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Question Number 215312 by ajfour last updated on 02/Jan/25

Commented by mr W last updated on 06/Jan/25

$$pleasetryQ215395$$

Answered by mr W last updated on 03/Jan/25

Commented by mr W last updated on 03/Jan/25

$$I=\frac{{ML}^2}{12}$$$$b(\tan \theta +\tan \varphi )=\frac{3L}{2}$$$$\Rightarrow \tan \varphi =\frac{3L}{2b}-\tan \theta$$$$mu\sin \theta =mU\sin \varphi$$$$\Rightarrow U=\frac{u\sin \theta }{\sin \varphi }$$$$mu\cos \theta -J=mU\cos \varphi$$$$\Rightarrow J=mu(\cos \theta -\frac{\sin \theta }{\tan \varphi })$$$$MV=J$$$$\Rightarrow V=\frac{mu}{M(\cos \theta -\frac{\sin \theta }{\tan \varphi })}$$$$\frac{{ML}^2\omega }{12}=J(\frac{L}{2}-b\tan \theta )$$$$\Rightarrow \omega =\frac{12mu(\frac{1}{2}-\frac{b\tan \theta }{L})}{ML(\cos \theta -\frac{\sin \theta }{\tan \varphi })}$$$$\frac{{mU}^2}{2}+\frac{{MV}^2}{2}+\frac{1}{2}\times \frac{{ML}^2{\omega }^2}{12}=\frac{{mu}^2}{2}$$$$\frac{{\sin }^2\theta }{{\sin }^2\varphi }+\frac{m}{M{(\cos \theta -\frac{\sin \theta }{\tan \varphi })}^2}+\frac{12m{(\frac{1}{2}-\frac{b\tan \theta }{L})}^2}{M{(\cos \theta -\frac{\sin \theta }{\tan \varphi })}^2}=1$$$$\frac{\mu [1+12{(\frac{1}{2}-\frac{b\tan \theta }{L})}^2]}{{(\cos \theta -\frac{\sin \theta }{\tan \varphi })}^2}=1-\frac{{\sin }^2\theta }{{\sin }^2\varphi }$$$$with\mu =\frac{m}{M},\xi =\frac{b}{L}$$$$\Rightarrow \frac{\mu [1+12{(\frac{1}{2}-\xi \tan \theta )}^2]{(1+{\tan }^2\theta )}^2}{{(1-\frac{\tan \theta }{\tan \varphi })}^2}+{\left(\frac{\tan \theta }{\tan \varphi }\right)}^2=1$$$$with\tan \varphi =\frac{3}{2\xi }-\tan \theta$$$$==================$$$${\theta }_1=\omega t=\frac{12\mu (\frac{1}{2}-\xi \tan \theta )ut}{L(\cos \theta -\frac{\sin \theta }{\tan \varphi })}$$$$x_1=\frac{L}{2}$$$$y_1=b+Vt=\xi L+\frac{\mu ut}{\cos \theta -\frac{\sin \theta }{\tan \varphi }}$$$$x_2=b\tan \theta +Ut\sin \varphi =\xi L\tan \theta +ut\sin \theta$$$$y_2=b-Ut\cos \varphi =\xi L-\frac{ut\sin \theta }{\tan \varphi }$$

Commented by mr W last updated on 03/Jan/25

Commented by mr W last updated on 03/Jan/25

Commented by mr W last updated on 03/Jan/25

Commented by mr W last updated on 03/Jan/25

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