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Question-136984

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Question Number 136984 by mr W last updated on 28/Mar/21
Commented by mr W last updated on 28/Mar/21
see Q136382
$${see}\:{Q}\mathrm{136382} \\ $$
Commented by mr W last updated on 30/Mar/21
r=(b−y)tan γ dx=rdθ dy=tan φ dx dz=dr=−tan γ dy=−tan γ tan φ dx dz=−tan γ tan φ rdθ ω=(dθ/dt) I=((3Ma^2 )/(10)) E=(1/2)Iω^2 dE=−Mgdz=Mgtan γ tan φ rdθ (1/2)×((3Ma^2 )/(10))×2ω(dω/dt)=Mgtan γ tan φ r(dθ/dt) (dω/dt)=α=((10g tan φ )/(3ab))×r=λr with λ=((10g tan φ )/(3ab)) v_x =rω a_x =(dv_x /dt)=ω(dr/dt)+r(dω/dt) =−rω^2 tan φ tan γ+λr^2 at t=0: ω=0 ⇒a_x =λr^2 =((10g tan φ r^2 )/(3ab)) with r=(b−x tan φ)(a/b) =================== ω(dω/dθ)=λr ωdω=λdx (ω^2 /2)=λ(x−x_0 ) ω=(dx/(rdt))=(√(2λ(x−x_0 ))) (dx/((b−x tan φ)tan γ dt))=(√(2λ(x−x_0 ))) (dx/( (√(2λ)) tan φ tan γ((b/(tan φ))−x)(√(x−x_0 ))))=dt ⇒t=(1/( ((2 tan φ)/b)(√((5ga tan φ )/(3b)))))∫_x_0 ^x (dx/( ((b/(tan φ))−x)(√(x−x_0 )))) ⇒t=(1/( ((2 tan φ)/b)(√(((5ag)/3)(1−((x_0 tan φ)/b)))))) ln (((√((b/(tan φ))−x_0 ))+(√(x−x_0 )))/( (√((b/(tan φ))−x_0 ))−(√(x−x_0 )))) ⇒t=(1/( 2(√((5g tan γ tan φ)/(3x_m )))))×(1/( (√(1−ξ_0 )))) ln (((√(1−ξ_0 ))+(√(ξ−ξ_0 )))/( (√(1−ξ_0 ))−(√(ξ−ξ_0 )))) ⇒t=(√((3x_m )/(20g tan φ tan γ))) Φ ⇒Φ=(1/( (√(1−ξ_0 )))) ln (((√(1−ξ_0 ))+(√(ξ−ξ_0 )))/( (√(1−ξ_0 ))−(√(ξ−ξ_0 )))) x_m =(b/(tan φ)) ξ=(x/x_m ), ξ_0 =(x_0 /x_m )
$${r}=\left({b}−{y}\right)\mathrm{tan}\:\gamma \\ $$$${dx}={rd}\theta \\ $$$${dy}=\mathrm{tan}\:\phi\:{dx} \\ $$$${dz}={dr}=−\mathrm{tan}\:\gamma\:{dy}=−\mathrm{tan}\:\gamma\:\mathrm{tan}\:\phi\:{dx} \\ $$$${dz}=−\mathrm{tan}\:\gamma\:\mathrm{tan}\:\phi\:{rd}\theta \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$${I}=\frac{\mathrm{3}{Ma}^{\mathrm{2}} }{\mathrm{10}} \\ $$$${E}=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \\ $$$${dE}=−{Mgdz}={Mg}\mathrm{tan}\:\gamma\:\mathrm{tan}\:\phi\:{rd}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{Ma}^{\mathrm{2}} }{\mathrm{10}}×\mathrm{2}\omega\frac{{d}\omega}{{dt}}={Mg}\mathrm{tan}\:\gamma\:\mathrm{tan}\:\phi\:{r}\frac{{d}\theta}{{dt}} \\ $$$$\frac{{d}\omega}{{dt}}=\alpha=\frac{\mathrm{10}{g}\:\mathrm{tan}\:\phi\:}{\mathrm{3}{ab}}×{r}=\lambda{r} \\ $$$${with}\:\lambda=\frac{\mathrm{10}{g}\:\mathrm{tan}\:\phi\:}{\mathrm{3}{ab}} \\ $$$${v}_{{x}} ={r}\omega \\ $$$${a}_{{x}} =\frac{{dv}_{{x}} }{{dt}}=\omega\frac{{dr}}{{dt}}+{r}\frac{{d}\omega}{{dt}} \\ $$$$\:\:\:\:=−{r}\omega^{\mathrm{2}} \mathrm{tan}\:\phi\:\mathrm{tan}\:\gamma+\lambda{r}^{\mathrm{2}} \\ $$$${at}\:{t}=\mathrm{0}:\:\omega=\mathrm{0} \\ $$$$\Rightarrow{a}_{{x}} =\lambda{r}^{\mathrm{2}} =\frac{\mathrm{10}{g}\:\mathrm{tan}\:\phi\:{r}^{\mathrm{2}} \:}{\mathrm{3}{ab}} \\ $$$${with}\:{r}=\left({b}−{x}\:\mathrm{tan}\:\phi\right)\frac{{a}}{{b}} \\ $$$$=================== \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=\lambda{r} \\ $$$$\omega{d}\omega=\lambda{dx} \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\lambda\left({x}−{x}_{\mathrm{0}} \right) \\ $$$$\omega=\frac{{dx}}{{rdt}}=\sqrt{\mathrm{2}\lambda\left({x}−{x}_{\mathrm{0}} \right)} \\ $$$$\frac{{dx}}{\left({b}−{x}\:\mathrm{tan}\:\phi\right)\mathrm{tan}\:\gamma\:{dt}}=\sqrt{\mathrm{2}\lambda\left({x}−{x}_{\mathrm{0}} \right)} \\ $$$$\frac{{dx}}{\:\sqrt{\mathrm{2}\lambda}\:\mathrm{tan}\:\phi\:\mathrm{tan}\:\gamma\left(\frac{{b}}{\mathrm{tan}\:\phi}−{x}\right)\sqrt{{x}−{x}_{\mathrm{0}} }}={dt} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\:\frac{\mathrm{2}\:\mathrm{tan}\:\phi}{{b}}\sqrt{\frac{\mathrm{5}{ga}\:\mathrm{tan}\:\phi\:}{\mathrm{3}{b}}}}\int_{{x}_{\mathrm{0}} } ^{{x}} \frac{{dx}}{\:\left(\frac{{b}}{\mathrm{tan}\:\phi}−{x}\right)\sqrt{{x}−{x}_{\mathrm{0}} }} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\:\frac{\mathrm{2}\:\mathrm{tan}\:\phi}{{b}}\sqrt{\frac{\mathrm{5}{ag}}{\mathrm{3}}\left(\mathrm{1}−\frac{{x}_{\mathrm{0}} \:\mathrm{tan}\:\phi}{{b}}\right)}}\:\mathrm{ln}\:\frac{\sqrt{\frac{{b}}{\mathrm{tan}\:\phi}−{x}_{\mathrm{0}} }+\sqrt{{x}−{x}_{\mathrm{0}} }}{\:\sqrt{\frac{{b}}{\mathrm{tan}\:\phi}−{x}_{\mathrm{0}} }−\sqrt{{x}−{x}_{\mathrm{0}} }} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\frac{\mathrm{5}{g}\:\mathrm{tan}\:\gamma\:\mathrm{tan}\:\phi}{\mathrm{3}{x}_{{m}} }}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\xi_{\mathrm{0}} }}\:\mathrm{ln}\:\frac{\sqrt{\mathrm{1}−\xi_{\mathrm{0}} }+\sqrt{\xi−\xi_{\mathrm{0}} }}{\:\sqrt{\mathrm{1}−\xi_{\mathrm{0}} }−\sqrt{\xi−\xi_{\mathrm{0}} }} \\ $$$$\Rightarrow{t}=\sqrt{\frac{\mathrm{3}{x}_{{m}} }{\mathrm{20}{g}\:\mathrm{tan}\:\phi\:\mathrm{tan}\:\gamma}}\:\Phi \\ $$$$\Rightarrow\Phi=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\xi_{\mathrm{0}} }}\:\mathrm{ln}\:\frac{\sqrt{\mathrm{1}−\xi_{\mathrm{0}} }+\sqrt{\xi−\xi_{\mathrm{0}} }}{\:\sqrt{\mathrm{1}−\xi_{\mathrm{0}} }−\sqrt{\xi−\xi_{\mathrm{0}} }} \\ $$$${x}_{{m}} =\frac{{b}}{\mathrm{tan}\:\phi} \\ $$$$\xi=\frac{{x}}{{x}_{{m}} },\:\xi_{\mathrm{0}} =\frac{{x}_{\mathrm{0}} }{{x}_{{m}} } \\ $$
Commented by ajfour last updated on 29/Mar/21
Superb solution sir. Thanks.
$${Superb}\:{solution}\:{sir}.\:{Thanks}. \\ $$
Commented by mr W last updated on 30/Mar/21

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