Question Number 137986 by ajfour last updated on 08/Apr/21
Commented by ajfour last updated on 08/Apr/21
$${Find}\:{maximum}\:{speed}\:{gained} \\ $$$${by}\:{cylinder}\:{if}\:{motion}\:{is} \\ $$$${initiated}.\:{Bottom}\:{corner}\:{is} \\ $$$${hinged}\:{to}\:{ground}. \\ $$
Answered by mr W last updated on 08/Apr/21
Commented by mr W last updated on 11/Apr/21
![let ω=−(dθ/dt) sin φ=((r−(√2)r sin θ)/r)=1−(√2) sin θ s=(√2)r cos θ+r cos φ =(√2)r cos θ+r(√(1−(1−(√2) sin θ)^2 )) =(√2)r cos θ+r(√(2 sin θ((√2)−sin θ))) (ds/dθ)=−(√2)r sin θ+r((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ))))) v=(ds/dt)=−ω(ds/dθ)=ωr[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))] I_A =((m((√2)r)^2 )/6)+mr^2 =((4mr^2 )/3) Iα=mgrsin ((π/4)−θ)+N sin φ s−N cos φ r ((4r)/(3g))α−sin ((π/2)−θ)=(N/(mg))(1−(√2) sin θ)((√2) cos θ) ⇒(N/(mg))=((((4r)/(3g))α−sin ((π/4)−θ))/( (√2)cos θ (1−(√2) sin θ))) when N=0: α=((3g)/(4r)) sin ((π/4)−θ) (1/2)I_A ω^2 +(1/2)Mv^2 =mgr[1−cos ((π/4)−θ)] ((4mr^2 )/3)ω^2 +Mω^2 r^2 [(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 =2mgr[1−cos ((π/4)−θ)] ω^2 {(4/3)+(M/m)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 }=((2g)/r)[1−cos ((π/4)−θ)] ω=(√(g/r))×(√((1−cos ((π/4)−θ))/((1/3)+((2M)/m)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 ))) with Φ(θ)=(√((1−cos ((π/4)−θ))/((1/3)+((2M)/m)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))]^2 ))) ⇒ω=(√(g/r))×Φ(θ) α=(dω/dt)=−ω(dω/dθ)=−(g/r)Φ(θ)Φ′(θ) −(g/r)Φ(θ)Φ′(θ)=((3g)/(4r)) sin ((π/4)−θ) ⇒Φ(θ)Φ′(θ)+(3/4) sin ((π/4)−θ)=0 v=(√(gr))×Φ(θ)[(√2) sin θ−((cos θ(1−(√2) sin θ))/( (√(sin θ((√2)−sin θ)))))] examples: (M/m)=1 we get N=0 at θ≈0.5795 (=32.20°) v_(max) ≈0.082(√(gr)) (M/m)=10 we get N=0 at θ≈0.5373 (=30.79°) v_(max) ≈0.056(√(gr))](https://www.tinkutara.com/question/Q138001.png)
$${let}\:\omega=−\frac{{d}\theta}{{dt}} \\ $$$$\mathrm{sin}\:\phi=\frac{{r}−\sqrt{\mathrm{2}}{r}\:\mathrm{sin}\:\theta}{{r}}=\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta \\ $$$${s}=\sqrt{\mathrm{2}}{r}\:\mathrm{cos}\:\theta+{r}\:\mathrm{cos}\:\phi \\ $$$$\:\:\:=\sqrt{\mathrm{2}}{r}\:\mathrm{cos}\:\theta+{r}\sqrt{\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\sqrt{\mathrm{2}}{r}\:\mathrm{cos}\:\theta+{r}\sqrt{\mathrm{2}\:\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)} \\ $$$$\frac{{ds}}{{d}\theta}=−\sqrt{\mathrm{2}}{r}\:\mathrm{sin}\:\theta+{r}\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}} \\ $$$${v}=\frac{{ds}}{{dt}}=−\omega\frac{{ds}}{{d}\theta}=\omega{r}\left[\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right] \\ $$$${I}_{{A}} =\frac{{m}\left(\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} }{\mathrm{6}}+{mr}^{\mathrm{2}} =\frac{\mathrm{4}{mr}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${I}\alpha={mgr}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)+{N}\:\mathrm{sin}\:\phi\:{s}−{N}\:\mathrm{cos}\:\phi\:{r} \\ $$$$\frac{\mathrm{4}{r}}{\mathrm{3}{g}}\alpha−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\frac{{N}}{{mg}}\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\frac{\frac{\mathrm{4}{r}}{\mathrm{3}{g}}\alpha−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)}{\:\sqrt{\mathrm{2}}\mathrm{cos}\:\theta\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)} \\ $$$${when}\:{N}=\mathrm{0}: \\ $$$$\alpha=\frac{\mathrm{3}{g}}{\mathrm{4}{r}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right) \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{A}} \omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{Mv}^{\mathrm{2}} ={mgr}\left[\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)\right] \\ $$$$\frac{\mathrm{4}{mr}^{\mathrm{2}} }{\mathrm{3}}\omega^{\mathrm{2}} +{M}\omega^{\mathrm{2}} {r}^{\mathrm{2}} \left[\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right]^{\mathrm{2}} =\mathrm{2}{mgr}\left[\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)\right] \\ $$$$\omega^{\mathrm{2}} \left\{\frac{\mathrm{4}}{\mathrm{3}}+\frac{{M}}{{m}}\left[\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right]^{\mathrm{2}} \right\}=\frac{\mathrm{2}{g}}{{r}}\left[\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)\right] \\ $$$$\omega=\sqrt{\frac{{g}}{{r}}}×\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)}{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}{M}}{{m}}\left[\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right]^{\mathrm{2}} }} \\ $$$${with}\:\Phi\left(\theta\right)=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)}{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}{M}}{{m}}\left[\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right]^{\mathrm{2}} }} \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}}{{r}}}×\Phi\left(\theta\right) \\ $$$$\alpha=\frac{{d}\omega}{{dt}}=−\omega\frac{{d}\omega}{{d}\theta}=−\frac{{g}}{{r}}\Phi\left(\theta\right)\Phi'\left(\theta\right) \\ $$$$−\frac{{g}}{{r}}\Phi\left(\theta\right)\Phi'\left(\theta\right)=\frac{\mathrm{3}{g}}{\mathrm{4}{r}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right) \\ $$$$\Rightarrow\Phi\left(\theta\right)\Phi'\left(\theta\right)+\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)=\mathrm{0} \\ $$$${v}=\sqrt{{gr}}×\Phi\left(\theta\right)\left[\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right] \\ $$$${examples}: \\ $$$$\frac{{M}}{{m}}=\mathrm{1} \\ $$$${we}\:{get}\:{N}=\mathrm{0}\:{at}\:\theta\approx\mathrm{0}.\mathrm{5795}\:\left(=\mathrm{32}.\mathrm{20}°\right) \\ $$$${v}_{{max}} \approx\mathrm{0}.\mathrm{082}\sqrt{{gr}} \\ $$$$\frac{{M}}{{m}}=\mathrm{10} \\ $$$${we}\:{get}\:{N}=\mathrm{0}\:{at}\:\theta\approx\mathrm{0}.\mathrm{5373}\:\left(=\mathrm{30}.\mathrm{79}°\right) \\ $$$${v}_{{max}} \approx\mathrm{0}.\mathrm{056}\sqrt{{gr}} \\ $$
Commented by ajfour last updated on 12/Apr/21
![Ncos φ=MA mgrsin (π/4−θ)−Nr(√2)sin (θ−φ) =(((4mr^2 )/3))α when contact breaks N=0, A=0, 3gsin (π/4−θ)=4αr αsin (θ−φ)=ω^2 cos (θ−φ) 3gsin (π/4−θ)sin (θ−φ)=4ω^2 rcos (θ−φ) .(i) ωr(√2)sin (θ−φ)=Vcos φ ...(ii) (1/2)(((4mr^2 )/3))ω^2 +(1/2)MV^2 =mgr[1−cos (π/4−θ)] ....(iii) 3g(cos θ−sin θ)sin (θ−φ)=4(√2)ω^2 rcos (θ−φ) ⇒ ((2/3)+((Msin^2 (θ−φ))/(mcos^2 φ)))[((3(cos θ−sin θ)sin (θ−φ))/(4cos (θ−φ)))] = (√2)−cos θ−sin θ with (√2)sin θ=1−sin φ θ is obtained from above eq. Now V=((ωr(√2)sin (θ−φ))/(cos φ)) , hence V^( 2) ={((3(cos θ−sin θ)sin^3 (θ−φ))/( 2(√2)cos^2 φcos (θ−φ)))}gr](https://www.tinkutara.com/question/Q138263.png)
$${N}\mathrm{cos}\:\phi={MA} \\ $$$${mgr}\mathrm{sin}\:\left(\pi/\mathrm{4}−\theta\right)−{Nr}\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\theta−\phi\right) \\ $$$$=\left(\frac{\mathrm{4}{mr}^{\mathrm{2}} }{\mathrm{3}}\right)\alpha \\ $$$${when}\:{contact}\:{breaks} \\ $$$${N}=\mathrm{0},\:{A}=\mathrm{0}, \\ $$$$\mathrm{3}{g}\mathrm{sin}\:\left(\pi/\mathrm{4}−\theta\right)=\mathrm{4}\alpha{r}\:\: \\ $$$$\alpha\mathrm{sin}\:\left(\theta−\phi\right)=\omega^{\mathrm{2}} \mathrm{cos}\:\left(\theta−\phi\right) \\ $$$$\mathrm{3}{g}\mathrm{sin}\:\left(\pi/\mathrm{4}−\theta\right)\mathrm{sin}\:\left(\theta−\phi\right)=\mathrm{4}\omega^{\mathrm{2}} {r}\mathrm{cos}\:\left(\theta−\phi\right)\:\:.