Question Number 131071 by mr W last updated on 01/Feb/21
Commented by mr W last updated on 01/Feb/21
$${a}\:{small}\:{ball}\:{of}\:{mass}\:{m}\:{is}\:{fixed}\:{on} \\ $$$${the}\:{inner}\:{wall}\:{of}\:{a}\:{hollow}\:{cylinder} \\ $$$${of}\:{radius}\:{R}\:{and}\:{mass}\:{M}. \\ $$$${the}\:{cylinder}\:{is}\:{released}\:{from}\:{rest} \\ $$$${at}\:{the}\:{position}\:{as}\:{shown}\:{and}\:{rolls} \\ $$$${on}\:{the}\:{ground}. \\ $$$${find}\:{the}\:{contact}\:{force}\:{N}\:{in}\:{terms}\:{of} \\ $$$${x}. \\ $$
Answered by ajfour last updated on 01/Feb/21
Commented by ajfour last updated on 01/Feb/21
![mgR(1−cos θ)=Mu^2 +K K=(1/2)m[(ωRcos θ+u)^2 +(ωRsin θ)^2 ] N−(M+m)g=m(d^2 y/dt^2 ) y=R+Rcos θ (dy/dt)=−ωRsin θ (d^2 y/dt^2 )=−ω^2 Rcos θ−αRsin θ θ=(x/R) ωR=u mgR(1−cos θ)=Mu^2 +K K=mu^2 (1+cos θ) u^2 =((mgR(1−cos θ))/(M+m(1+cos θ)))=ω^2 R^2 N=(M+m)g+m(d^2 y/dt^2 ) (d^2 y/dt^2 )=−ω^2 Rcos θ−αRsin θ u^2 =((mgR(1−cos θ))/(M+m(1+cos θ)))=ω^2 R^2 α=((ωdω)/dθ) ; θ=(x/R)](https://www.tinkutara.com/question/Q131118.png)
$${mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)={Mu}^{\mathrm{2}} +{K} \\ $$$$\:\:\:{K}=\frac{\mathrm{1}}{\mathrm{2}}{m}\left[\left(\omega{R}\mathrm{cos}\:\theta+{u}\right)^{\mathrm{2}} +\left(\omega{R}\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right] \\ $$$${N}−\left({M}+{m}\right){g}={m}\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} } \\ $$$${y}={R}+{R}\mathrm{cos}\:\theta \\ $$$$\frac{{dy}}{{dt}}=−\omega{R}\mathrm{sin}\:\theta \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }=−\omega^{\mathrm{2}} {R}\mathrm{cos}\:\theta−\alpha{R}\mathrm{sin}\:\theta \\ $$$$\theta=\frac{{x}}{{R}} \\ $$$$\omega{R}={u} \\ $$$${mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)={Mu}^{\mathrm{2}} +{K} \\ $$$${K}={mu}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${u}^{\mathrm{2}} =\frac{{mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{M}+{m}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}=\omega^{\mathrm{2}} {R}^{\mathrm{2}} \\ $$$${N}=\left({M}+{m}\right){g}+{m}\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }=−\omega^{\mathrm{2}} {R}\mathrm{cos}\:\theta−\alpha{R}\mathrm{sin}\:\theta \\ $$$${u}^{\mathrm{2}} =\frac{{mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{M}+{m}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}=\omega^{\mathrm{2}} {R}^{\mathrm{2}} \\ $$$$\alpha=\frac{\omega{d}\omega}{{d}\theta}\:\:\:\:;\:\:\:\theta=\frac{{x}}{{R}} \\ $$$$ \\ $$
Commented by ajfour last updated on 01/Feb/21
$$\frac{\mathrm{1}}{\mathrm{2}}{Mu}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({MR}^{\mathrm{2}} \right)\omega^{\mathrm{2}} ={Mu}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 01/Feb/21
$${yes}.\:{i}\:{see}. \\ $$
Answered by mr W last updated on 01/Feb/21
Commented by mr W last updated on 02/Feb/21
![x=Rθ ω=(dθ/dt) α=(dω/dt) OS=f=(m/(M+m))×R=(R/((M/m)+1)) let λ=(M/m)+1 ⇒f=(R/λ) I_B =MR^2 +MR^2 +m(2R cos (θ/2))^2 =2MR^2 +2mR^2 (cos θ+1) =2R^2 [M+m(1+cos θ)] (1/2)I_B ω^2 =mgR(1−cos θ) R^2 [M+m(1+cos θ)]ω^2 =mgR(1−cos θ) ω^2 =((g(1−cos θ))/(R((M/m)+1+cos θ)))=((g(1−cos θ))/(R(λ+cos θ))) ⇒ω=(√(g/R))(√((1−cos θ)/(λ+cos θ))) (dθ/dt)=(√(g/R))(√((1−cos θ)/(λ+cos θ))) ∫(√((λ+cos θ)/(1−cos θ)))dθ=(√(g/R))∫dt ⇒t=(√(R/g))∫_0 ^θ (√((λ+cos θ)/(1−cos θ)))dθ period T T=2(√(R/g))∫_0 ^π (√((λ+cos θ)/(1−cos θ)))dθ (d/dt)(I_B ω)=(M+m)gfsin θ I_B α+ω^2 (dI_B /dθ)=((gR)/λ)(M+m)sin θ 2R^2 [M+m(1+cos θ)]α−2ω^2 R^2 msin θ=((gR)/λ)(M+m)sin θ (λ+1+cos θ)α=((g/(2R))+ω^2 )sin θ α=(g/(2R))×(((λ+2−cos θ)sin θ)/((λ+cos θ)(λ+1+cos θ))) v_(y,S) =ωfsin θ=(R/λ)ω sin θ a_(y,S) =(dv_y /dt)=(R/λ)(α sin θ+ω^2 cos θ) a_(y,S) =((g(1−cos θ)[cos^2 θ+3(λ+1)cos θ+λ+2])/(2λ(λ+cos θ)(λ+1+cos θ))) (M+m)g−N=(M+m)a_(y,S) N=(M+m)(g−a_(y,S) ) ⇒(N/((M+m)g))=1−(((1−cos θ)[cos^2 θ+3(λ+1)cos θ+λ+2])/(2λ(λ+cos θ)(λ+1+cos θ))) (N_(max) /((M+m)g))=1+(2/(λ(λ−1)))](https://www.tinkutara.com/question/Q131129.png)
