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 165296 by ajfour last updated on 28/Jan/22

Commented by ajfour last updated on 28/Jan/22

$$Ifwhenthediscrollsdownandis$$$$abouttotohitthepivot,therod$$$$getshorizontal,findinitialangle$$$$\theta atwhichsystemisreleased.$$$$\left(assumepurerollingthroughout\right)$$

Answered by TheSupreme last updated on 29/Jan/22

$$lockingx$$$$\mathrm{\Delta }\theta =\mathrm{\Delta }\phi$$$$locking\theta$$$$x=L-\mathrm{\Delta }\phi R$$$$x_F=0$$$$\mathrm{\Delta }{\phi }_F=\frac{L}{R}=\mathrm{\Delta }\theta$$$${\theta }_0={\theta }_F-\mathrm{\Delta }\theta =\frac{L}{R}$$

Answered by mr W last updated on 30/Jan/22

Commented by mr W last updated on 29/Jan/22

$$thecommoncenterofmassofboth$$$$objects\left(green\right)remainsunchanged$$$$inhorizontaldirection.$$$$M\times \frac{L}{2}=M\times \frac{L}{2}\times \cos \theta +m\times (L\cos \theta -R\sin \theta )$$$$1=\cos \theta +\frac{2m}{M}\times (\cos \theta -\frac{R}{L}\sin \theta )$$$$let\xi =1+\frac{2m}{M},\lambda =\frac{2mR}{ML}$$$$1=\xi \cos \theta -\lambda \sin \theta$$$$\frac{1}{\sqrt{{\xi }^2+{\lambda }^2}}=\frac{\xi }{\sqrt{{\xi }^2+{\lambda }^2}}\times \cos \theta -\frac{\lambda }{\sqrt{{\xi }^2+{\lambda }^2}}\times \sin \theta$$$$\frac{1}{\sqrt{{\xi }^2+{\lambda }^2}}=\cos (\theta +\varphi )$$$$\theta ={\cos }^{-1}\frac{1}{\sqrt{{\xi }^2+{\lambda }^2}}-{\tan }^{-1}\frac{\lambda }{\xi }$$$$\Rightarrow \theta ={\cos }^{-1}\frac{1}{\sqrt{1+\frac{4m}{M}+(1+\frac{R^2}{L^2})\frac{4m^2}{M^2}}}-{\tan }^{-1}\frac{R}{(1+\frac{M}{2m})L}$$$$example:M=m,L=5R$$$$\theta ={\cos }^{-1}\frac{5}{\sqrt{229}}-{\tan }^{-1}\frac{2}{15}\approx 63.11°$$

Commented by Tawa11 last updated on 29/Jan/22

$$\mathrm{Weldone}\mathrm{sir}.$$

Commented by ajfour last updated on 29/Jan/22

$$Reallygreatlydone,finenotice$$$$Sir.Thanksimmensely!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com