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Question Number 186692 by ajfour last updated on 08/Feb/23

Commented by ajfour last updated on 08/Feb/23

$$\left(i\right)Findu,giventherest.$$$$\left(ii\right)Repeat\left(i\right)if{R}_1=a,R_2=b.$$

Answered by mr W last updated on 08/Feb/23

Commented by mr W last updated on 09/Feb/23

$$g_l=g\sin \theta$$$$g_r=g\cos \theta$$$$L=vt+\frac{g_l{t}^2}{2}$$$$t=\frac{-v+\sqrt{v^2+2g_{l}L}}{g_l}$$$$u_1^2=u^2-2g_{r}R(1-\cos \phi )$$$$N={mg}_r\cos \phi +\frac{{mu}_1^2}{R}={mg}_r\cos \phi +\frac{{mu}^2-2{mg}_{r}R(1-\cos \phi )}{R}$$$$N={mg}_r(3\cos \phi -2)+\frac{{mu}^2}{R}⩾0$$$$\frac{u^2}{R}⩾5g_r=5g\cos \theta \Rightarrow u⩾\sqrt{5gR\cos \theta }$$$$u_1=\frac{Rd\phi }{dt}=\sqrt{u^2-2g_{r}R(1-\cos \phi )}$$$$\sqrt{\frac{R}{2g_r}}\times \frac{d\phi }{\sqrt{\frac{u^2}{2g_{r}R}-1+\cos \phi }}=dt$$$$\sqrt{\frac{2R}{g_r}}{\int }_0^{\pi }\frac{d\phi }{\sqrt{\frac{u^2}{2g_{r}R}-1+\cos \phi }}=t$$$$\sqrt{\frac{2R}{g_r}}{\int }_0^{\pi }\frac{d\phi }{\sqrt{\frac{u^2}{2g_{r}R}-1+\cos \phi }}=\frac{-v+\sqrt{v^2+2g_{l}L}}{g_l}$$$$\Rightarrow \sqrt{\frac{2R}{g\cos \theta }}{\int }_0^{\pi }\frac{d\phi }{\sqrt{\frac{u^2}{2gR\cos \theta }-1+\cos \phi }}=\frac{-v+\sqrt{v^2+2gL\sin \theta }}{g\sin \theta }$$$$\Rightarrow \frac{4gR\sin \theta }{u}F(\frac{\pi }{2}\mid \frac{4gR\cos \theta }{u^2})=\sqrt{v^2+2gL\sin \theta }-v$$

Commented by ajfour last updated on 09/Feb/23

$$Iget$$$${\int }_0^{2\pi }\frac{\sqrt{R\tan \theta }}{\sqrt{(\frac{u_1^2}{2g_{R}R}-1)+\cos \phi }}d\phi$$$$={\int }_0^L\frac{dx}{\sqrt{x+\frac{v_0^2}{2g_L}}}$$$$butiwanttodealthisusing$$$$Torque-AngularMomentumways$$$$too!$$

Commented by ajfour last updated on 09/Feb/23

Thank you Sir. What is this F(theta) Fourier Integral?

Commented by mr W last updated on 10/Feb/23

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