u t r y t h i s q u e s t i o n . .$
Commented by mr W last updated on 23/Dec/21

$i^{'} l l t r y h o w f a r i c a n g o .$
	Answered by mr W last updated on 23/Dec/21

	Commented by mr W last updated on 23/Dec/21

	$x_{1} = s$ $φ = \frac{s}{R}$ $l e t ω = \frac{d θ}{d t}$ $x_{2} = s - R \sin θ$ $y_{2} = R (1 + \cos θ)$ $V = \frac{d s}{d t} = ω \frac{d s}{d θ}$ $A = \frac{d V}{d t} = ω \frac{d V}{d θ}$ $N \sin θ R = (\frac{{M R}^{2}}{2} + {M R}^{2}) (\frac{A}{R})$ $\Rightarrow N = \frac{3 M A}{2 \sin θ} = \frac{3 M}{2 \sin θ} \times \frac{ω d V}{d θ}$ $N = 0 \Leftrightarrow \frac{d V}{d θ} = 0$ $v_{x} = \frac{{d x}_{2}}{d t} = V - R \cos θ ω$ $v_{y} = \frac{{d y}_{2}}{d t} = - R \sin θ ω$ $\frac{m}{2} (v_{x}^{2} + v_{y}^{2}) + \frac{1}{2} \times \frac{3 {M R}^{2}}{2} \times {(\frac{V}{R})}^{2} = m g R (1 - \cos θ)$ $v_{x}^{2} + v_{y}^{2} + \frac{3 {M V}^{2}}{2 m} = 2 g R (1 - \cos θ)$ $V^{2} + R^{2} ω^{2} - 2 V R ω \cos θ + \frac{3 {M V}^{2}}{2 m} = 2 g R (1 - \cos θ)$ $ω^{2} - \frac{2 g (1 - \cos θ)}{R} - 2 ω \cos θ (\frac{V}{R}) + (\frac{3 M}{2 m} + 1) {(\frac{V}{R})}^{2} = 0$ $\Rightarrow \frac{V}{R} = \frac{ω \cos θ + \sqrt{ω^{2} (\cos^{2} θ - μ) + \frac{2 μ g (1 - \cos θ)}{R}}}{μ}$ $w i t h μ = \frac{3 M}{2 m} + 1$ $\frac{ω d s}{R d θ} = \frac{ω \cos θ + \sqrt{ω^{2} (\cos^{2} θ - μ) + \frac{2 μ g (1 - \cos θ)}{R}}}{μ}$ $\frac{d s}{d θ} = R \times \frac{\cos θ + \sqrt{\cos^{2} θ - μ + \frac{2 μ g (1 - \cos θ)}{R ω^{2}}}}{μ}$ $\Rightarrow s = R \int_{0}^{θ} \frac{\cos θ + \sqrt{\cos^{2} θ - μ + \frac{2 μ g (1 - \cos θ)}{R ω^{2}}}}{μ} d θ$ $a_{x} = \frac{{d v}_{x}}{d t} = \frac{ω d}{d θ} (V - R \cos θ ω) = ω (\frac{d V}{d θ} - R \cos θ \frac{d ω}{d θ} + R \sin θ ω)$ $a_{y} = \frac{{d v}_{y}}{d t} = \frac{ω d}{d θ} (- R \sin θ ω) = ω (- R \sin θ \frac{d ω}{d θ} - R \cos θ ω)$ ${m a}_{x} = - N \sin θ$ $m ω (\frac{d V}{d θ} - R \cos θ \frac{d ω}{d θ} + R \sin θ ω) = - \frac{3 M}{2} \times \frac{ω d V}{d θ}$ $\Rightarrow μ \frac{d}{d θ} (\frac{V}{R}) - \cos θ \frac{d ω}{d θ} + \sin θ ω = 0$ ${m a}_{y} = N \cos θ - m g$ $m ω (- R \sin θ \frac{d ω}{d θ} - R \cos θ ω) = \frac{3 M}{2 \sin θ} \times \frac{ω d V}{d θ} \cos θ - m g$ $\Rightarrow \frac{3 M}{2 m} \times \frac{d}{d θ} (\frac{V}{R}) = \frac{g}{ω R} - (\sin θ \frac{d ω}{d θ} + \cos θ ω) \tan θ$ $\frac{3 M}{2 m μ} \cos θ \frac{d ω}{d θ} - \frac{3 M}{2 m μ} \sin θ ω = \frac{g}{ω R} - \frac{\sin^{2} θ}{\cos θ} \frac{d ω}{d θ} - \sin θ ω$ $\begin{matrix} (\frac{3 M}{3 M + 2 m} + \tan^{2} θ) \cos θ \frac{d ω}{d θ} - (\frac{2 m}{3 M + 2 m}) \sin θ ω - \frac{g}{ω R} = 0 \end{matrix}$ $s o l v e t h i s d . e . (h a r d t o d o!) f o r ω = f (θ)$ $u n d e r ω = 0 a t θ = 0$ $. . . . . . .$ $o n c e w e h a v e ω = f (θ), w e a l s o g e t \frac{V}{R} .$ $f r o m \frac{d}{d θ} (\frac{V}{R}) = 0 w e g e t θ_{1} a t w h i c h t h e$ $s m a l l b a l l l o s e s c o n t a c t t o c y l i n d e r .$ $w e a l s o g e t t h e v_{x} a n d v_{y} a t t h i s i n s t a n t .$ $u p o n n o w t h e s m a l l b a l l h a s t h e$ $m o t i o n o f p r o j e c t i l e a n d w e c a n f i n d$ $t h e p o s i t i o n w h e r e i t h i t s t h e g r o u n d .$
	Commented by mr W last updated on 23/Dec/21

