Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 23472 by ajfour last updated on 31/Oct/17

Commented by ajfour last updated on 31/Oct/17

Q.23390 (continued..)

$${Q}.\mathrm{23390}\:\left({continued}..\right) \\ $$

Answered by ajfour last updated on 31/Oct/17

 Applying ΣF_x =Σma_x from the  ground frame:    f=(M+m)(αR)+((m(αR)cos θ)/2)                          −mω^2 ((R/2))sin θ   ...(i)  Applying Στ = I_(net) α from the  frame of centre of ring:  ((mgRsin θ)/2)−fR−(mαR)((R/2)cos θ)                =(MR^2 +((mR^2 )/3))α   ...(ii)  eliminating f from (i) and (ii):  ((mgsin θ)/2)−(M+m)(αR)  −((m(αR)cos θ)/2)+((mω^2 Rsin θ)/2)      −((m(αR)cos θ)/2) =(M+(m/3))(αR)  ⇒ (αR)[M+(m/3)+M+m+mcos θ]               =((msin θ)/2)(ω^2 R+g)  ⇒  α=((m(ω^2 +g/R)sin θ)/(2[2M+((4m)/3)+mcos θ]))  ⇒  ((ωdω)/dθ) =((m(ω^2 +(g/R))sin θ)/(2[2M+((4m)/3)+mcos θ]))  ∫_0 ^(  ω) ((2ωdω)/(ω^2 +(g/R))) = ∫_0 ^(  θ) ((msin θ)/(2M+((4m)/3)+mcos θ))   ln (((ω^2 +(g/R))/(g/R)))=ln (((2M+((4m)/3)+m)/(2M+((4m)/3)+mcos θ)))  ((ω^2 R)/g)+1 =1−((m(1−cos θ))/(2M+((4m)/3)+mcos θ))      𝛚^2  =(g/R)(((1−cos 𝛉)/(((2M)/m)+(4/3)+cos 𝛉)))  And   𝛂 =(1/2)((d(𝛚^2 ))/d𝛉) .  To obtain the time for the return,  ⇒ 𝛚= (d𝛉/dt)=(√(g/R))[((1−cos θ)/(b^2 +cos θ))]^(1/2)   ⇒ T =4(√(R/g))∫_0 ^(  π)   [((b^2 +cos θ)/(1−cos θ))]^(1/2) dθ      where b^2 =((2M)/m)+(4/3) .     .... might continue ...

