Question Number 159293 by ajfour last updated on 15/Nov/21
Commented by ajfour last updated on 15/Nov/21
$${Trying}\:{Q}.\mathrm{159085}\:{again}.. \\ $$
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)})) ......](Q159446.png)
$${let}\:{position},\:{velocity}\:{and}\:{acc}. \\ $$$$\:{of}\:{bottom}\:{wheel}\:{of}\:{box}\:{be}\:{x},{v},{a}. \\ $$$$\frac{{d}\theta}{{dt}}=\omega \\ $$$${let}\:{R}=\mathrm{2} \\ $$$${x}=\mathrm{8sin}\:\theta\:\:;\:\:{v}=\mathrm{8}\omega\mathrm{cos}\:\theta\:\:\:\:\:....\left({i}\right) \\ $$$${let}\:\:\:\mathrm{tan}^{−\mathrm{1}} \mathrm{2}=\beta \\ $$$$\theta+\beta−\mathrm{90}°=\phi\:\:\left({at}\:{start}\:\phi=\mathrm{0}\right) \\ $$$${let}\:{V}\:{be}\:{center}\:{of}\:{mass}\:{velocity}\:{of} \\ $$$${box}.\:\:\:\:{and}\:\:{q}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${V}_{{x}} ={v}−\omega{q}\mathrm{sin}\:\left(\theta+\beta\right) \\ $$$$\:\:\:\:\:=\mathrm{8}\omega\mathrm{cos}\:\theta−\omega{q}\mathrm{sin}\:\left(\theta+\beta\right) \\ $$$${V}_{{y}} =−\omega{q}\mathrm{cos}\:\left(\theta+\beta\right) \\ $$$${V}^{\:\mathrm{2}} =\mathrm{64}\omega^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+\omega^{\mathrm{2}} {q}^{\mathrm{2}} −\mathrm{2}\omega{qv}\mathrm{sin}\:\left(\theta+\beta\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:..\left({ii}\right) \\ $$$${let}\:{velocity}\:{of}\:{sphere}\:{be}\:{u}. \\ $$$${u}\mathrm{cos}\:\varphi={v}\mathrm{cos}\:\varphi−\mathrm{4}\omega\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:{u}^{\mathrm{2}} =\omega^{\mathrm{2}} \left(\mathrm{8cos}\:\theta−\frac{\mathrm{4sin}\:\theta}{\mathrm{cos}\:\varphi}\right)^{\mathrm{2}} \:\:\:\bigstar \\ $$$${u}^{\mathrm{2}} ={v}^{\mathrm{2}} +\frac{\mathrm{16}\omega^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:^{\mathrm{2}} \varphi}−\frac{\mathrm{8}\omega{v}\mathrm{sin}\:\theta}{\mathrm{cos}\:\varphi}\:\:..\left({iii}\right) \\ $$$$\&\:\:\mathrm{1}+\mathrm{sin}\:\varphi=\mathrm{2sin}\:\theta \\ $$$${When}\:{contact}\:{breaks}\:\:\:\frac{{du}}{{dt}}=\mathrm{0} \\ $$$$\&\:{from}\:{energy}\:{conservation} \\ $$$$\:\:\frac{\mathrm{2}\bigtriangleup{U}_{{g}} }{{m}}=\:\mathrm{2}{g}\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4cos}\:\theta−\mathrm{2sin}\:\theta\right) \\ $$$$\:\:\:\:\:=\:{V}^{\:\mathrm{2}} +{I}\omega^{\mathrm{2}} +{u}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:...\left({iii}\right) \\ $$$$\:\:\:\:\:={v}^{\mathrm{2}} +\omega^{\mathrm{2}} {q}^{\mathrm{2}} −\mathrm{2}\omega{qv}\mathrm{sin}\:\left(\theta+\beta\right) \\ $$$$\:\:+{v}^{\mathrm{2}} +\frac{\mathrm{16}\omega^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:^{\mathrm{2}} \varphi}−\frac{\mathrm{8}\omega{v}\mathrm{sin}\:\theta}{\mathrm{cos}\:\varphi}+\frac{\mathrm{80}\omega^{\mathrm{2}} }{\mathrm{12}} \\ $$$${now}\:\:{using}\:..\left({i}\right)\&\left({ii}\right)\:{in}\:{above}\:{eq}. \\ $$$$\:\frac{\mathrm{2}\bigtriangleup{U}_{{g}} }{{m}\omega^{\mathrm{2}} }=\:\left\{\mathrm{64cos}\:^{\mathrm{2}} \theta+\mathrm{20}\right. \\ $$$$\left.−\mathrm{32}\sqrt{\mathrm{5}}\mathrm{cos}\:\theta\mathrm{sin}\:\left(\theta+\beta\right)\right\}+\left\{\frac{\mathrm{20}}{\mathrm{3}}\right\} \\ $$$$\left\{\mathrm{64cos}\:^{\mathrm{2}} \theta+\frac{\mathrm{16sin}\:^{\mathrm{2}} \theta}{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{64sin}\:\theta\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }}\right\} \\ $$$$\:\:\:\:\Rightarrow \\ $$$$\mathrm{2}{g}\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4cos}\:\theta−\mathrm{2sin}\:\theta\right) \\ $$$$\:\:\:\:=\:\frac{{u}^{\mathrm{2}} }{\left(\mathrm{8cos}\:\theta−\frac{\mathrm{4sin}\:\theta}{\mathrm{cos}\:\varphi}\right)^{\mathrm{2}} }\left\{\frac{\mathrm{80}}{\mathrm{3}}+\mathrm{128cos}\:^{\mathrm{2}} \theta\right. \\ $$$$\:\:\:\:\:\:\:−\mathrm{32}\sqrt{\mathrm{5}}\mathrm{cos}\:\theta\mathrm{sin}\:\left(\theta+\beta\right) \\ $$$$\left.\:\:\:\:\:+\frac{\mathrm{16sin}\:^{\mathrm{2}} \theta}{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{64sin}\:\theta\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }}\right\} \\ $$$$ \\ $$$$\Rightarrow\:\:{finally}\:\:{u}^{\mathrm{2}} \left(\theta\right)\:\:{is}\:= \\ $$$$\:\frac{\mathrm{2}{g}\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4cos}\:\theta−\mathrm{2sin}\:\theta\right)\left[\mathrm{8cos}\:\theta−\frac{\mathrm{4sin}\:\theta}{\:\sqrt{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }}\right]^{\mathrm{2}} }{\left\{\mathrm{128cos}\:^{\mathrm{2}} \theta+\mathrm{32}\sqrt{\mathrm{5}}\mathrm{cos}\:\theta\mathrm{sin}\:\left(\theta+\beta\right)+\frac{\mathrm{16sin}\:^{\mathrm{2}} \theta}{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{64sin}\:\theta\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{1}−\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }}+\frac{\mathrm{80}}{\mathrm{3}}\right\}}\:\:\:\:\: \\ $$$$...... \\ $$