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Question-80648

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Question Number 80648 by mr W last updated on 05/Feb/20
Commented by mr W last updated on 05/Feb/20
three thin−wall tubes with identical size and mass are released from the state as shown. find the time which the upper tube takes to hit the ground. there is no friction between the tubes, but the ground is rough such that the tubes can′t slip over it.
$${three}\:{thin}−{wall}\:{tubes}\:{with}\:{identical} \\ $$$${size}\:{and}\:{mass}\:{are}\:{released}\:{from}\:{the} \\ $$$${state}\:{as}\:{shown}. \\ $$$${find}\:{the}\:{time}\:{which}\:{the}\:{upper}\:{tube} \\ $$$${takes}\:{to}\:{hit}\:{the}\:{ground}. \\ $$$${there}\:{is}\:{no}\:{friction}\:{between}\:{the}\:{tubes}, \\ $$$${but}\:{the}\:{ground}\:{is}\:{rough}\:{such}\:{that}\:{the} \\ $$$${tubes}\:{can}'{t}\:{slip}\:{over}\:{it}. \\ $$
Commented by ajfour last updated on 05/Feb/20
Commented by ajfour last updated on 05/Feb/20
V=−(dy/dt) , A=(dV/dt) y=2rsin θ ⇒ −V=2rcos θ((dθ/dt)) friction on right tube is f^← fr=(mr^2 )(a/r) ⇒ f=ma Ncos θ−f=m(dv/dt) = ma ⇒ N=((2ma)/(cos θ)) mg−2Nsin θ=((mVdV)/dy) = mA ⇒ mg−((4masin θ)/(cos θ))=mA ⇒ A=g−4atan θ y=2rsin θ , x=2rcos θ Vsin θ=vcos θ .......(I) V=((vcos θ)/(sin θ))=−2rcos θ((dθ/dt)) ⇒ v=−2rsin θ((dθ/dt)) .....(II) differentiating (I) ⇒ V=vcot θ ⇒ Asin θ+Vcos θ(dθ/dt) = acos θ−vsin θ(dθ/dt) ⇒ gsin θ−4asin θtan θ +vcos θcot θ((dθ/dt)) = acos θ−vsin θ((dθ/dt)) ⇒ gsin θ−4(((sin^2 θ)/(cos θ)))(dv/dt) +((vcos^2 θ)/(sin θ))((dθ/dt))=(dv/dt)cos θ−vsin θ(dθ/dt) using (II) mgr(√3)−2mgrsin θ = 2mv^2 +((mV^( 2) )/2) gr(√3)−2grsin θ=2v^2 +((v^2 cos^2 θ)/(2sin^2 θ)) v^2 =((2gr((√3)−2sin θ)sin^2 θ)/(3+sin^2 θ)) ⇒ 2v(dv/dt)=2gr{(((3+sin^2 θ)(2(√3)sin θcos θ−6sin^2 θcos θ)−2sin θcos θ((√3)−2sin θ)sin^2 θ)/((1+sin^2 θ)^2 ))} a=(dv/dt)= 0 ⇒ N=0 ⇒ (3+sin^2 θ)(2(√3)cos θ−6sin θcos θ)=2cos θ((√3)−2sin θ)sin^2 θ θ ≈ 33.92046° .....
$${V}=−\frac{{dy}}{{dt}}\:,\:\:{A}=\frac{{dV}}{{dt}} \\ $$$${y}=\mathrm{2}{r}\mathrm{sin}\:\theta\:\:\Rightarrow\:−{V}=\mathrm{2}{r}\mathrm{cos}\:\theta\left(\frac{{d}\theta}{{dt}}\right) \\ $$$${friction}\:{on}\:{right}\:{tube}\:{is}\:\overset{\leftarrow} {{f}} \\ $$$$\:{fr}=\left({mr}^{\mathrm{2}} \right)\frac{{a}}{{r}}\:\:\:\Rightarrow\:\:{f}={ma} \\ $$$${N}\mathrm{cos}\:\theta−{f}={m}\frac{{dv}}{{dt}}\:=\:{ma} \\ $$$$\Rightarrow\:\:{N}=\frac{\mathrm{2}{ma}}{\mathrm{cos}\:\theta} \\ $$$${mg}−\mathrm{2}{N}\mathrm{sin}\:\theta=\frac{{mVdV}}{{dy}}\:=\:{mA} \\ $$$$\Rightarrow\:\:{mg}−\frac{\mathrm{4}{ma}\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}={mA} \\ $$$$\Rightarrow\:\:{A}={g}−\mathrm{4}{a}\mathrm{tan}\:\theta \\ $$$${y}=\mathrm{2}{r}\mathrm{sin}\:\theta\:\:,\:\:{x}=\mathrm{2}{r}\mathrm{cos}\:\theta \\ $$$${V}\mathrm{sin}\:\theta={v}\mathrm{cos}\:\theta\:\:\:\:\:\:\:\:\:…….\left({I}\right) \\ $$$${V}=\frac{{v}\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=−\mathrm{2}{r}\mathrm{cos}\:\theta\left(\frac{{d}\theta}{{dt}}\right) \\ $$$$\Rightarrow\:\:\:{v}=−\mathrm{2}{r}\mathrm{sin}\:\theta\left(\frac{{d}\theta}{{dt}}\right)\:\:\:\:…..