$${x}={R}\theta \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$$\alpha=\frac{{d}\omega}{{dt}} \\ $$$${OS}={f}=\frac{{m}}{{M}+{m}}×{R}=\frac{{R}}{\frac{{M}}{{m}}+\mathrm{1}} \\ $$$${let}\:\lambda=\frac{{M}}{{m}}+\mathrm{1} \\ $$$$\Rightarrow{f}=\frac{{R}}{\lambda} \\ $$$${I}_{{B}} ={MR}^{\mathrm{2}} +{MR}^{\mathrm{2}} +{m}\left(\mathrm{2}{R}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}{MR}^{\mathrm{2}} +\mathrm{2}{mR}^{\mathrm{2}} \left(\mathrm{cos}\:\theta+\mathrm{1}\right) \\ $$$$=\mathrm{2}{R}^{\mathrm{2}} \left[{M}+{m}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{B}} \omega^{\mathrm{2}} ={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${R}^{\mathrm{2}} \left[{M}+{m}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\right]\omega^{\mathrm{2}} ={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\omega^{\mathrm{2}} =\frac{{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\frac{{M}}{{m}}+\mathrm{1}+\mathrm{cos}\:\theta\right)}=\frac{{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\lambda+\mathrm{cos}\:\theta\right)} \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}}{{R}}}\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\lambda+\mathrm{cos}\:\theta}} \\ $$$$\frac{{d}\theta}{{dt}}=\sqrt{\frac{{g}}{{R}}}\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\lambda+\mathrm{cos}\:\theta}} \\ $$$$\int\sqrt{\frac{\lambda+\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}}{d}\theta=\sqrt{\frac{{g}}{{R}}}\int{dt} \\ $$$$\Rightarrow{t}=\sqrt{\frac{{R}}{{g}}}\int_{\mathrm{0}} ^{\theta} \sqrt{\frac{\lambda+\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}}{d}\theta \\ $$$${period}\:{T} \\ $$$${T}=\mathrm{2}\sqrt{\frac{{R}}{{g}}}\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\lambda+\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}}{d}\theta \\ $$$$ \\ $$$$\frac{{d}}{{dt}}\left({I}_{{B}} \omega\right)=\left({M}+{m}\right){gf}\mathrm{sin}\:\theta \\ $$$${I}_{{B}} \alpha+\omega^{\mathrm{2}} \frac{{dI}_{{B}} }{{d}\theta}=\frac{{gR}}{\lambda}\left({M}+{m}\right)\mathrm{sin}\:\theta \\ $$$$\mathrm{2}{R}^{\mathrm{2}} \left[{M}+{m}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\right]\alpha−\mathrm{2}\omega^{\mathrm{2}} {R}^{\mathrm{2}} {m}\mathrm{sin}\:\theta=\frac{{gR}}{\lambda}\left({M}+{m}\right)\mathrm{sin}\:\theta \\ $$$$\left(\lambda+\mathrm{1}+\mathrm{cos}\:\theta\right)\alpha=\left(\frac{{g}}{\mathrm{2}{R}}+\omega^{\mathrm{2}} \right)\mathrm{sin}\:\theta \\ $$$$\alpha=\frac{{g}}{\mathrm{2}{R}}×\frac{\left(\lambda+\mathrm{2}−\mathrm{cos}\:\theta\right)\mathrm{sin}\:\theta}{\left(\lambda+\mathrm{cos}\:\theta\right)\left(\lambda+\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$ \\ $$$${v}_{{y},{S}} =\omega{f}\mathrm{sin}\:\theta=\frac{{R}}{\lambda}\omega\:\mathrm{sin}\:\theta \\ $$$${a}_{{y},{S}} =\frac{{dv}_{{y}} }{{dt}}=\frac{{R}}{\lambda}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \mathrm{cos}\:\theta\right) \\ $$$${a}_{{y},{S}} =\frac{{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left[\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{3}\left(\lambda+\mathrm{1}\right)\mathrm{cos}\:\theta+\lambda+\mathrm{2}\right]}{\mathrm{2}\lambda\left(\lambda+\mathrm{cos}\:\theta\right)\left(\lambda+\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\left({M}+{m}\right){g}−{N}=\left({M}+{m}\right){a}_{{y},{S}} \\ $$$${N}=\left({M}+{m}\right)\left({g}−{a}_{{y},{S}} \right) \\ $$$$\Rightarrow\frac{{N}}{\left({M}+{m}\right){g}}=\mathrm{1}−\frac{\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left[\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{3}\left(\lambda+\mathrm{1}\right)\mathrm{cos}\:\theta+\lambda+\mathrm{2}\right]}{\mathrm{2}\lambda\left(\lambda+\mathrm{cos}\:\theta\right)\left(\lambda+\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\frac{{N}_{{max}} }{\left({M}+{m}\right){g}}=\mathrm{1}+\frac{\mathrm{2}}{\lambda\left(\lambda−\mathrm{1}\right)} \\ $$
Commented by mr W last updated on 02/Feb/21
