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Question Number 214178 by ajfour last updated on 30/Nov/24
Commented by ajfour last updated on 01/Dec/24
Q. 214132 But with left roller fixed.
$${Q}.\:\mathrm{214132}\:\:{But}\:{with}\:{left}\:{roller}\:{fixed}. \\ $$
Answered by ajfour last updated on 30/Nov/24
Answered by mr W last updated on 01/Dec/24
Commented by mr W last updated on 01/Dec/24
let ξ=(a/b) ω=(dθ/dt) ω_1 =(dϕ/dt)=ω(dϕ/dθ) ((sin θ)/b)=((sin ϕ)/a)=((sin (θ+ϕ))/(AB)) ⇒sin ϕ=((a sin θ)/b) cos ϕ (dϕ/dθ)=((a cos θ)/b) (dϕ/dθ)=((a cos θ cos ϕ)/b)=((a cos θ)/b)(√(1−((a^2 sin^2 θ)/b^2 ))) (dϕ/dθ)=ξ cos θ(√(1−ξ^2 sin^2 θ)) AB=((b sin (θ+ϕ))/(sin θ))=b(cos ϕ+((sin ϕ)/(tan θ))) AB=b((√(1−((a^2 sin^2 θ)/b^2 )))+((a cos θ)/b)) x_D =((b cos θ)/2)+(b/2)((√(1−((a^2 sin^2 θ)/b^2 )))+((a cos θ)/b)) x_D =(b/2)[(√(1−((a^2 sin^2 θ)/b^2 )))+(1+(a/b))cos θ] x_D =(b/2)[(√(1−ξ^2 sin^2 θ))+(1+ξ)cos θ] y_D =((b sin θ)/2) v_(Dx) =(dx_D /dt)=ω(dx_D /dθ)=((ωb cos θ)/2)[((ξ^2 sin θ)/( (√(1−ξ^2 sin^2 θ))))−(1+ξ)] v_(Dy) =(dy_D /dt)=ω(dy_D /dθ)=((ωb cos θ)/2) h=b sin θ loss in P.E.=gain in K.E. (M+m)g(((h_0 −h)/2))=(1/2)×((Mb^2 ω^2 )/3)+(1/2)×((ma^2 ω_1 ^2 )/(12))+(m/2)(((ωb cos θ)/2))^2 {1+[((ξ^2 sin θ)/( (√(1−ξ^2 sin^2 θ))))−(1+ξ)]^2 } (M+m)g(h_0 −b sin θ)=((Mb^2 ω^2 )/3)+((ma^2 ω^2 ξ^2 cos^2 θ (1−ξ^2 sin^2 θ))/(12))+((mω^2 b^2 cos^2 θ)/4){1+[((ξ^2 sin θ)/( (√(1−ξ^2 sin^2 θ))))−(1+ξ)]^2 } (M+m)g(h_0 −b sin θ)=((Mb^2 ω^2 )/3)+((mb^2 ω^2 ξ^4 cos^2 θ (1−ξ^2 sin^2 θ))/(12))+((mω^2 b^2 cos^2 θ)/4){1+[((ξ^2 sin θ)/( (√(1−ξ^2 sin^2 θ))))−(1+ξ)]^2 } ⇒ω^2 =(((M+m)g((h_0 /b)−sin θ))/(((Mb)/3)+((mbξ^4 cos^2 θ (1−ξ^2 sin^2 θ))/(12))+((mb cos^2 θ)/4){1+[((ξ^2 sin θ)/( (√(1−ξ^2 sin^2 θ))))−(1+ξ)]^2 })) ⇒ω=(√(((M+m)g((h_0 /b)−sin θ))/(((Mb)/3)+((mbξ^4 cos^2 θ (1−ξ^2 sin^2 θ))/(12))+((mb cos^2 θ)/4){1+[((ξ^2 sin θ)/( (√(1−ξ^2 sin^2 θ))))−(1+ξ)]^2 }))) T=∫_0 ^θ_0 (dθ/ω) v_(Py) =ωb cos θ at t=T: ω=(√((12(M+m)gh_0 )/([4M+mξ^2 (3+ξ^2 )]b^2 ))) v_(Py) =(√((12(M+m)gh_0 )/(4M+mξ^2 (3+ξ^2 )))) case: M=m, a=b v_(Py) =(√(3gh_0 )) > (√(2gh_0 ))
$${let}\:\xi=\frac{{a}}{{b}} \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$$\omega_{\mathrm{1}} =\frac{{d}\varphi}{{dt}}=\omega\frac{{d}\varphi}{{d}\theta} \\ $$$$\frac{\mathrm{sin}\:\theta}{{b}}=\frac{\mathrm{sin}\:\varphi}{{a}}=\frac{\mathrm{sin}\:\left(\theta+\varphi\right)}{{AB}} \\ $$$$\Rightarrow\mathrm{sin}\:\varphi=\frac{{a}\:\mathrm{sin}\:\theta}{{b}} \\ $$$$\mathrm{cos}\:\varphi\:\frac{{d}\varphi}{{d}\theta}=\frac{{a}\:\mathrm{cos}\:\theta}{{b}} \\ $$$$\frac{{d}\varphi}{{d}\theta}=\frac{{a}\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\varphi}{{b}}=\frac{{a}\:\mathrm{cos}\:\theta}{{b}}\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }} \\ $$$$\frac{{d}\varphi}{{d}\theta}=\xi\:\mathrm{cos}\:\theta\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$${AB}=\frac{{b}\:\mathrm{sin}\:\left(\theta+\varphi\right)}{\mathrm{sin}\:\theta}={b}\left(\mathrm{cos}\:\varphi+\frac{\mathrm{sin}\:\varphi}{\mathrm{tan}\:\theta}\right) \\ $$$${AB}={b}\left(\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }}+\frac{{a}\:\mathrm{cos}\:\theta}{{b}}\right) \\ $$$${x}_{{D}} =\frac{{b}\:\mathrm{cos}\:\theta}{\mathrm{2}}+\frac{{b}}{\mathrm{2}}\left(\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }}+\frac{{a}\:\mathrm{cos}\:\theta}{{b}}\right) \\ $$$${x}_{{D}} =\frac{{b}}{\mathrm{2}}\left[\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }}+\left(\mathrm{1}+\frac{{a}}{{b}}\right)\mathrm{cos}\:\theta\right] \\ $$$${x}_{{D}} =\frac{{b}}{\mathrm{2}}\left[\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}+\left(\mathrm{1}+\xi\right)\mathrm{cos}\:\theta\right] \\ $$$${y}_{{D}} =\frac{{b}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${v}_{{Dx}} =\frac{{dx}_{{D}} }{{dt}}=\omega\frac{{dx}_{{D}} }{{d}\theta}=\frac{\omega{b}\:\mathrm{cos}\:\theta}{\mathrm{2}}\left[\frac{\xi^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}−\left(\mathrm{1}+\xi\right)\right] \\ $$$${v}_{{Dy}} =\frac{{dy}_{{D}} }{{dt}}=\omega\frac{{dy}_{{D}} }{{d}\theta}=\frac{\omega{b}\:\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$${h}={b}\:\mathrm{sin}\:\theta \\ $$$${loss}\:{in}\:{P}.{E}.={gain}\:{in}\:{K}.{E}. \\ $$$$\left({M}+{m}\right){g}\left(\frac{{h}_{\mathrm{0}} −{h}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{Mb}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{ma}^{\mathrm{2}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{12}}+\frac{{m}}{\mathrm{2}}\left(\frac{\omega{b}\:\mathrm{cos}\:\theta}{\mathrm{2}}\right)^{\mathrm{2}} \left\{\mathrm{1}+\left[\frac{\xi^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}−\left(\mathrm{1}+\xi\right)\right]^{\mathrm{2}} \right\} \\ $$$$\left({M}+{m}\right){g}\left({h}_{\mathrm{0}} −{b}\:\mathrm{sin}\:\theta\right)=\frac{{Mb}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{3}}+\frac{{ma}^{\mathrm{2}} \omega^{\mathrm{2}} \xi^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)}{\mathrm{12}}+\frac{{m}\omega^{\mathrm{2}} {b}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}}\left\{\mathrm{1}+\left[\frac{\xi^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}−\left(\mathrm{1}+\xi\right)\right]^{\mathrm{2}} \right\} \\ $$$$\left({M}+{m}\right){g}\left({h}_{\mathrm{0}} −{b}\:\mathrm{sin}\:\theta\right)=\frac{{Mb}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{3}}+\frac{{mb}^{\mathrm{2}} \omega^{\mathrm{2}} \xi^{\mathrm{4}} \:\mathrm{cos}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)}{\mathrm{12}}+\frac{{m}\omega^{\mathrm{2}} {b}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}}\left\{\mathrm{1}+\left[\frac{\xi^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}−\left(\mathrm{1}+\xi\right)\right]^{\mathrm{2}} \right\} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\left({M}+{m}\right){g}\left(\frac{{h}_{\mathrm{0}} }{{b}}−\mathrm{sin}\:\theta\right)}{\frac{{Mb}}{\mathrm{3}}+\frac{{mb}\xi^{\mathrm{4}} \mathrm{cos}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)}{\mathrm{12}}+\frac{{mb}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}}\left\{\mathrm{1}+\left[\frac{\xi^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}−\left(\mathrm{1}+\xi\right)\right]^{\mathrm{2}} \right\}} \\ $$$$\Rightarrow\omega=\sqrt{\frac{\left({M}+{m}\right){g}\left(\frac{{h}_{\mathrm{0}} }{{b}}−\mathrm{sin}\:\theta\right)}{\frac{{Mb}}{\mathrm{3}}+\frac{{mb}\xi^{\mathrm{4}} \mathrm{cos}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)}{\mathrm{12}}+\frac{{mb}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}}\left\{\mathrm{1}+\left[\frac{\xi^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}−\left(\mathrm{1}+\xi\right)\right]^{\mathrm{2}} \right\}}} \\ $$$${T}=\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \frac{{d}\theta}{\omega} \\ $$$${v}_{{Py}} =\omega{b}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${at}\:{t}={T}: \\ $$$$\omega=\sqrt{\frac{\mathrm{12}\left({M}+{m}\right){gh}_{\mathrm{0}} }{\left[\mathrm{4}{M}+{m}\xi^{\mathrm{2}} \left(\mathrm{3}+\xi^{\mathrm{2}} \right)\right]{b}^{\mathrm{2}} }} \\ $$$${v}_{{Py}} =\sqrt{\frac{\mathrm{12}\left({M}+{m}\right){gh}_{\mathrm{0}} }{\mathrm{4}{M}+{m}\xi^{\mathrm{2}} \left(\mathrm{3}+\xi^{\mathrm{2}} \right)}} \\ $$$${case}:\:{M}={m},\:{a}={b} \\ $$$${v}_{{Py}} =\sqrt{\mathrm{3}{gh}_{\mathrm{0}} }\:>\:\sqrt{\mathrm{2}{gh}_{\mathrm{0}} } \\ $$
Commented by ajfour last updated on 01/Dec/24
how you do all these so quick sir, astonishes me. Gime time to understand both. Thanks enormous! And I made a video:
$${how}\:{you}\:{do}\:{all}\:{these}\:{so}\:{quick}\:{sir},\: \\ $$$${astonishes}\:{me}.\:{Gime}\:{time}\:{to} \\ $$$${understand}\:{both}.\:{Thanks}\:{enormous}! \\ $$$${And}\:{I}\:{made}\:{a}\:{video}: \\ $$
Commented by ajfour last updated on 01/Dec/24
https://youtu.be/ioBdOk1IO-8?si=tvSoMjDmslyv2Rjk
Commented by mr W last updated on 01/Dec/24
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