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Question Number 133730 by mr W last updated on 24/Feb/21

Commented by mr W last updated on 23/Feb/21

$$asQ133275,butwithcoefficientof$$$$restitution\mathit{e}fortheimpacts.$$

Commented by Abdoulaye last updated on 23/Feb/21

$$howtotakeithere?$$

Answered by mr W last updated on 23/Feb/21

Commented by mr W last updated on 26/Feb/21

$$let\eta =\frac{h}{b},\xi =\frac{a}{b},\lambda =\frac{p}{b},\mu =\frac{q}{b}$$$$equationofparabola:$$$$y=h-\frac{{hx}^2}{b^2}$$$$\frac{dy}{dx}=-\frac{2hx}{b^2}$$$$sayB(p,y_B)with{y}_B=h-\frac{{hp}^2}{b^2}$$$$sayC(q,y_B)with{y}_C=h-\frac{{hq}^2}{b^2}$$$$\tan \theta =\frac{2hp}{b^2}=2\eta \lambda$$$$\tan \phi =-\frac{2hq}{b^2}=-2\eta \mu$$$$\tan \alpha =\frac{y_B}{p-a}=\frac{h-\frac{{hp}^2}{b^2}}{p-a}=\frac{\eta (1-{\lambda }^2)}{\lambda -\xi }$$$$\tan \beta =\frac{y_C}{a-q}=\frac{h-\frac{{hq}^2}{b^2}}{a-q}=\frac{\eta (1-{\mu }^2)}{\xi -\mu }$$$$\tan \varphi =\frac{y_C-y_B}{p-q}=\frac{-\frac{{hq}^2}{b^2}+\frac{{hp}^2}{b^2}}{p-q}=\eta (\lambda +\mu )$$$$\gamma =\pi -\alpha -\theta$$$$\delta =\pi -\beta -\phi$$$$$$$$\sigma =\theta -\varphi$$$$\tan \sigma =e\tan \gamma$$$$\tan (\theta -\varphi )=-e\tan (\theta +\alpha )$$$$\frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }=-e\frac{\tan \theta +\tan \alpha }{1-\tan \theta \tan \alpha }$$$$\frac{2\eta \lambda -\eta (\lambda +\mu )}{1+2\eta \lambda \eta (\lambda +\mu )}=-e\frac{2\eta \lambda +\frac{\eta (1-{\lambda }^2)}{\lambda -\xi }}{1-2\eta \lambda \frac{\eta (1-{\lambda }^2)}{\lambda -\xi }}$$$$\frac{\lambda -\mu }{1+2{\eta }^2\lambda (\lambda +\mu )}=-e\frac{{\lambda }^2-2\xi \lambda +1}{\lambda -\xi -2{\eta }^2\lambda (1-{\lambda }^2)}$$$$\Rightarrow \frac{\lambda -\mu }{1+2{\eta }^2\lambda (\lambda +\mu )}+e\times \frac{{\lambda }^2-2\xi \lambda +1}{\lambda -\xi -2{\eta }^2\lambda (1-{\lambda }^2)}=0...\left(i\right)$$$$$$$$\rho =\phi +\varphi$$$$e\tan \rho =\tan \delta$$$$\tan (\phi +\varphi )=-\frac{1}{e}\tan (\phi +\beta )$$$$\frac{\tan \phi +\tan \varphi }{1-\tan \phi \tan \varphi }=-\frac{1}{e}\frac{\tan \phi +\tan \beta }{1-\tan \phi \tan \beta }$$$$\frac{-2\eta \mu +\eta (\lambda +\mu )}{1+2\eta \mu \eta (\lambda +\mu )}=-\frac{1}{e}\frac{-2\eta \mu +\frac{\eta (1-{\mu }^2)}{\xi -\mu }}{1+2\eta \mu \frac{\eta (1-{\mu }^2)}{\xi -\mu }}$$$$\Rightarrow \frac{\lambda -\mu }{1+2{\eta }^2\mu (\lambda +\mu )}-\frac{1}{e}\times \frac{{\mu }^2-2\xi \mu +1}{\mu -\xi -2{\eta }^2\mu (1-{\mu }^2)}=0...\left(ii\right)$$$$$$$$from\left(i\right)and\left(ii\right)weget\lambda and\mu .$$$$$$$$seeexamples.$$$$$$$$suchthattheballafterimpactatA$$$$continuestomovetoB,Cetc.$$$$\tan \alpha =e\tan \beta$$$$\frac{\eta (1-{\lambda }^2)}{\lambda -\xi }=e\frac{\eta (1-{\mu }^2)}{\xi -\mu }$$$$(1-{\lambda }^2)(\xi -\mu )=e(1-{\mu }^2)(\lambda -\xi )$$$$\xi =\frac{(1-{\lambda }^2)\mu +e(1-{\mu }^2)\lambda }{(1-{\lambda }^2)+e(1-{\mu }^2)}...\left(iii\right)$$

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by mr W last updated on 24/Feb/21

Commented by Abdoulaye last updated on 24/Feb/21

$$$$$$\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{do}\mathrm{these}\mathrm{kinds}\mathrm{of}\mathrm{figures}$$$$\mathrm{onthis}\mathrm{app}?$$

Commented by mr W last updated on 24/Feb/21

$$imakethegraphsusingtheapp$$$$Grapher.$$

Commented by Abdoulaye last updated on 24/Feb/21

$$okaythankyou!$$

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