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Question Number 159293 by ajfour last updated on 15/Nov/21

Commented by ajfour last updated on 15/Nov/21

$T r y i n g Q . 159085 a g a i n . .$
	Answered by ajfour last updated on 17/Nov/21
	$let position, velocity and acc. of bottom wheel of box be x,v,a. (dθ/dt)=ω let R=2 x=8sin θ ; v=8ωcos θ ....(i) let tan^(−1) 2=β θ+β−90°=φ (at start φ=0) let V be center of mass velocity of box. and q=2(√5) V_x =v−ωqsin (θ+β) =8ωcos θ−ωqsin (θ+β) V_y =−ωqcos (θ+β) V^( 2) =64ω^2 cos^2 θ+ω^2 q^2 −2ωqvsin (θ+β) ..(ii) let velocity of sphere be u. ucos ϕ=vcos ϕ−4ωsin θ ⇒ u^2 =ω^2 (8cos θ−((4sin θ)/(cos ϕ)))^2 ★ u^2 =v^2 +((16ω^2 sin^2 θ)/(cos^2 ϕ))−((8ωvsin θ)/(cos ϕ)) ..(iii) & 1+sin ϕ=2sin θ When contact breaks (du/dt)=0 & from energy conservation ((2△U_g )/m)= 2g(2(√5)−4cos θ−2sin θ) = V^( 2) +Iω^2 +u^2 ...(iii) =v^2 +ω^2 q^2 −2ωqvsin (θ+β) +v^2 +((16ω^2 sin^2 θ)/(cos^2 ϕ))−((8ωvsin θ)/(cos ϕ))+((80ω^2 )/(12)) now using ..(i)&(ii) in above eq. ((2△U_g )/(mω^2 ))= {64cos^2 θ+20 −32(√5)cos θsin (θ+β)}+{((20)/3)} {64cos^2 θ+((16sin^2 θ)/(1−(2sin θ−1)^2 ))−((64sin θcos θ)/( (√(1−(2sin θ−1)^2 ))))} ⇒ 2g(2(√5)−4cos θ−2sin θ) = (u^2 /((8cos θ−((4sin θ)/(cos ϕ)))^2 )){((80)/3)+128cos^2 θ −32(√5)cos θsin (θ+β) +((16sin^2 θ)/(1−(2sin θ−1)^2 ))−((64sin θcos θ)/( (√(1−(2sin θ−1)^2 ))))} ⇒ finally u^2 (θ) is = ((2g(2(√5)−4cos θ−2sin θ)[8cos θ−((4sin θ)/( (√(1−(2sin θ−1)^2 ))))]^2 )/({128cos^2 θ+32(√5)cos θsin (θ+β)+((16sin^2 θ)/(1−(2sin θ−1)^2 ))−((64sin θcos θ)/( (√(1−(2sin θ−1)^2 ))))+((80)/3)})) ......$
	$l e t p o s i t i o n, v e l o c i t y a n d a c c .$ $o f b o t t o m w h e e l o f b o x b e x, v, a .$ $\frac{d θ}{d t} = ω$ $l e t R = 2$ $x = 8 \sin θ; v = 8 ω \cos θ . . . . (i)$ $l e t \tan^{- 1} 2 = β$ $θ + β - 90 ° = ϕ (a t s t a r t ϕ = 0)$ $l e t V b e c e n t e r o f m a s s v e l o c i t y o f$ $b o x . a n d q = 2 \sqrt{5}$ $V_{x} = v - ω q \sin (θ + β)$ $= 8 ω \cos θ - ω q \sin (θ + β)$ $V_{y} = - ω q \cos (θ + β)$ $V^{2} = 64 ω^{2} \cos^{2} θ + ω^{2} q^{2} - 2 ω q v \sin (θ + β)$ $. . (i i)$ $l e t v e l o c i t y o f s p h e r e b e u .$ $u \cos φ = v \cos φ - 4 ω \sin θ$ $\Rightarrow u^{2} = ω^{2} {(8 \cos θ - \frac{4 \sin θ}{\cos φ})}^{2} ★$ $u^{2} = v^{2} + \frac{16 ω^{2} \sin^{2} θ}{\cos^{2} φ} - \frac{8 ω v \sin θ}{\cos φ} . . (i i i)$ $& 1 + \sin φ = 2 \sin θ$ $W h e n c o n t a c t b r e a k s \frac{d u}{d t} = 0$ $& f r o m e n e r g y c o n s e r v a t i o n$ $\frac{2 △ U_{g}}{m} = 2 g (2 \sqrt{5} - 4 \cos θ - 2 \sin θ)$ $= V^{2} + I ω^{2} + u^{2} . . . (i i i)$ $= v^{2} + ω^{2} q^{2} - 2 ω q v \sin (θ + β)$ $+ v^{2} + \frac{16 ω^{2} \sin^{2} θ}{\cos^{2} φ} - \frac{8 ω v \sin θ}{\cos φ} + \frac{80 ω^{2}}{12}$ $n o w u s i n g . . (i) & (i i) i n a b o v e e q .$ $\frac{2 △ U_{g}}{m ω^{2}} = {64 \cos^{2} θ + 20$ $- 32 \sqrt{5} \cos θ \sin (θ + β)} + {\frac{20}{3}}$ ${64 \cos^{2} θ + \frac{16 \sin^{2} θ}{1 - {(2 \sin θ - 1)}^{2}} - \frac{64 \sin θ \cos θ}{\sqrt{1 - {(2 \sin θ - 1)}^{2}}}}$ $\Rightarrow$ $2 g (2 \sqrt{5} - 4 \cos θ - 2 \sin θ)$ $= \frac{u^{2}}{{(8 \cos θ - \frac{4 \sin θ}{\cos φ})}^{2}} {\frac{80}{3} + 128 \cos^{2} θ$ $- 32 \sqrt{5} \cos θ \sin (θ + β)$ $+ \frac{16 \sin^{2} θ}{1 - {(2 \sin θ - 1)}^{2}} - \frac{64 \sin θ \cos θ}{\sqrt{1 - {(2 \sin θ - 1)}^{2}}}}$ $\Rightarrow f i n a l l y u^{2} (θ) i s =$ $\frac{2 g (2 \sqrt{5} - 4 \cos θ - 2 \sin θ) {[8 \cos θ - \frac{4 \sin θ}{\sqrt{1 - {(2 \sin θ - 1)}^{2}}}]}^{2}}{{128 \cos^{2} θ + 32 \sqrt{5} \cos θ \sin (θ + β) + \frac{16 \sin^{2} θ}{1 - {(2 \sin θ - 1)}^{2}} - \frac{64 \sin θ \cos θ}{\sqrt{1 - {(2 \sin θ - 1)}^{2}}} + \frac{80}{3}}}$ $. . . . . .$
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