Question Number 214578 by ajfour last updated on 12/Dec/24
Commented by ajfour last updated on 12/Dec/24
$${If}\:{released}\:{as}\:{shown},\:{after}\:{what}\:{time} \\ $$$${does}\:{the}\:{other}\:{end}\:{of}\:{seesaw}\:{hits} \\ $$$${the}\:{ground}.\:\left({b}>{a}\right) \\ $$
Answered by mr W last updated on 13/Dec/24
Commented by mr W last updated on 13/Dec/24
![say μ=(b/a), ξ=(h/a) e=((a+b)/2)−a=((b−a)/2)=(((μ−1)a)/2) I=((m(a+b)^2 )/(12))+me^2 =((ma^2 [(μ+1)^2 +3(μ−1)^2 ])/(12)) =((ma^2 (μ^2 −μ+1))/3) or I=((m(a^3 +b^3 ))/(3(a+b)))=((ma^2 (μ^3 +1))/(3(μ+1))) at t=0: sin α=(h/a)=ξ ⇒α=sin^(−1) ξ θ=−α=−sin^(−1) ξ at t=T: sin β=(h/b)=(ξ/μ) ⇒β=sin^(−1) (ξ/μ) θ=β=sin^(−1) (ξ/μ) at t: ω=(dθ/dt) mge(sin α+sin θ)=((Iω^2 )/2) mg(((μ−1)/2))a(ξ+sin θ)=((ma^2 (μ^2 −μ+1)ω^2 )/6) g(μ−1)(ξ+sin θ)=((a(μ^2 −μ+1)ω^2 )/3) ω^2 =((3g(μ^2 −1)(ξ+sin θ))/(a(μ^3 +1))) (dt/dθ)=(1/ω)=(√((a(μ^3 +1))/(3g(μ^2 −1)(ξ+sin θ)))) ⇒T=(√((a(μ^3 +1))/(3g(μ^2 −1))))∫_(−sin^(−1) ξ) ^(sin^(−1) (ξ/μ)) (dθ/( (√(ξ+sin θ)))) example: ξ=(h/a)=(1/3), μ=(b/a)=2 T=(√(a/g))∫_(−sin^(−1) (1/3)) ^(sin^(−1) (1/6)) (dθ/( (√((1/3)+sin θ)))) ≈1.451940097932(√(a/g))](Q214588.png)
$${say}\:\mu=\frac{{b}}{{a}},\:\xi=\frac{{h}}{{a}} \\ $$$${e}=\frac{{a}+{b}}{\mathrm{2}}−{a}=\frac{{b}−{a}}{\mathrm{2}}=\frac{\left(\mu−\mathrm{1}\right){a}}{\mathrm{2}} \\ $$$${I}=\frac{{m}\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{12}}+{me}^{\mathrm{2}} =\frac{{ma}^{\mathrm{2}} \left[\left(\mu+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left(\mu−\mathrm{1}\right)^{\mathrm{2}} \right]}{\mathrm{12}} \\ $$$$\:\:=\frac{{ma}^{\mathrm{2}} \left(\mu^{\mathrm{2}} −\mu+\mathrm{1}\right)}{\mathrm{3}} \\ $$$${or}\:{I}=\frac{{m}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)}{\mathrm{3}\left({a}+{b}\right)}=\frac{{ma}^{\mathrm{2}} \left(\mu^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{3}\left(\mu+\mathrm{1}\right)} \\ $$$${at}\:{t}=\mathrm{0}: \\ $$$$\mathrm{sin}\:\alpha=\frac{{h}}{{a}}=\xi\:\Rightarrow\alpha=\mathrm{sin}^{−\mathrm{1}} \xi \\ $$$$\theta=−\alpha=−\mathrm{sin}^{−\mathrm{1}} \xi \\ $$$${at}\:{t}={T}: \\ $$$$\mathrm{sin}\:\beta=\frac{{h}}{{b}}=\frac{\xi}{\mu}\:\Rightarrow\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\mu} \\ $$$$\theta=\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\mu} \\ $$$${at}\:{t}: \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$${mge}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\theta\right)=\frac{{I}\omega^{\mathrm{2}} }{\mathrm{2}} \\ $$$${mg}\left(\frac{\mu−\mathrm{1}}{\mathrm{2}}\right){a}\left(\xi+\mathrm{sin}\:\theta\right)=\frac{{ma}^{\mathrm{2}} \left(\mu^{\mathrm{2}} −\mu+\mathrm{1}\right)\omega^{\mathrm{2}} }{\mathrm{6}} \\ $$$${g}\left(\mu−\mathrm{1}\right)\left(\xi+\mathrm{sin}\:\theta\right)=\frac{{a}\left(\mu^{\mathrm{2}} −\mu+\mathrm{1}\right)\omega^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}\left(\mu^{\mathrm{2}} −\mathrm{1}\right)\left(\xi+\mathrm{sin}\:\theta\right)}{{a}\left(\mu^{\mathrm{3}} +\mathrm{1}\right)} \\ $$$$\frac{{dt}}{{d}\theta}=\frac{\mathrm{1}}{\omega}=\sqrt{\frac{{a}\left(\mu^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{3}{g}\left(\mu^{\mathrm{2}} −\mathrm{1}\right)\left(\xi+\mathrm{sin}\:\theta\right)}} \\ $$$$\Rightarrow{T}=\sqrt{\frac{{a}\left(\mu^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{3}{g}\left(\mu^{\mathrm{2}} −\mathrm{1}\right)}}\int_{−\mathrm{sin}^{−\mathrm{1}} \xi} ^{\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\mu}} \frac{{d}\theta}{\:\sqrt{\xi+\mathrm{sin}\:\theta}} \\ $$$$ \\ $$$${example}: \\ $$$$\xi=\frac{{h}}{{a}}=\frac{\mathrm{1}}{\mathrm{3}},\:\mu=\frac{{b}}{{a}}=\mathrm{2} \\ $$$${T}=\sqrt{\frac{{a}}{{g}}}\int_{−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}} \frac{{d}\theta}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{sin}\:\theta}} \\ $$$$\:\:\:\:\approx\mathrm{1}.\mathrm{451940097932}\sqrt{\frac{{a}}{{g}}} \\ $$
Commented by ajfour last updated on 13/Dec/24
$${mg}\left(\frac{{b}−{a}}{\mathrm{2}}\right)\mathrm{cos}\:\theta=\frac{{m}}{\mathrm{3}}\left(\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}+{b}}\right)\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dt}}\left(\frac{{d}\theta}{{dt}}\right)=\frac{\omega{d}\omega}{{d}\theta}=\frac{\mathrm{3}{g}}{\mathrm{2}}\left(\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{3}} +{a}^{\mathrm{3}} }\right)\mathrm{cos}\:\theta=\lambda\mathrm{cos}\:\theta \\ $$$$\frac{\omega}{\:\sqrt{\mathrm{2}}}=\sqrt{\lambda}\sqrt{\mathrm{sin}\:\theta+\mathrm{sin}\:\alpha} \\ $$$${T}=\sqrt{\frac{{b}^{\mathrm{3}} +{a}^{\mathrm{3}} }{\mathrm{3}{g}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}}\int_{−\alpha} ^{\:\beta} \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\alpha+\mathrm{sin}\:\theta}} \\ $$$$ \\ $$
Commented by ajfour last updated on 13/Dec/24
$${Thanks}\:{sir}.\:{meticulously}\:{done}. \\ $$