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Question Number 80057 by ajfour last updated on 30/Jan/20

Commented by ajfour last updated on 30/Jan/20

If released as shown in upper diagram, after what time does the upper sphere loose contact, with the lower ones?

$${If}\:{released}\:{as}\:{shown}\:{in}\:{upper} \\ $$$${diagram},\:{after}\:{what}\:{time}\:{does} \\ $$$${the}\:{upper}\:{sphere}\:{loose}\:{contact}, \\ $$$${with}\:{the}\:{lower}\:{ones}? \\ $$

Commented by ajfour last updated on 30/Jan/20

(assume frictionless surfaces).

$$\left({assume}\:{frictionless}\:{surfaces}\right). \\ $$

Answered by mr W last updated on 30/Jan/20

Commented by mr W last updated on 04/Feb/20

R=radius of circles y=2R sin θ x=2R cos θ let ω=(dθ/dt), α=(dω/dt)=ω(dω/dθ) V, A=velocity and acceleration ↓ v, a=velocity and acceleration ← → V=−(dy/dt)=−2R cos θ (dθ/dt)=−2R cos θ ω v=(dx/dt)=−2R sin θ (dθ/dt)=−2R sin θ ω (1/2)mV^2 +2×(1/2)mv^2 =mg((√3)R−2R sin θ) 2R(1+sin^2 θ)ω^2 =g((√3)−2 sin θ) ⇒ω^2 =(g/(2R))×(((√3)−2 sin θ)/(1+sin^2 θ)) ⇒ω=−(√(g/(2R)))(√(((√3)−2 sin θ)/(1+sin^2 θ))) A=(dV/dt)=2R(sin θ ω^2 −cos θ α) a=(dv/dt)=2R(−cos θ ω^2 −sin θ α) mg−2N sin θ=mA mg−2N sin θ=2mR(sin θ ω^2 −cos θ α) ...(i) N cos θ=ma N cos θ=2mR(−cos θ ω^2 −sin θ α) ...(ii) (i) cos θ+2×(ii) sin θ: (g/(2R)) cos θ=−ω^2 sin θcos θ−(1+sin^2 θ)α α=−((((g/(2R)) +ω^2 sin θ)cos θ)/((1+sin^2 θ))) ⇒α=−(g/(2R))×(((1+(√3)sin θ−sin^2 θ)cos θ)/((1+sin^2 θ)^2 )) put ω^2 and α into (ii): ⇒(N/(mg))=((sin^3 θ+3 sin θ−(√3))/((1+sin^2 θ)^2 )) for N=0 (loss of contact): ⇒sin^3 θ+3 sin θ−(√3)=0 ⇒sin θ=((((√7)+(√3))/2))^(1/3) −(((√(7−(√3)))/2))^(1/3) ⇒θ_1 =sin^(−1) (((((√7)+(√3))/2))^(1/3) −(((√(7−(√3)))/2))^(1/3) ) ⇒θ_1 ≈0.556506 rad (≈31.885421°) ω=(dθ/dt)=−(√(g/(2R)))(√(((√3)−2 sin θ)/(1+sin^2 θ))) ∫_0 ^( t) dt=−(√((2R)/g))∫_(π/3) ^( θ) (√((1+sin^2 θ)/((√3)−2 sin θ))) dθ ⇒t(θ)=(√((2R)/g))∫_θ ^( (π/3)) (√((1+sin^2 θ)/((√3)−2 sin θ))) dθ t_1 =t(θ_1 )=(√((2R)/g))∫_θ_1 ^( (π/3)) (√((1+sin^2 θ)/((√3)−2 sin θ))) dθ ≈1.670696(√((2R)/g)) that means at time t_1 =1.670696(√((2R)/g)) and at θ_1 =31.885421° the contact gets lost. V=(√(2Rg)) cos θ(√(((√3)−2 sin θ)/(1+sin^2 θ))) at θ=θ_1 : (V_1 /(√(2Rg)))=cos θ_1 (√(((√3)−2 sin θ_1 )/(1+sin^2 θ_1 )))≈0.617121 that means at the moment of loss of contact the velocity downwards is V_1 =0.617121 (√(2Rg)).