	${t h a t}^{'} s t h e p o i n t i c a n m a x i m a l l y c o m e$ $t o .$
	Commented by Tawa11 last updated on 23/Dec/21

	$Great sir$
	Commented by ajfour last updated on 23/Dec/21
	$mg(2R)=(1/2)(((mR^2 )/2))ω^2 +(1/2)m(v_x ^2 +v_y ^2 ) +mgR(1+cos θ) ...(i) (energy conservation) mv_x =mωR (linear momentum ..(ii) conservation) mx=ms ..(iii) (center of mass ) 2s=Rcos θ ..)iv) mv_x R(1+cos θ)+mv_y x =((mR^2 )/2)ω+mR^2 ω ...(v) (angular momentum conservation) now v_y t+(1/2)gt^2 =R(1+cos θ) ..(vi) v_x t=(b−x) ..(vii) ................................................ so v_x =ωR (b−((Rcos θ)/2))((v_y /(ωR)))+(1/2)(g/(ω^2 R^2 ))(b−((Rcos θ)/2))^2 =R(1+cos θ) say (b/R)=λ ; (v_y /(ωR))=μ 4(2λ−cos θ)μ+(g/(ω^2 R))(2λ−cos θ)^2 μ^2 =8(1+cos θ) .....(A) λ =? (so we need μ and cos θ) and from ..(v) 2(1+cos θ)+μcos θ=3 ⇒ cos θ=(1/(2+μ)) And from ..(i) ((8g)/R)=ω^2 +2ω^2 (1+μ^2 )+((4g)/R)(1+cos θ) ⇒ ω^2 =((4g)/R)(((1−cos θ)/(3+2μ^2 ))) ...(I) ⇒ ω^2 R=4g(((1−(1/(2+μ)))/(3+2μ^2 ))) also (2ωRcos θ)^2 +(μωRsin θ)^2 =gRcos θ ⇒ ω^2 = (g/R)(((cos θ)/(4cos^2 θ+μ^2 sin^2 θ))) ..(II) from ...(I) & (II) 4(1−cos θ)(4cos^2 θ+μ^2 sin^2 θ) =cos θ(3+2μ^2 ) but cos θ=(1/(2+μ)) hence we obtain 𝛍 from 4(1−(1/(2+μ))){((2/(2+μ)))^2 +(μ^2 /(1−(1/((2+μ)^2 ))))} =(((3+2μ^2 )/(2+μ))) and now eq..(A) gives us λ=(b/R)$
	$m g (2 R) = \frac{1}{2} (\frac{{m R}^{2}}{2}) ω^{2} + \frac{1}{2} m (v_{x}^{2} + v_{y}^{2})$ $+ m g R (1 + \cos θ) . . . (i)$ $(e n e r g y c o n s e r v a t i o n)$ ${m v}_{x} = m ω R (l i n e a r m o m e n t u m$ $. . (i i) c o n s e r v a t i o n)$ $m x = m s . . (i i i) (c e n t e r o f m a s s)$ $2 s = R \cos θ . .) i v)$ ${m v}_{x} R (1 + \cos θ) + {m v}_{y} x$ $= \frac{{m R}^{2}}{2} ω + {m R}^{2} ω . . . (v)$ $(a n g u l a r m o m e n t u m c o n s e r v a t i o n)$ $n