Question Number 226104 by fantastic2 last updated on 19/Nov/25
$${Q}\mathrm{226042} \\ $$
Answered by fantastic2 last updated on 19/Nov/25
Commented by fantastic2 last updated on 19/Nov/25
$${v}={v}_{{b}} +{v}_{{com}} \\ $$$${v}_{{b}} =\omega_{\mathrm{0}} {r}\Rightarrow\omega_{\mathrm{0}} =\frac{{v}_{{b}} }{{r}} \\ $$$${v}_{{com}} =\omega_{\mathrm{0}} {R} \\ $$$${v}={v}_{{b}} +{v}_{{b}} \left(\frac{{R}}{{r}}\right)\Rightarrow{v}_{{b}} =\frac{{vr}}{{R}+{r}} \\ $$$${v}_{{com}} =\frac{{v}_{{b}} {R}}{{r}}=\frac{{vR}}{{R}+{r}} \\ $$$${l}={distance}\:{fromB}\:{to}\:{center} \\ $$$${l}=\frac{{R}}{\mathrm{sin}\:\beta} \\ $$$${arc}\:{length}\:{l}_{{s}} =\frac{{R}}{\mathrm{sin}\:\beta}{d}\beta \\ $$$${l}_{{s}} ={dx}\mathrm{sin}\:\beta \\ $$$$\Rightarrow{Rd}\beta={dx}\mathrm{sin}\:^{\mathrm{2}} \beta \\ $$$$\Rightarrow\frac{{R}}{\mathrm{2}}{d}\alpha={dx}\mathrm{sin}\:^{\mathrm{2}} \beta \\ $$$$\Rightarrow{R}\frac{{d}\alpha}{{dt}}=\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\frac{{dx}}{{dt}} \\ $$$$\Rightarrow{R}\omega=\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right){v}_{{com}} \\ $$$$\Rightarrow\omega=\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)}{{R}+{r}}{v} \\ $$
Commented by fantastic2 last updated on 19/Nov/25
