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Question Number 60133 by ajfour last updated on 18/May/19

Commented by ajfour last updated on 18/May/19

$Find angle α when m separates$ $from M . (m slides down from top$ $of M, motion being initiated$ $somehow with a slight impulse) .$
	Answered by mr W last updated on 22/May/19
	$position of M: x_1 (+ve to right) velocity of M: v_1 acceleration of M: a ω=(dθ/dt) α=(dω/dt)=ω (dω/dθ) position of m: x_2 =x_1 +R sin θ y_2 =R cos θ v_(2x) =v_1 +R cos θ ω v_(2y) =−R sin θ ω a_(2x) =a_1 +R cos θ α−R sin θ ω^2 a_(2y) =−R sin θ α−R cos θ ω^2 ma_(2x) =N sin θ ⇒m(a_1 +R cos θ α−R sin θ ω^2 )=N sin θ ...(i) ma_(2y) =N cos θ−mg ⇒mR(sin θ α+cos θ ω^2 )=mg−N cos θ ...(ii) Ma_1 =−N sin θ ...(iii) ⇒Rω((dω/dθ) cos θ−ω sin θ)=(1+(m/M))[g−Rω(sin θ (dω/dθ)+ω cos θ )] tan θ with μ=(m/M) ⇒ω (dω/dθ)=((sin θ)/(1+μ sin^2 θ))[(1+μ)(g/R)−μω^2 cos θ] ⇒(dω^2 /dθ)=((2 sin θ)/(1+μ sin^2 θ))[(1+μ)(g/R)−μω^2 cos θ] with Φ=ω^2 ⇒(dΦ/dθ)+((μ sin 2θ)/(1+μ sin^2 θ))Φ=((2(1+μ)g)/R)×((sin θ)/(1+μ sin^2 θ)) ω=(√Φ)=f(θ) ... but not possible to solve.... try ing an other way: the CoM of both masses should be unchanged in horizontal direction during the motion, because no external force is acting in horizontal direction. −x_1 M=(x_1 +R sin θ)m ⇒−v_1 M=(v_1 +R cos θ ω)m ⇒v_1 =−((mR cos θ ω)/(m+M))=−((μR cos θ ω)/(1+μ)) ⇒a_1 =((μR)/(1+μ))(sin θ ω^2 −cos θ α) (1/2)Mv_1 ^2 +(1/2)m(v_(2x) ^2 +v_(2y) ^2 )=mgR(1−cos θ) v_1 ^2 +μ(v_(2x) ^2 +v_(2y) ^2 )=2gμR(1−cos θ) (1+μ)v_1 ^2 +2μRω cos θ v_1 +μR^2 ω^2 =2gμR(1−cos θ) −((μ^2 R^2 cos^2 θ ω^2 )/(1+μ))+μR^2 ω^2 =2gμR(1−cos θ) (((1+μ sin^2 θ)/(1+μ)))Rω^2 =2g(1−cos θ) ⇒ω^2 =((2g(1+μ)(1−cos θ))/(R(1+μ sin^2 θ))) ⇒2ωα=((2g(1+μ))/R){((sin θ)/(1+μ sin^2 θ))−(((1−cos θ)2μ sin θ cos θ)/((1+μ sin^2 θ)^2 ))}ω ⇒α=((g(1+μ) sin θ)/(R(1+μ sin^2 θ)^2 )){1+μ sin^2 θ−(1−cos θ)2μcos θ} ⇒α=((g(1+μ) sin θ)/(R(1+μ sin^2 θ)^2 )){1+μ(1−cos θ)^2 } N=−((Ma_1 )/(sin θ))=−((μRM)/((1+μ)sin θ))(sin θ ω^2 −cos θ α) N=−((μRM)/((1+μ)sin θ)){((2g(1+μ)sin θ(1−cos θ))/(R(1+μ sin^2 θ)))−((g(1+μ) sin θ cos θ [1+μ(1−cos θ)^2 ])/(R(1+μ sin^2 θ)^2 ))} N=−((μMg)/((1+μ sin^2 θ)^2 )){2(1−cos θ)(1+μ sin^2 θ)−cos θ [1+μ(1−cos θ)^2 ]} N=−((μMg)/((1+μ sin^2 θ)^2 )){2−3cos θ+μ(2+cos θ)(1−cos θ)^2 } ⇒(N/(Mg))=((μ[3cos θ−μ(2+cos θ)(1−cos θ)^2 −2])/((1+μ sin^2 θ)^2 )) for N=0: 2−3cos θ+μ(2+cos θ)(1−cos θ)^2 =0 2−3cos θ+μ(2+cos θ)(1−2cos θ+cos^2 θ)=0 ⇒(μ/(1+μ)) cos^3 θ−3 cos θ+2=0 ⇒λ cos^3 α−3 cos α+2=0 ⇒α=cos^(−1) {((((√(1−λ))−1)/λ))^(1/3) −((((√(1−λ))+1)/λ))^(1/3) }$
	$p o s i t i o n o f M : x_{1} (+ v e t o r i g h t)$ $v e l o c i t y o f M : v_{1}$ $a c c e l e r a t i o n o f M : a$ $ω = \frac{d θ}{d t}$ $α = \frac{d ω}{d t} = ω \frac{d ω}{d θ}$ $p o s i t i o n o f m :$ $x_{2} = x_{1} + R \sin θ$ $y_{2} = R \cos θ$ $v_{2 x} = v_{1} + R \cos θ ω$ $v_{2 y} = - R \sin θ ω$ $a_{2 x} = a_{1} + R \cos θ α - R \sin θ ω^{2}$ $a_{2 y} = - R \sin θ α - R \cos θ ω^{2}$ ${m a}_{2 x} = N \sin θ$ $\Rightarrow m (a_{1} + R \cos θ α - R \sin θ ω^{2}) = N \sin θ . . . (i)$ ${m a}_{2 y} = N \cos θ - m g$ $\Rightarrow