o w v_{y} t + \frac{1}{2} {g t}^{2} = R (1 + \cos θ) . . (v i)$ $v_{x} t = (b - x) . . (v i i)$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $s o v_{x} = ω R$ $(b - \frac{R \cos θ}{2}) (\frac{v_{y}}{ω R}) + \frac{1}{2} \frac{g}{ω^{2} R^{2}} {(b - \frac{R \cos θ}{2})}^{2}$ $= R (1 + \cos θ)$ $s a y \frac{b}{R} = λ; \frac{v_{y}}{ω R} = μ$ $4 (2 λ - \cos θ) μ + \frac{g}{ω^{2} R} {(2 λ - \cos θ)}^{2} μ^{2}$ $= 8 (1 + \cos θ) . . . . . (A)$ $λ = ? (s o w e n e e d μ a n d \cos θ)$ $a n d f r o m . . (v)$ $2 (1 + \cos θ) + μ \cos θ = 3$ $\Rightarrow \cos θ = \frac{1}{2 + μ}$ $A n d f r o m . . (i)$ $\frac{8 g}{R} = ω^{2} + 2 ω^{2} (1 + μ^{2}) + \frac{4 g}{R} (1 + \cos θ)$ $\Rightarrow ω^{2} = \frac{4 g}{R} (\frac{1 - \cos θ}{3 + 2 μ^{2}}) . . . (I)$ $\Rightarrow$ $ω^{2} R = 4 g (\frac{1 - \frac{1}{2 + μ}}{3 + 2 μ^{2}})$ $a l s o {(2 ω R \cos θ)}^{2} + {(μ ω R \sin θ)}^{2}$ $= g R \cos θ$ $\Rightarrow ω^{2} = \frac{g}{R} (\frac{\cos θ}{4 \cos^{2} θ + μ^{2} \sin^{2} θ}) . . (I I)$ $f r o m . . . (I) & (I I)$ $4 (1 - \cos θ) (4 \cos^{2} θ + μ^{2} \sin^{2} θ)$ $= \cos θ (3 + 2 μ^{2})$ $b u t \cos θ = \frac{1}{2 + μ}$ $h e n c e w e o b t a i n μ f r o m$ $4 (1 - \frac{1}{2 + μ}) {{(\frac{2}{2 + μ})}^{2} + \frac{μ^{2}}{1 - \frac{1}{{(2 + μ)}^{2}}}}$ $= (\frac{3 + 2 μ^{2}}{2 + μ})$ $a n d n o w e q . . (A) g i v e s u s λ = \frac{b}{R}$
	Commented by mr W last updated on 23/Dec/21

	$g r e a t!$ $i n e e d s o m e t i m e t o f o l l o w i t .$ $i t h i n k w i t h ω y o u {d o n}^{'} t m e a n$ $ω = \frac{d θ}{d t}, r i g h t ?$
	Commented by ajfour last updated on 23/Dec/21

	$y e a h i t s n o t \frac{d θ}{d t}, i t s t h e a n g u l a r$ $v e l o c i t y o f r o t a t i o n o f t h e c y l i n d e r .$
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