Question Number 226278 by mr W last updated on 24/Nov/25
Answered by mr W last updated on 24/Nov/25
Commented by mr W last updated on 24/Nov/25
![d=2R cos (θ/2) let V=Rω_1 in tangential direction: u cos θ=u_1 sin θ−ω_1 d cos (θ/2) u cos θ=u_1 sin θ−2V cos^2 (θ/2) ...(i) in normal direction: eu sin θ=u_1 cos θ+ω_1 d sin (θ/2) eu sin θ=u_1 cos θ+V sin θ ...(ii) I_A =((MR^2 )/2)+MR^2 =((3MR^2 )/2) conservation of angular momentum about point A: I_A ω_1 +mu_1 R sin θ=muR(1+cos θ) ((3MV)/(2m))+u_1 sin θ=u(1+cos θ) ...(iii) from (i): u_1 sin θ=u cos θ+2V cos^2 (θ/2) from (ii): u_1 cos θ=(eu−V )sin θ (u cos θ+2V cos^2 (θ/2))cos θ=(eu−V )sin^2 θ (1+cos θ)V=u[(e+1)sin^2 θ−1] V=((u[(e+1)sin^2 θ−1])/(1+cos θ)) let μ=((3M)/(2m)) from (iii): μV+u cos θ+2V cos^2 (θ/2)=u(1+cos θ) V=(u/(μ+1+cos θ)) ((u[(e+1)sin^2 θ−1])/(1+cos θ))=(u/(μ+1+cos θ)) (e+1) sin^2 θ+(μ/(μ+1+cos θ))=2 we can solve this eqn. to get θ. V=Rω_1 =(u/(μ+1+cos θ)) ⇒ω_1 =(u/(R(μ+1+cos θ))) u_1 =(eu−V )tan θ ⇒u_1 =u(e−(1/(μ+1+cos θ)) )tan θ example: M=3m ⇒μ=((3M)/(2m))=4.5 e=0.5 1.5 sin^2 θ+((4.5)/(5.5+cos θ))=2 ⇒θ≈65.369° ⇒ω_1 ≈(u/(5.9168R)) ⇒u_1 ≈0.7219u](Q226281.png)
$${d}=\mathrm{2}{R}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$${let}\:{V}={R}\omega_{\mathrm{1}} \\ $$$${in}\:{tangential}\:{direction}: \\ $$$${u}\:\mathrm{cos}\:\theta={u}_{\mathrm{1}} \mathrm{sin}\:\theta−\omega_{\mathrm{1}} {d}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$${u}\:\mathrm{cos}\:\theta={u}_{\mathrm{1}} \mathrm{sin}\:\theta−\mathrm{2}{V}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\:\:\:...\left({i}\right) \\ $$$${in}\:{normal}\:{direction}: \\ $$$${eu}\:\mathrm{sin}\:\theta={u}_{\mathrm{1}} \mathrm{cos}\:\theta+\omega_{\mathrm{1}} {d}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$${eu}\:\mathrm{sin}\:\theta={u}_{\mathrm{1}} \mathrm{cos}\:\theta+{V}\:\mathrm{sin}\:\theta\:\:\:...\left({ii}\right) \\ $$$${I}_{{A}} =\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}+{MR}^{\mathrm{2}} =\frac{\mathrm{3}{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${conservation}\:{of}\:{angular}\:{momentum} \\ $$$${about}\:{point}\:{A}: \\ $$$${I}_{{A}} \omega_{\mathrm{1}} +{mu}_{\mathrm{1}} {R}\:\mathrm{sin}\:\theta={muR}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$\frac{\mathrm{3}{MV}}{\mathrm{2}{m}}+{u}_{\mathrm{1}} \:\mathrm{sin}\:\theta={u}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\:\:\:...\left({iii}\right) \\ $$$${from}\:\left({i}\right): \\ $$$${u}_{\mathrm{1}} \mathrm{sin}\:\theta={u}\:\mathrm{cos}\:\theta+\mathrm{2}{V}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$${from}\:\left({ii}\right): \\ $$$${u}_{\mathrm{1}} \:\mathrm{cos}\:\theta=\left({eu}−{V}\:\right)\mathrm{sin}\:\theta \\ $$$$\left({u}\:\mathrm{cos}\:\theta+\mathrm{2}{V}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\:\theta=\left({eu}−{V}\:\right)\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\left(\mathrm{1}+\mathrm{cos}\:\theta\right){V}={u}\left[\left({e}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \:\theta−\mathrm{1}\right] \\ $$$${V}=\frac{{u}\left[\left({e}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \:\theta−\mathrm{1}\right]}{\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$${let}\:\mu=\frac{\mathrm{3}{M}}{\mathrm{2}{m}} \\ $$$${from}\:\left({iii}\right): \\ $$$$\mu{V}+{u}\:\mathrm{cos}\:\theta+\mathrm{2}{V}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}={u}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${V}=\frac{{u}}{\mu+\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\frac{{u}\left[\left({e}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \:\theta−\mathrm{1}\right]}{\mathrm{1}+\mathrm{cos}\:\theta}=\frac{{u}}{\mu+\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\left({e}+\mathrm{1}\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta+\frac{\mu}{\mu+\mathrm{1}+\mathrm{cos}\:\theta}=\mathrm{2} \\ $$$${we}\:{can}\:{solve}\:{this}\:{eqn}.\:{to}\:{get}\:\theta. \\ $$$${V}={R}\omega_{\mathrm{1}} =\frac{{u}}{\mu+\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\omega_{\mathrm{1}} =\frac{{u}}{{R}\left(\mu+\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$${u}_{\mathrm{1}} =\left({eu}−{V}\:\right)\mathrm{tan}\:\theta \\ $$$$\Rightarrow{u}_{\mathrm{1}} ={u}\left({e}−\frac{\mathrm{1}}{\mu+\mathrm{1}+\mathrm{cos}\:\theta}\:\right)\mathrm{tan}\:\theta \\ $$$$ \\ $$$${example}: \\ $$$${M}=\mathrm{3}{m}\:\Rightarrow\mu=\frac{\mathrm{3}{M}}{\mathrm{2}{m}}=\mathrm{4}.\mathrm{5} \\ $$$${e}=\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{1}.\mathrm{5}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\frac{\mathrm{4}.\mathrm{5}}{\mathrm{5}.\mathrm{5}+\mathrm{cos}\:\theta}=\mathrm{2} \\ $$$$\Rightarrow\theta\approx\mathrm{65}.\mathrm{369}° \\ $$$$\Rightarrow\omega_{\mathrm{1}} \approx\frac{{u}}{\mathrm{5}.\mathrm{9168}{R}} \\ $$$$\Rightarrow{u}_{\mathrm{1}} \approx\mathrm{0}.\mathrm{7219}{u} \\ $$