m R (\sin θ α + \cos θ ω^{2}) = m g - N \cos θ . . . (i i)$ ${M a}_{1} = - N \sin θ . . . (i i i)$ $\Rightarrow R ω (\frac{d ω}{d θ} \cos θ - ω \sin θ) = (1 + \frac{m}{M}) [g - R ω (\sin θ \frac{d ω}{d θ} + ω \cos θ)] \tan θ$ $w i t h μ = \frac{m}{M}$ $\Rightarrow ω \frac{d ω}{d θ} = \frac{\sin θ}{1 + μ \sin^{2} θ} [(1 + μ) \frac{g}{R} - μ ω^{2} \cos θ]$ $\Rightarrow \frac{d ω^{2}}{d θ} = \frac{2 \sin θ}{1 + μ \sin^{2} θ} [(1 + μ) \frac{g}{R} - μ ω^{2} \cos θ]$ $w i t h Φ = ω^{2}$ $\Rightarrow \frac{d Φ}{d θ} + \frac{μ \sin 2 θ}{1 + μ \sin^{2} θ} Φ = \frac{2 (1 + μ) g}{R} \times \frac{\sin θ}{1 + μ \sin^{2} θ}$ $ω = \sqrt{Φ} = f (θ)$ $. . . b u t n o t p o s s i b l e t o s o l v e . . . .$ $t r y i n g a n o t h e r w a y :$ $t h e C o M o f b o t h m a s s e s s h o u l d b e$ $u n c h a n g e d i n h o r i z o n t a l d i r e c t i o n$ $d u r i n g t h e m o t i o n, b e c a u s e n o e x t e r n a l$ $f o r c e i s a c t i n g i n h o r i z o n t a l d i r e c t i o n .$ $- x_{1} M = (x_{1} + R \sin θ) m$ $\Rightarrow - v_{1} M = (v_{1} + R \cos θ ω) m$ $\Rightarrow v_{1} = - \frac{m R \cos θ ω}{m + M} = - \frac{μ R \cos θ ω}{1 + μ}$ $\Rightarrow a_{1} = \frac{μ R}{1 + μ} (\sin θ ω^{2} - \cos θ α)$ $\frac{1}{2} {M v}_{1}^{2} + \frac{1}{2} m (v_{2 x}^{2} + v_{2 y}^{2}) = m g R (1 - \cos θ)$ $v_{1}^{2} + μ (v_{2 x}^{2} + v_{2 y}^{2}) = 2 g μ R (1 - \cos θ)$ $(1 + μ) v_{1}^{2} + 2 μ R ω \cos θ v_{1} + μ R^{2} ω^{2} = 2 g μ R (1 - \cos θ)$ $- \frac{μ^{2} R^{2} \cos^{2} θ ω^{2}}{1 + μ} + μ R^{2} ω^{2} = 2 g μ R (1 - \cos θ)$ $(\frac{1 + μ \sin^{2} θ}{1 + μ}) R ω^{2} = 2 g (1 - \cos θ)$ $\Rightarrow ω^{2} = \frac{2 g (1 + μ) (1 - \cos θ)}{R (1 + μ \sin^{2} θ)}$ $\Rightarrow 2 ω α = \frac{2 g (1 + μ)}{R} {\frac{\sin θ}{1 + μ \sin^{2} θ} - \frac{(1 - \cos θ) 2 μ \sin θ \cos θ}{{(1 + μ \sin^{2} θ)}^{2}}} ω$ $\Rightarrow α = \frac{g (1 + μ) \sin θ}{R {(1 + μ \sin^{2} θ)}^{2}} {1 + μ \sin^{2} θ - (1 - \cos θ) 2 μ \cos θ}$ $\Rightarrow α = \frac{g (1 + μ) \sin θ}{R {(1 + μ \sin^{2} θ)}^{2}} {1 + μ {(1 - \cos θ)}^{2}}$ $N = - \frac{{M a}_{1}}{\sin θ} = - \frac{μ R M}{(1 + μ) \sin θ} (\sin θ ω^{2} - \cos θ α)$ $N = - \frac{μ R M}{(1 + μ) \sin θ} {\frac{2 g (1 + μ) \sin θ (1 - \cos θ)}{R (1 + μ \sin^{2} θ)} - \frac{g (1 + μ) \sin θ \cos θ [1 + μ {(1 - \cos θ)}^{2}]}{R {(1 + μ \sin^{2} θ)}^{2}}}$ $N = - \frac{μ M g}{{(1 + μ \sin^{2} θ)}^{2}} {2 (1 - \cos θ) (1 + μ \sin^{2} θ) - \cos θ [1 + μ {(1 - \cos θ)}^{2}]}$ $N = - \frac{μ M g}{{(1 + μ \sin^{2} θ)}^{2}} {2 - 3 \cos θ + μ (2 + \cos θ) {(1 - \cos θ)}^{2}}$ $\Rightarrow \frac{N}{M g} = \frac{μ [3 \cos θ - μ (2 + \cos θ) {(1 - \cos θ)}^{2} - 2]}{{(1 + μ \sin^{2} θ)}^{2}}$ $f o r N = 0 :$ $2 - 3 \cos θ + μ (2 + \cos θ) {(1 - \cos θ)}^{2} = 0$ $2 - 3 \cos θ + μ (2 + \cos θ) (1 - 2 \cos θ + \cos^{2} θ) = 0$ $\Rightarrow \frac{μ}{1 + μ} \cos^{3} θ - 3 \cos θ + 2 = 0$ $\Rightarrow λ \cos^{3} α - 3 \cos α + 2 = 0$ $\Rightarrow α = \cos^{- 1} {\sqrt[3]{\frac{\sqrt{1 - λ} - 1}{λ}} - \sqrt[3]{\frac{\sqrt{1 - λ} + 1}{λ}}}$
	Commented by ajfour last updated on 19/May/19

