Question Number 155381 by ajfour last updated on 29/Sep/21
Commented by mr W last updated on 30/Sep/21
Commented by mr W last updated on 30/Sep/21
https://en.m.wikipedia.org/wiki/Double_pendulum
Commented by ajfour last updated on 30/Sep/21
$${lets}\:{first}\:{try}\:{my}\:{Q}\:{ourselves} \\ $$$${sir},\:{thanks}\:{for}\:{the}\:{link},\:{anyway}. \\ $$
Commented by TheHoneyCat last updated on 30/Sep/21
to Mr W:
the double pendulum is a chaotic system that cannot be solved, however, I think that the question here is not that.
"Find Φ(θ)" seems to imply we are imposing θ(t).
The problem is therefore no longer chaotic.
This is of course only true if we actually look for Φ(θ, dθ/dt)...
Commented by mr W last updated on 30/Sep/21
$${also}\:{here}\:{it}\:{is}\:{the}\:{motion}\:{of}\:{a} \\ $$$${double}\:{pendulum}. \\ $$
Answered by ajfour last updated on 30/Sep/21
![Energy conservation overall_(−) Mg((a/2))(1−cos θ) +mg((a/2))(3−2cos θ−cos φ) =((Ma^2 ω_θ ^2 )/3)+((ma^2 ω_φ ^2 )/(12)) +((ma^2 )/2)[(ω_θ −((ω_φ cos φ)/2))^2 +((ω_φ ^2 sin^2 φ)/4)] ...........(i) Mgsin θ−F_r =((Mω_θ ^2 a)/2) ....(ii) (F_r sin φ+F_t cos φ)(a/2)=(((((ma^2 )/(12)))(ω_θ −ω_φ )d(ω_θ −ω_φ ))/(d(θ−φ))) .......(iii) (Mgcos θ−F_t )a=((Mω_θ dω_θ )/dθ) ......(iv) ⇒ if m=kM (gsin θ−((ω_θ ^2 a)/2))sin φ+(gcos θ−((ω_θ dω_θ )/dθ))cos φ =(((ka)/6))((d(ω_θ ^2 −ω_φ ^2 ))/(d(θ−φ))) ....(I) now from ..(i) 4ω_θ ^2 +kω_φ ^2 +6k(ω_θ ^2 +ω_φ ^2 −ω_θ ω_φ cos φ) =((6g)/a){(1−cos θ+3k−2kcos θ−kcos φ) ....(II) let (ω_φ /ω_θ )=β ⇒ (φ/θ)=β ω_θ ^2 =(((1−2k)cos θ−kcos (kθ)+3k+1)/(4+6k+7kβ^2 +−6kβcos (kθ))) ⇒ ω_θ ^2 =f(θ,β) ∀ φ=βθ ...(A) & ((g/a))cos [θ(1−β)]−((ω_θ ^2 sin (βθ))/2)−(((cos (βθ))/2))((d(ω_θ ^2 ))/dθ) =((d[ω_θ ^2 (1−kβ^2 )])/(d[θ(1−β)])) ....(B) from (A) & (B) θ and β=(φ/θ) can be related using φ=0, when θ=0 ★](https://www.tinkutara.com/question/Q155458.png)
$$\underset{−} {{Energy}\:{conservation}\:{overall}} \\ $$$${Mg}\left(\frac{{a}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\:\:+{mg}\left(\frac{{a}}{\mathrm{2}}\right)\left(\mathrm{3}−\mathrm{2cos}\:\theta−\mathrm{cos}\:\phi\right) \\ $$$$=\frac{{Ma}^{\mathrm{2}} \omega_{\theta} ^{\mathrm{2}} }{\mathrm{3}}+\frac{{ma}^{\mathrm{2}} \omega_{\phi} ^{\mathrm{2}} }{\mathrm{12}} \\ $$$$+\frac{{ma}^{\mathrm{2}} }{\mathrm{2}}\left[\left(\omega_{\theta} −\frac{\omega_{\phi} \mathrm{cos}\:\phi}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\omega_{\phi} ^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \phi}{\mathrm{4}}\right] \\ $$$$\:\:\:\:\:………..\left({i}\right) \\ $$$${Mg}\mathrm{sin}\:\theta−{F}_{{r}} =\frac{{M}\omega_{\theta} ^{\mathrm{2}} {a}}{\mathrm{2}}\:\:\:\:….