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Question Number 144420 by ajfour last updated on 25/Jun/21
Commented by ajfour last updated on 25/Jun/21
Two cylinders one on top of another are released and motion initiated with a mild disturbance. If all surfaces are frictionless find the speeds u and v of the cylinders just when they loose contact with each other.
$${Two}\:{cylinders}\:{one}\:{on}\:{top}\:{of} \\ $$$${another}\:{are}\:{released}\:{and} \\ $$$${motion}\:{initiated}\:{with}\:{a}\:{mild} \\ $$$${disturbance}.\:{If}\:{all}\:{surfaces} \\ $$$${are}\:{frictionless}\:{find}\:{the} \\ $$$${speeds}\:{u}\:{and}\:{v}\:{of}\:{the}\:{cylinders} \\ $$$${just}\:{when}\:{they}\:{loose}\:{contact} \\ $$$${with}\:{each}\:{other}. \\ $$
Answered by ajfour last updated on 25/Jun/21
Commented by ajfour last updated on 25/Jun/21
energy conserva^n : ....(i) mg(2R)(1−cos θ) =(1/2)m{v^2 +(u_r cos θ−v)^2 +u_r ^2 sin^2 θ} momentum conserva^n : mv=m(u_r cos θ−v) ⇒ u_r cos θ=2v ...(ii) m(centripetal acc.)=F_r ((mu_r ^2 )/(2R))=mgcos θ ....(iii) ⇒ 2gR=(u_r ^2 /(cos θ)) ....(A) now using (ii) (v^2 /u_r ^2 )=((cos^2 θ)/4) .....(B) using all (A),(B) in ..(i) ((2(1−cos θ))/(cos θ))=((cos^2 θ)/2)+1−cos^2 θ ⇒ 4−4cos θ=−cos^3 θ+2cos θ ⇒ cos^3 θ−6cos θ+4=0 ⇒ cos θ=(√3)−1 now v^2 =((u_r ^2 cos^2 θ)/4)=((gRcos^3 θ)/2) v^2 =((gR((√3)−1)^3 )/2)=gR(3(√3)−5) v=(√(gR(3(√3)−5)))
$${energy}\:{conserva}^{{n}} :\:\:\:….\left({i}\right) \\ $$$${mg}\left(\mathrm{2}{R}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{m}\left\{{v}^{\mathrm{2}} +\left({u}_{{r}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +{u}_{{r}} ^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right\} \\ $$$${momentum}\:{conserva}^{{n}} :\: \\ $$$${mv}={m}\left({u}_{{r}} \mathrm{cos}\:\theta−{v}\right)\:\:\:\: \\ $$$$\Rightarrow\:\:\:{u}_{{r}} \mathrm{cos}\:\theta=\mathrm{2}{v}\:\:\:\:…\left({ii}\right) \\ $$$${m}\left({centripetal}\:{acc}.\right)={F}_{{r}} \\ $$$$\frac{{mu}_{{r}} ^{\mathrm{2}} }{\mathrm{2}{R}}={mg}\mathrm{cos}\:\theta\:\:\:\:\:….\left({iii}\right) \\ $$$$\Rightarrow\:\mathrm{2}{gR}=\frac{{u}_{{r}} ^{\mathrm{2}} }{\mathrm{cos}\:\theta}\:\:\:\:….\left({A}\right) \\ $$$${now}\:\:{using}\:\:\left({ii}\right) \\ $$$$\frac{{v}^{\mathrm{2}} }{{u}_{{r}} ^{\mathrm{2}} }=\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{4}}\:\:\:\:\:\:…..\left({B}\right) \\ $$$${using}\:{all}\:\left({A}\right),\left({B}\right)\:{in}\:..\left({i}\right) \\ $$$$\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta}=\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}}+\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:\mathrm{4}−\mathrm{4cos}\:\theta=−\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{2cos}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{cos}\:^{\mathrm{3}} \theta−\mathrm{6cos}\:\theta+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$${now}\:\:{v}^{\mathrm{2}} =\frac{{u}_{{r}} ^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{4}}=\frac{{gR}\mathrm{cos}\:^{\mathrm{3}} \theta}{\mathrm{2}} \\ $$$${v}^{\mathrm{2}} =\frac{{gR}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{2}}={gR}\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}\right) \\ $$$$\:{v}=\sqrt{{gR}\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}\right)} \\ $$$$ \\ $$
Answered by mr W last updated on 25/Jun/21
Commented by mr W last updated on 25/Jun/21
x_B =x_A −2R sin θ y_B =R+2R cos θ ω=(dθ/dt) v=(dx_A /dt) a=(dv/dt) u_(Bx) =(dx_B /dt)=v−2R cos θ ω=−v ⇒v=R cos θ ω u_(By) =(dy_B /dt)=−2R sin θ ω (1/2)m(v^2 +u_(Bx) ^2 +u_(By) ^2 )=mg2R(1−cos θ) v^2 +(v−2R cos θ ω)^2 +(−2R sin θ ω)^2 =4gR(1−cos θ) R^2 cos^2 θ ω^2 +R^2 cos^2 θ ω^2 +4R^2 sin^2 θ ω^2 =4gR(1−cos θ) (1+sin^2 θ)ω^2 =2g(1−cos θ) ⇒ω^2 =((2g(1−cos θ))/(R(1+sin^2 θ))) ⇒ω=(√((2g(1−cos θ))/(R(2−cos^2 θ)))) 2ω(dω/dθ)=((2g)/R)[((sin θ)/(1+sin^2 θ))−((2sin θ cos θ(1−cos θ))/((1+sin^2 θ)^2 ))] ω(dω/dθ)=((g sin θ(cos^2 θ−2 cos θ+2))/(R(1+sin^2 θ)^2 )) a=(dv/dt)=Rω(−sin θ ω+cos θ (dω/dθ)) ma=Nsin θ N=0 ⇒a=0 −sin θ ω+cos θ (dω/dθ)=0 tan θ ω^2 =ω (dω/dθ)=((g sin θ(cos^2 θ−2 cos θ+2))/(R(1+sin^2 θ)^2 )) tan θ ((2g(1−cos θ))/(R(1+sin^2 θ)))=((g sin θ(cos^2 θ−2 cos θ+2))/((1+sin^2 θ)^2 )) ((2(1−cos θ))/(cos θ))=((cos^2 θ−2 cos θ+2)/(1+sin^2 θ)) ((2−2cos θ)/(cos θ))=((cos^2 θ−2 cos θ+2)/(2−cos^2 θ)) cos^3 θ−6 cos θ+4=0 (cos θ−2)(cos^2 θ+2 cos θ−2)=0 ⇒cos θ=(√3)−1 ⇒θ=cos^(−1) ((√3)−1)≈42.9° v= (√((2gR(1−cos θ)cos^2 θ)/(2−cos^2 θ))) ⇒v=(2−(√3)) (√(((√3)+1)gR))≈0.443(√(gR))