	$thanks sir, excellent!$ $Your soution shall determine all$ $quantities as a function of θ .$
	Commented by mr W last updated on 19/May/19

	$ω = \frac{d θ}{d t} = \sqrt{\frac{2 g (1 + μ)}{R}} \times \sqrt{\frac{1 - \cos θ}{1 + μ \sin^{2} θ}} = κ \sqrt{\frac{1 - \cos θ)}{1 + μ \sin^{2} θ}}$ $κ \int_{0}^{t} d t = \int_{0}^{θ} \sqrt{\frac{1 + μ \sin^{2} θ}{1 - \cos θ}} d θ$ $κ t = \int_{0}^{θ} \sqrt{\frac{1 + μ \sin^{2} θ}{1 - \cos θ}} d θ = . . . . .$ $s e e m s i m p o s s i b l e t o i n t e g r a t e . . .$
	Commented by mr W last updated on 19/May/19

	Commented by mr W last updated on 19/May/19

	$f o r m / M = 1 / 3, a t θ = 46.0052 °, b l o c k m$ $s e p a r a t e s f r o m b l o c k M .$
	Commented by mr W last updated on 19/May/19

	$w e c a n d e t e r m i n e w h e r e t h e s m a l l b l o c k$ $s t r i k e s t h e g r o u n d, b u t w e c a n n o t$ $d e t e r m i n e t h e t i m e i t t a k e s .$
	Answered by ajfour last updated on 18/May/19