$$\:{Applying}\:\Sigma{F}_{{x}} =\Sigma{ma}_{{x}} {from}\:{the} \\ $$$${ground}\:{frame}: \\ $$$$\:\:{f}=\left({M}+{m}\right)\left(\alpha{R}\right)+\frac{{m}\left(\alpha{R}\right)\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{m}\omega^{\mathrm{2}} \left(\frac{{R}}{\mathrm{2}}\right)\mathrm{sin}\:\theta\:\:\:...\left({i}\right) \\ $$$${Applying}\:\Sigma\tau\:=\:{I}_{{net}} \alpha\:{from}\:{the} \\ $$$${frame}\:{of}\:{centre}\:{of}\:{ring}: \\ $$$$\frac{{mgR}\mathrm{sin}\:\theta}{\mathrm{2}}−{fR}−\left({m}\alpha{R}\right)\left(\frac{{R}}{\mathrm{2}}\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({MR}^{\mathrm{2}} +\frac{{mR}^{\mathrm{2}} }{\mathrm{3}}\right)\alpha\:\:\:...\left({ii}\right) \\ $$$${eliminating}\:\boldsymbol{{f}}\:{from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\frac{{mg}\mathrm{sin}\:\theta}{\mathrm{2}}−\left({M}+{m}\right)\left(\alpha{R}\right) \\ $$$$−\frac{{m}\left(\alpha{R}\right)\mathrm{cos}\:\theta}{\mathrm{2}}+\frac{{m}\omega^{\mathrm{2}} {R}\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:−\frac{{m}\left(\alpha{R}\right)\mathrm{cos}\:\theta}{\mathrm{2}}\:=\left({M}+\frac{{m}}{\mathrm{3}}\right)\left(\alpha{R}\right) \\ $$$$\Rightarrow\:\left(\alpha{R}\right)\left[{M}+\frac{{m}}{\mathrm{3}}+{M}+{m}+{m}\mathrm{cos}\:\theta\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{m}\mathrm{sin}\:\theta}{\mathrm{2}}\left(\omega^{\mathrm{2}} {R}+{g}\right) \\ $$$$\Rightarrow\:\:\alpha=\frac{{m}\left(\omega^{\mathrm{2}} +{g}/{R}\right)\mathrm{sin}\:\theta}{\mathrm{2}\left[\mathrm{2}{M}+\frac{\mathrm{4}{m}}{\mathrm{3}}+{m}\mathrm{cos}\:\theta\right]} \\ $$$$\Rightarrow\:\:\frac{\omega{d}\omega}{{d}\theta}\:=\frac{{m}\left(\omega^{\mathrm{2}} +\frac{{g}}{{R}}\right)\mathrm{sin}\:\theta}{\mathrm{2}\left[\mathrm{2}{M}+\frac{\mathrm{4}{m}}{\mathrm{3}}+{m}\mathrm{cos}\:\theta\right]} \\ $$$$\int_{\mathrm{0}} ^{\:\:\omega} \frac{\mathrm{2}\omega{d}\omega}{\omega^{\mathrm{2}} +\frac{{g}}{{R}}}\:=\:\int_{\mathrm{0}} ^{\:\:\theta} \frac{{m}\mathrm{sin}\:\theta}{\mathrm{2}{M}+\frac{\mathrm{4}{m}}{\mathrm{3}}+{m}\mathrm{cos}\:\theta} \\ $$$$\:\mathrm{ln}\:\left(\frac{\omega^{\mathrm{2}} +\frac{{g}}{{R}}}{\frac{{g}}{{R}}}\right)=\mathrm{ln}\:\left(\frac{\mathrm{2}{M}+\frac{\mathrm{4}{m}}{\mathrm{3}}+{m}}{\mathrm{2}{M}+\frac{\mathrm{4}{m}}{\mathrm{3}}+{m}\mathrm{cos}\:\theta}\right) \\ $$$$\frac{\omega^{\mathrm{2}} {R}}{{g}}+\mathrm{1}\:=\mathrm{1}−\frac{{m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}{M}+\frac{\mathrm{4}{m}}{\mathrm{3}}+{m}\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:\boldsymbol{\omega}^{\mathrm{2}} \:=\frac{\boldsymbol{{g}}}{\boldsymbol{{R}}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\boldsymbol{\theta}}{\frac{\mathrm{2}\boldsymbol{{M}}}{\boldsymbol{{m}}}+\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{cos}\:\boldsymbol{\theta}}\right) \\ $$$${And}\:\:\:\boldsymbol{\alpha}\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\boldsymbol{{d}}\left(\boldsymbol{\omega}^{\mathrm{2}} \right)}{\boldsymbol{{d}\theta}}\:. \\ $$$${To}\:{obtain}\:{the}\:{time}\:{for}\:{the}\:{return}, \\ $$$$\Rightarrow\:\boldsymbol{\omega}=\:\frac{\boldsymbol{{d}\theta}}{\boldsymbol{{dt}}}=\sqrt{\frac{{g}}{{R}}}\left[\frac{\mathrm{1}−\mathrm{cos}\:\theta}{{b}^{\mathrm{2}} +\mathrm{cos}\:\theta}\right]^{\mathrm{1}/\mathrm{2}} \\ $$$$\Rightarrow\:{T}\:=\mathrm{4}\sqrt{\frac{{R}}{{g}}}\int_{\mathrm{0}} ^{\:\:\pi} \:\:\left[\frac{{b}^{\mathrm{2}} +\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}\right]^{\mathrm{1}/\mathrm{2}} {d}\theta \\ $$$$\:\:\:\:{where}\:{b}^{\mathrm{2}} =\frac{\mathrm{2}{M}}{{m}}+\frac{\mathrm{4}}{\mathrm{3}}\:. \\ $$$$\:\:\:....\:{might}\:{continue}\:... \\ $$

Commented by mrW1 last updated on 31/Oct/17

how is it to explain, that the time   from θ=0 is not convergent?

$$\mathrm{how}\:\mathrm{is}\:\mathrm{it}\:\mathrm{to}\:\mathrm{explain},\:\mathrm{that}\:\mathrm{the}\:\mathrm{time}\: \\ $$$$\mathrm{from}\:\theta=\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{convergent}? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com