$${x}_{{B}} ={x}_{{A}} −\mathrm{2}{R}\:\mathrm{sin}\:\theta \\ $$$${y}_{{B}} ={R}+\mathrm{2}{R}\:\mathrm{cos}\:\theta \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$${v}=\frac{{dx}_{{A}} }{{dt}} \\ $$$${a}=\frac{{dv}}{{dt}} \\ $$$${u}_{{Bx}} =\frac{{dx}_{{B}} }{{dt}}={v}−\mathrm{2}{R}\:\mathrm{cos}\:\theta\:\omega=−{v} \\ $$$$\Rightarrow{v}={R}\:\mathrm{cos}\:\theta\:\omega \\ $$$${u}_{{By}} =\frac{{dy}_{{B}} }{{dt}}=−\mathrm{2}{R}\:\mathrm{sin}\:\theta\:\omega \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{m}\left({v}^{\mathrm{2}} +{u}_{{Bx}} ^{\mathrm{2}} +{u}_{{By}} ^{\mathrm{2}} \right)={mg}\mathrm{2}{R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${v}^{\mathrm{2}} +\left({v}−\mathrm{2}{R}\:\mathrm{cos}\:\theta\:\omega\right)^{\mathrm{2}} +\left(−\mathrm{2}{R}\:\mathrm{sin}\:\theta\:\omega\right)^{\mathrm{2}} =\mathrm{4}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${R}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta\:\omega^{\mathrm{2}} +{R}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta\:\omega^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\:\omega^{\mathrm{2}} =\mathrm{4}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)\omega^{\mathrm{2}} =\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)} \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \:\theta\right)}} \\ $$$$\mathrm{2}\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{2}{g}}{{R}}\left[\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta}−\frac{\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }\right] \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=\frac{{g}\:\mathrm{sin}\:\theta\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\:\mathrm{cos}\:\theta+\mathrm{2}\right)}{{R}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$${a}=\frac{{dv}}{{dt}}={R}\omega\left(−\mathrm{sin}\:\theta\:\omega+\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}\right) \\ $$$$ \\ $$$${ma}={N}\mathrm{sin}\:\theta \\ $$$${N}=\mathrm{0}\:\Rightarrow{a}=\mathrm{0} \\ $$$$−\mathrm{sin}\:\theta\:\omega+\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}=\mathrm{0} \\ $$$$\mathrm{tan}\:\theta\:\omega^{\mathrm{2}} =\omega\:\frac{{d}\omega}{{d}\theta}=\frac{{g}\:\mathrm{sin}\:\theta\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\:\mathrm{cos}\:\theta+\mathrm{2}\right)}{{R}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\theta\:\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)}=\frac{{g}\:\mathrm{sin}\:\theta\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\:\mathrm{cos}\:\theta+\mathrm{2}\right)}{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta}=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\:\mathrm{cos}\:\theta+\mathrm{2}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\frac{\mathrm{2}−\mathrm{2cos}\:\theta}{\mathrm{cos}\:\theta}=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\:\mathrm{cos}\:\theta+\mathrm{2}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{6}\:\mathrm{cos}\:\theta+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{cos}\:\theta−\mathrm{2}\right)\left(\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{2}\:\mathrm{cos}\:\theta−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\approx\mathrm{42}.\mathrm{9}° \\ $$$${v}=\:\sqrt{\frac{\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \:\theta}} \\ $$$$\Rightarrow{v}=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:\sqrt{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){gR}}\approx\mathrm{0}.\mathrm{443}\sqrt{{gR}} \\ $$
Commented by ajfour last updated on 25/Jun/21
Thanks Sir, right answer!

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