Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 209352 by Spillover last updated on 07/Jul/24

Answered by mr W last updated on 08/Jul/24

$$L={mr}^2\omega =constant$$$${\omega }_{max}=\frac{L}{{mr}_{min}^2}=\frac{L}{m{(a-c)}^2}$$$${\omega }_{min}=\frac{L}{{mr}_{max}^2}=\frac{L}{m{(a+c)}^2}$$$$n=\frac{{\omega }_{max}}{{\omega }_{min}}={\left(\frac{a+c}{a-c}\right)}^2={\left(\frac{1+e}{1-e}\right)}^2$$$$\Rightarrow \frac{1+e}{1-e}=\sqrt{n}$$$$\Rightarrow e=\frac{\sqrt{n}-1}{\sqrt{n}+1}✓$$

Answered by Spillover last updated on 12/Jul/24

$$fromconservationofangularmomentum$$$$L={mr}^2\theta$$$${\theta }_{max}=\frac{L}{{mr}_{max}^2}{\theta }_{min}=\frac{L}{{mr}_{min}^2}$$$$where{r}_{min}=a(1-e)r_{max}=a(1+e)$$$$ratioofangularvelocity\left(n\right)$$$$n=\frac{{\theta }_{max}=\frac{L}{{mr}_{max}^2}}{{\theta }_{min}=\frac{L}{{mr}_{min}^2}}=\frac{{\theta }_{max}}{{\theta }_{min}}=\frac{r_{max}^2}{r_{min}^2}$$$$n={\left(\frac{1+e}{1-e}\right)}^2\sqrt{n}=\frac{1+e}{1-e}$$$$\sqrt{n}(1-e)=1+e$$$$e=\frac{\sqrt{n}-1}{\sqrt{n}+1}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com