Menu Close

Question-144420




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!

Leave a Reply

Your email address will not be published. Required fields are marked *