	Commented by ajfour last updated on 19/May/19
	$let u be velocity of block m with respect to M. Nsin θ=MA mgcos θ−N−mAsin θ=((mu^2 )/R) [when θ=α, u^2 =Rgcos α ...(I) since then N=0, A=0 ] (mgcos θ−((mu^2 )/R))sin θ=A(M+msin^2 θ) (cos θ−(u^2 /(Rg)))sin θ=(A/g)(λ+sin^2 θ) mgsin θ+mAcos θ=((mudu)/(Rdθ)) ....(II) ⇒ (1/(Rg))((udu)/dθ)=sin θ+cos θsin θ{((cos θ−(u^2 /(Rg)))/(λ+sin^2 θ))} (where λ=(M/m)) let (u^2 /(2Rg))=s (ds/dθ)=sin θ+((cos θsin θ)/(λ+sin^2 θ))(cos θ−s) ....$
	$let u be velocity of block m$ $with respect to M .$ $Nsin θ = MA$ $mgcos θ - N - mAsin θ = \frac{{mu}^{2}}{R}$ $[when θ = α, u^{2} = Rgcos α . . . (I)$ $since then N = 0, A = 0]$ $(mgcos θ - \frac{{mu}^{2}}{R}) \sin θ = A (M + msin^{2} θ)$ $(\cos θ - \frac{u^{2}}{Rg}) \sin θ = \frac{A}{g} (λ + \sin^{2} θ)$ $mgsin θ + mAcos θ = \frac{mudu}{Rd θ} . . . . (II)$ $\Rightarrow \frac{1}{Rg} \frac{udu}{d θ} = \sin θ + \cos θ \sin θ {\frac{\cos θ - \frac{u^{2}}{Rg}}{λ + \sin^{2} θ}}$ $(where λ = \frac{M}{m})$ $let \frac{u^{2}}{2 Rg} = s$ $\frac{ds}{d θ} = \sin θ + \frac{\cos θ \sin θ}{λ + \sin^{2} θ} (\cos θ - s)$ $. . . .$
	Commented by ajfour last updated on 19/May/19

	$mgR (1 - \cos α) = \frac{1}{2} {MV}^{2} + \frac{1}{2} m [{(ucos α - V)}^{2} + u^{2} \sin^{2} α]$ $(energy conservation)$ $mucos α = (M + m) V$ $(momentum conservation$ $along horizontal)$ $u^{2} = Rgcos α$ $\Rightarrow 2 mgR (1 - \cos α) = M {(\frac{m}{M + m})}^{2} (Rgcos α) \cos^{2} α$ $+ m [Rgcos^{3} α {(1 - \frac{m}{M + m})}^{2} + Rgcos α \sin^{2} α]$ $\Rightarrow$ $2 (1 - \cos α) = \frac{λ \cos^{3} α}{{(λ + 1)}^{2}} + {(\frac{λ}{λ + 1})}^{2} \cos^{3} α$ $+ \cos α \sin^{2} α$ $let \cos α = x$ $3 x - \frac{1}{λ + 1} x^{3} - 2 = 0$ $say if λ >> 1$ $\Rightarrow x = \cos α = \frac{2}{3}, α = \cos^{- 1} \frac{2}{3} \approx 48.19 °$ $if M = 3 m \Rightarrow λ = 3$ $3 x - \frac{x^{3}}{4} - 2 = 0$ $or \cos^{3} α - 12 \cos α + 8 = 0$ $\Rightarrow α \approx 46.005 ° .$
	Commented by mr W last updated on 19/May/19

	$y o u r m e t h o d i s a b s o l u t e l y c o r r e c t s i r!$
	Commented by ajfour last updated on 19/May/19

	$t h a n k y o u S i r!$
	Commented by mr W last updated on 19/May/19

	$y o u u s e d i r e c t l y t h e m o t i o n e q n . i n$ $r a d i a l a n d t a n g e n t i a l d i r e c t i o n s,$ $w h i l e i u s u a l l y d e r i v e t h e v e l o c i t y$ $a n d a c c e l e r a t i o n f r o m t h e x - a n d$ $y - c o o r d i n a t e s w h i c h i s m o s t l y$ $l e n g t h y a n d c o m p l i c a t e d . y o u r$ $m e t h o d i s t h u s s h o r t e r a n d b e t t e r .$
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