$${R}={radius}\:{of}\:{circles} \\ $$$${y}=\mathrm{2}{R}\:\mathrm{sin}\:\theta \\ $$$${x}=\mathrm{2}{R}\:\mathrm{cos}\:\theta \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}},\:\alpha=\frac{{d}\omega}{{dt}}=\omega\frac{{d}\omega}{{d}\theta} \\ $$$${V},\:{A}={velocity}\:{and}\:{acceleration}\:\downarrow \\ $$$${v},\:{a}={velocity}\:{and}\:{acceleration}\:\leftarrow\:\rightarrow \\ $$$$ \\ $$$${V}=−\frac{{dy}}{{dt}}=−\mathrm{2}{R}\:\mathrm{cos}\:\theta\:\frac{{d}\theta}{{dt}}=−\mathrm{2}{R}\:\mathrm{cos}\:\theta\:\omega \\ $$$${v}=\frac{{dx}}{{dt}}=−\mathrm{2}{R}\:\mathrm{sin}\:\theta\:\frac{{d}\theta}{{dt}}=−\mathrm{2}{R}\:\mathrm{sin}\:\theta\:\omega \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mg}\left(\sqrt{\mathrm{3}}{R}−\mathrm{2}{R}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{2}{R}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)\omega^{\mathrm{2}} ={g}\left(\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{{g}}{\mathrm{2}{R}}×\frac{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow\omega=−\sqrt{\frac{{g}}{\mathrm{2}{R}}}\sqrt{\frac{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$$ \\ $$$${A}=\frac{{dV}}{{dt}}=\mathrm{2}{R}\left(\mathrm{sin}\:\theta\:\omega^{\mathrm{2}} −\mathrm{cos}\:\theta\:\alpha\right) \\ $$$${a}=\frac{{dv}}{{dt}}=\mathrm{2}{R}\left(−\mathrm{cos}\:\theta\:\omega^{\mathrm{2}} −\mathrm{sin}\:\theta\:\alpha\right) \\ $$$$ \\ $$$${mg}−\mathrm{2}{N}\:\mathrm{sin}\:\theta={mA} \\ $$$${mg}−\mathrm{2}{N}\:\mathrm{sin}\:\theta=\mathrm{2}{mR}\left(\mathrm{sin}\:\theta\:\omega^{\mathrm{2}} −\mathrm{cos}\:\theta\:\alpha\right)\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${N}\:\mathrm{cos}\:\theta={ma} \\ $$$${N}\:\mathrm{cos}\:\theta=\mathrm{2}{mR}\left(−\mathrm{cos}\:\theta\:\omega^{\mathrm{2}} −\mathrm{sin}\:\theta\:\alpha\right)\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)\:\mathrm{cos}\:\theta+\mathrm{2}×\left({ii}\right)\:\mathrm{sin}\:\theta: \\ $$$$\frac{{g}}{\mathrm{2}{R}}\:\mathrm{cos}\:\theta=−\omega^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta−\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)\alpha \\ $$$$\alpha=−\frac{\left(\frac{{g}}{\mathrm{2}{R}}\:+\omega^{\mathrm{2}} \mathrm{sin}\:\theta\right)\mathrm{cos}\:\theta}{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)} \\ $$$$\Rightarrow\alpha=−\frac{{g}}{\mathrm{2}{R}}×\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−\mathrm{sin}^{\mathrm{2}} \:\theta\right)\mathrm{cos}\:\theta}{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$${put}\:\omega^{\mathrm{2}} \:{and}\:\alpha\:{into}\:\left({ii}\right): \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\frac{\mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{3}\:\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}}{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${for}\:{N}=\mathrm{0}\:\left({loss}\:{of}\:{contact}\right): \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{3}\:\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7}−\sqrt{\mathrm{3}}}}{\mathrm{2}}} \\ $$$$\Rightarrow\theta_{\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \left(\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7}−\sqrt{\mathrm{3}}}}{\mathrm{2}}}\right) \\ $$$$\Rightarrow\theta_{\mathrm{1}} \approx\mathrm{0}.\mathrm{556506}\:{rad}\:\left(\approx\mathrm{31}.\mathrm{885421}°\right) \\ $$$$ \\ $$$$\omega=\frac{{d}\theta}{{dt}}=−\sqrt{\frac{{g}}{\mathrm{2}{R}}}\sqrt{\frac{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$$\int_{\mathrm{0}} ^{\:{t}} {dt}=−\sqrt{\frac{\mathrm{2}{R}}{{g}}}\int_{\frac{\pi}{\mathrm{3}}} ^{\:\theta} \sqrt{\frac{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}}\:{d}\theta \\ $$$$\Rightarrow{t}\left(\theta\right)=\sqrt{\frac{\mathrm{2}{R}}{{g}}}\int_{\theta} ^{\:\frac{\pi}{\mathrm{3}}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}}\:{d}\theta \\ $$$${t}_{\mathrm{1}} ={t}\left(\theta_{\mathrm{1}} \right)=\sqrt{\frac{\mathrm{2}{R}}{{g}}}\int_{\theta_{\mathrm{1}} } ^{\:\frac{\pi}{\mathrm{3}}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}}\:{d}\theta \\ $$$$\approx\mathrm{1}.\mathrm{670696}\sqrt{\frac{\mathrm{2}{R}}{{g}}} \\ $$$$ \\ $$$${that}\:{means}\:{at}\:{time}\:{t}_{\mathrm{1}} =\mathrm{1}.\mathrm{670696}\sqrt{\frac{\mathrm{2}{R}}{{g}}} \\ $$$${and}\:{at}\:\theta_{\mathrm{1}} =\mathrm{31}.\mathrm{885421}°\:{the}\:{contact}\:{gets}\:{lost}. \\ $$$$ \\ $$$${V}=\sqrt{\mathrm{2}{Rg}}\:\mathrm{cos}\:\theta\sqrt{\frac{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$${at}\:\theta=\theta_{\mathrm{1}} : \\ $$$$\frac{{V}_{\mathrm{1}} }{\sqrt{\mathrm{2}{Rg}}}=\mathrm{cos}\:\theta_{\mathrm{1}} \sqrt{\frac{\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{1}} }}\approx\mathrm{0}.\mathrm{617121} \\ $$$${that}\:{means}\:{at}\:{the}\:{moment}\:{of}\:{loss}\:{of} \\ $$$${contact}\:{the}\:{velocity}\:{downwards}\:{is} \\ $$$${V}_{\mathrm{1}} =\mathrm{0}.\mathrm{617121}\:\sqrt{\mathrm{2}{Rg}}. \\ $$

Commented by mr W last updated on 31/Jan/20

Commented by ajfour last updated on 31/Jan/20

your method is straight and clear, Sir, must be true, unless there is something, that we need to amend in our concepts..

$${your}\:{method}\:{is}\:{straight}\:{and} \\ $$$${clear},\:{Sir},\:{must}\:{be}\:{true},\:{unless} \\ $$$${there}\:{is}\:{something},\:{that}\:{we} \\ $$$${need}\:{to}\:{amend}\:{in}\:{our}\:{concepts}.. \\ $$

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