\left({i}\right) \\ $$$$\omega{r}\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\theta−\phi\right)={V}\mathrm{cos}\:\phi\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}{mr}^{\mathrm{2}} }{\mathrm{3}}\right)\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{MV}^{\mathrm{2}} \\ $$$$\:\:\:={mgr}\left[\mathrm{1}−\mathrm{cos}\:\left(\pi/\mathrm{4}−\theta\right)\right]\:\:\:\:\:….\left({iii}\right) \\ $$$$\mathrm{3}{g}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\mathrm{sin}\:\left(\theta−\phi\right)=\mathrm{4}\sqrt{\mathrm{2}}\omega^{\mathrm{2}} {r}\mathrm{cos}\:\left(\theta−\phi\right) \\ $$$$\Rightarrow \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}+\frac{{M}\mathrm{sin}\:^{\mathrm{2}} \left(\theta−\phi\right)}{{m}\mathrm{cos}\:^{\mathrm{2}} \phi}\right)\left[\frac{\mathrm{3}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{4cos}\:\left(\theta−\phi\right)}\right] \\ $$$$\:\:=\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$${with} \\ $$$$\sqrt{\mathrm{2}}\mathrm{sin}\:\theta=\mathrm{1}−\mathrm{sin}\:\phi \\ $$$$\theta\:{is}\:{obtained}\:{from}\:{above}\:{eq}. \\ $$$${Now}\:\:\:{V}=\frac{\omega{r}\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{cos}\:\phi}\:\:,\:\:{hence} \\ $$$$\:\:{V}^{\:\mathrm{2}} \:=\left\{\frac{\mathrm{3}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\mathrm{sin}\:^{\mathrm{3}} \left(\theta−\phi\right)}{\:\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}\:^{\mathrm{2}} \phi\mathrm{cos}\:\left(\theta−\phi\right)}\right\}{gr}\: \\ $$$$ \\ $$
Commented by mr W last updated on 12/Apr/21
$${thanks}\:{for}\:{reviewing}\:{sir}! \\ $$
Commented by ajfour last updated on 12/Apr/21
Commented by mr W last updated on 12/Apr/21
Commented by mr W last updated on 12/Apr/21
$${case}\:\mathrm{2}: \\ $$$${v}_{{x},\:{C}_{\mathrm{1}} } ={v}_{{x},\:{C}_{\mathrm{2}} } ={V} \\ $$$${case}\:\mathrm{1}: \\ $$$${v}_{{x},\:{C}_{\mathrm{1}} } \neq{v}_{{x},\:{C}_{\mathrm{2}} } ={V} \\ $$
Commented by mr W last updated on 12/Apr/21
$${the}\:{corner}\:{C}_{\mathrm{1}} \:{doesn}'{t}\:{have}\:{the}\:{same} \\ $$$${v}_{{x}} \:{as}\:{the}\:{point}\:{C}_{\mathrm{2}} \:{on}\:{the}\:{cylinder}! \\ $$$${both}\:{points}\:{on}\:{the}\:{contact}\:{must}\:{have} \\ $$$${the}\:{same}\:{velocity}\:{in}\:{x}'−{direction}, \\ $$$${not}\:{in}\:{x}−{direction},\:{i}\:{think}. \\ $$
Commented by mr W last updated on 12/Apr/21
Commented by mr W last updated on 12/Apr/21
$${V}_{{C}\mathrm{2}} \mathrm{cos}\:\phi={V}_{{C}\mathrm{1}} \mathrm{sin}\:\left(\theta−\phi\right) \\ $$$$\Rightarrow{V}\mathrm{cos}\:\phi=\omega\sqrt{\mathrm{2}}{r}\mathrm{sin}\:\left(\theta−\phi\right) \\ $$$$\Rightarrow{V}=\frac{\omega{r}\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{cos}\:\phi} \\ $$$$\mathrm{sin}\:\phi=\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\phi=\sqrt{\mathrm{2sin}\:\theta\left(\sqrt{\mathrm{2}}\:−\mathrm{sin}\:\theta\right)} \\ $$$$\mathrm{sin}\:\left(\theta−\phi\right)=\mathrm{sin}\:\theta\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\mathrm{sin}\:\phi \\ $$$${V}=\omega{r}\sqrt{\mathrm{2}}\left(\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{2}\:\mathrm{sin}\:\theta\:\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right) \\ $$$${V}=\omega{r}\left(\sqrt{\mathrm{2}}\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{sin}\:\theta\:\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\theta\right)}}\right) \\ $$$${this}\:{is}\:{the}\:{same}\:{as}\:{i}\:{had}\:{above}. \\ $$
Commented by ajfour last updated on 12/Apr/21
$${Yes}\:{sir},\:{you}\:{are}\:{right};\:{I}\:'{ve}\:{edited}. \\ $$