\left({ii}\right) \\ $$$$\left({F}_{{r}} \mathrm{sin}\:\phi+{F}_{{t}} \mathrm{cos}\:\phi\right)\frac{{a}}{\mathrm{2}}=\frac{\left(\frac{{ma}^{\mathrm{2}} }{\mathrm{12}}\right)\left(\omega_{\theta} −\omega_{\phi} \right){d}\left(\omega_{\theta} −\omega_{\phi} \right)}{{d}\left(\theta−\phi\right)} \\ $$$$…….\left({iii}\right) \\ $$$$\left({Mg}\mathrm{cos}\:\theta−{F}_{{t}} \right){a}=\frac{{M}\omega_{\theta} {d}\omega_{\theta} }{{d}\theta} \\ $$$$\:\:\:……\left({iv}\right) \\ $$$$\Rightarrow\:\:\:{if}\:\:\:{m}={kM} \\ $$$$\left({g}\mathrm{sin}\:\theta−\frac{\omega_{\theta} ^{\mathrm{2}} {a}}{\mathrm{2}}\right)\mathrm{sin}\:\phi+\left({g}\mathrm{cos}\:\theta−\frac{\omega_{\theta} {d}\omega_{\theta} }{{d}\theta}\right)\mathrm{cos}\:\phi \\ $$$$\:\:\:=\left(\frac{{ka}}{\mathrm{6}}\right)\frac{{d}\left(\omega_{\theta} ^{\mathrm{2}} −\omega_{\phi} ^{\mathrm{2}} \right)}{{d}\left(\theta−\phi\right)}\:\:\:\:….\left({I}\right) \\ $$$$ \\ $$$${now}\:{from}\:..\left({i}\right) \\ $$$$\mathrm{4}\omega_{\theta} ^{\mathrm{2}} +{k}\omega_{\phi} ^{\mathrm{2}} +\mathrm{6}{k}\left(\omega_{\theta} ^{\mathrm{2}} +\omega_{\phi} ^{\mathrm{2}} −\omega_{\theta} \omega_{\phi} \mathrm{cos}\:\phi\right) \\ $$$$\:\:=\frac{\mathrm{6}{g}}{{a}}\left\{\left(\mathrm{1}−\mathrm{cos}\:\theta+\mathrm{3}{k}−\mathrm{2}{k}\mathrm{cos}\:\theta−{k}\mathrm{cos}\:\phi\right)\right. \\ $$$$\:\:\:….\left({II}\right) \\ $$$${let}\:\:\frac{\omega_{\phi} }{\omega_{\theta} }=\beta\:\:\:\:\Rightarrow\:\:\frac{\phi}{\theta}=\beta \\ $$$$\omega_{\theta} ^{\mathrm{2}} =\frac{\left(\mathrm{1}−\mathrm{2}{k}\right)\mathrm{cos}\:\theta−{k}\mathrm{cos}\:\left({k}\theta\right)+\mathrm{3}{k}+\mathrm{1}}{\mathrm{4}+\mathrm{6}{k}+\mathrm{7}{k}\beta^{\mathrm{2}} +−\mathrm{6}{k}\beta\mathrm{cos}\:\left({k}\theta\right)} \\ $$$$\Rightarrow\:\omega_{\theta} ^{\mathrm{2}} ={f}\left(\theta,\beta\right)\:\:\:\:\forall\:\:\phi=\beta\theta\:\:…\left({A}\right) \\ $$$$\& \\ $$$$\left(\frac{{g}}{{a}}\right)\mathrm{cos}\:\left[\theta\left(\mathrm{1}−\beta\right)\right]−\frac{\omega_{\theta} ^{\mathrm{2}} \mathrm{sin}\:\left(\beta\theta\right)}{\mathrm{2}}−\left(\frac{\mathrm{cos}\:\left(\beta\theta\right)}{\mathrm{2}}\right)\frac{{d}\left(\omega_{\theta} ^{\mathrm{2}} \right)}{{d}\theta} \\ $$$$\:=\frac{{d}\left[\omega_{\theta} ^{\mathrm{2}} \left(\mathrm{1}−{k}\beta^{\mathrm{2}} \right)\right]}{{d}\left[\theta\left(\mathrm{1}−\beta\right)\right]}\:\:\:\:\:\:….\left({B}\right) \\ $$$${from}\:\left({A}\right)\:\&\:\left({B}\right)\:\: \\ $$$$\theta\:\:{and}\:\beta=\frac{\phi}{\theta}\:\:\:{can}\:{be}\:{related} \\ $$$${using}\:\:\phi=\mathrm{0},\:\:{when}\:\theta=\mathrm{0} \\ $$$$\bigstar \\ $$$$ \\ $$