\left({II}\right) \\ $$$${differentiating}\:\left({I}\right) \\ $$$$\Rightarrow\:\:{V}={v}\mathrm{cot}\:\theta \\ $$$$\Rightarrow\:\:{A}\mathrm{sin}\:\theta+{V}\mathrm{cos}\:\theta\frac{{d}\theta}{{dt}}\:=\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\mathrm{cos}\:\theta−{v}\mathrm{sin}\:\theta\frac{{d}\theta}{{dt}} \\ $$$$\Rightarrow\:{g}\mathrm{sin}\:\theta−\mathrm{4}{a}\mathrm{sin}\:\theta\mathrm{tan}\:\theta \\ $$$$\:\:\:\:+{v}\mathrm{cos}\:\theta\mathrm{cot}\:\theta\left(\frac{{d}\theta}{{dt}}\right)\:\:=\: \\ $$$$\:\:\:\:\:\:\:\:{a}\mathrm{cos}\:\theta−{v}\mathrm{sin}\:\theta\left(\frac{{d}\theta}{{dt}}\right) \\ $$$$\Rightarrow\:{g}\mathrm{sin}\:\theta−\mathrm{4}\left(\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:\theta}\right)\frac{{dv}}{{dt}} \\ $$$$+\frac{{v}\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta}\left(\frac{{d}\theta}{{dt}}\right)=\frac{{dv}}{{dt}}\mathrm{cos}\:\theta−{v}\mathrm{sin}\:\theta\frac{{d}\theta}{{dt}} \\ $$$${using}\:\left({II}\right) \\ $$$${mgr}\sqrt{\mathrm{3}}−\mathrm{2}{mgr}\mathrm{sin}\:\theta \\ $$$$\:\:\:\:=\:\mathrm{2}{mv}^{\mathrm{2}} +\frac{{mV}^{\:\mathrm{2}} }{\mathrm{2}} \\ $$$${gr}\sqrt{\mathrm{3}}−\mathrm{2}{gr}\mathrm{sin}\:\theta=\mathrm{2}{v}^{\mathrm{2}} +\frac{{v}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2sin}\:^{\mathrm{2}} \theta} \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{2}{gr}\left(\sqrt{\mathrm{3}}−\mathrm{2sin}\:\theta\right)\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{3}+\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\Rightarrow\:\mathrm{2}{v}\frac{{dv}}{{dt}}=\mathrm{2}{gr}\left\{\frac{\left(\mathrm{3}+\mathrm{sin}\:^{\mathrm{2}} \theta\right)\left(\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\mathrm{cos}\:\theta−\mathrm{6sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta\right)−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta\left(\sqrt{\mathrm{3}}−\mathrm{2sin}\:\theta\right)\mathrm{sin}\:^{\mathrm{2}} \theta}{\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\right\} \\ $$$${a}=\frac{{dv}}{{dt}}=\:\mathrm{0}\:\:\Rightarrow\:{N}=\mathrm{0}\:\: \\ $$$$\Rightarrow\: \\ $$$$\left(\mathrm{3}+\mathrm{sin}\:^{\mathrm{2}} \theta\right)\left(\mathrm{2}\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{6sin}\:\theta\mathrm{cos}\:\theta\right)=\mathrm{2cos}\:\theta\left(\sqrt{\mathrm{3}}−\mathrm{2sin}\:\theta\right)\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\theta\:\approx\:\mathrm{33}.\mathrm{92046}° \\ $$$$….. \\ $$
Commented by mr W last updated on 05/Feb/20
super!
$${super}! \\ $$
Commented by mr W last updated on 06/Feb/20
final equation can be simplified to: sin^3 θ+9 sin θ−3(√3)=0 sin θ=(((3(√3)((√5)+1))/2))^(1/3) −(((3(√3)((√5)−1))/2))^(1/3) ⇒θ=33.920456°
$${final}\:{equation}\:{can}\:{be}\:{simplified}\:{to}: \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{9}\:\mathrm{sin}\:\theta−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$\Rightarrow\theta=\mathrm{33}.\mathrm{920456}° \\ $$
Commented by ajfour last updated on 06/Feb/20
Commented by ajfour last updated on 06/Feb/20
loss in potential energy is mg(r(√3)−2rsin θ) shouldn′t it be so, Sir ?
$${loss}\:{in}\:{potential}\:{energy}\:{is} \\ $$$${mg}\left({r}\sqrt{\mathrm{3}}−\mathrm{2}{r}\mathrm{sin}\:\theta\right) \\ $$$${shouldn}'{t}\:{it}\:{be}\:{so},\:{Sir}\:? \\ $$
Commented by mr W last updated on 06/Feb/20
certainly so. i had a short circuit. sorry!
$${certainly}\:{so}.\:{i}\:{had}\:{a}\:{short}\:{circuit}. \\ $$$${sorry}! \\ $$
Commented by zainal tanjung last updated on 06/Feb/20
$$ \\ $$$$ \\ $$

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