there-is-a-cylinder-on-the-horizontal-plane-and-there-is-a-little-ball-with-mass-m-at-the-center-of-a-circle-on-the-vertical-plane-and-there-is-a-little-hole-in-the-wall-of-the-cylinder-on-the-lift-of

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Question Number 196221 by liuxinnan last updated on 20/Aug/23

$$thereisacylinderonthehorizontalplane$$$$andthereisalittleballwithmassm$$$$atthecenterofacircleontheverticalplane$$$$andthereisalittleholeinthewallof$$$$thecylinderontheliftoftheballwhich$$$$justenoughfortheballtopassthrough$$$$thegravitationalaccelerationisg$$$$thecollisionbetweentheballand$$$$thecylinderwallcanberegardedas$$$$anelasticcoollision$$$$findallofthequondamspeed$$

Commented by liuxinnan last updated on 20/Aug/23

Commented by mr W last updated on 20/Aug/23

$$therearemanypossibilities.$$$$doyoumeanthattheballshouldfly$$$$throughtheholejustafteronetime$$$$collisionwiththecylinderwall?$$$$theballcanalsodirectlyflythrough$$$$theholewithoutcollisionorafter$$$$twotimescollisionetc.$$

Commented by liuxinnan last updated on 20/Aug/23

$$zeroonetwoandntimesareallok$$$$maybehavev=f\left(n\right)$$

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

Commented by mr W last updated on 23/Aug/23

$${let}^{\prime }sconsiderthecasethatthe$$$$ballflysthroughtheholeaftera$$$$singlecollisionwiththecylinder$$$$wall.$$

Commented by mr W last updated on 22/Aug/23

Commented by mr W last updated on 23/Aug/23

$$fromOtoA:$$$$t=\frac{R\cos \phi }{u\cos \theta }$$$$-R\sin \phi =u\sin \theta \times \frac{R\cos \phi }{u\cos \theta }-\frac{g}{2}\times {\left(\frac{R\cos \phi }{u\cos \theta }\right)}^2$$$$-\tan \phi =\tan \theta -\frac{gR\cos \phi (1+{\tan }^2\theta )}{2u^2}$$$$let\lambda =\frac{gR}{2u^2}$$$$\tan \phi +\tan \theta =\lambda \cos \phi (1+{\tan }^2\theta )$$$$\lambda \cos \phi {\tan }^2\theta -\tan \theta -\tan \phi +\lambda \cos \phi =0$$$$\Rightarrow \tan \theta =\frac{1+\sqrt{1+4\lambda (\sin \phi -\lambda {\cos }^2\phi )}}{2\lambda \cos \phi }$$$$atA:$$$$U_x=u\cos \theta$$$$U_y=u\sin \theta -g\times \frac{R\cos \phi }{u\cos \theta }$$$$=u(\sin \theta -\frac{2\lambda \cos \phi }{\cos \theta })$$$$=u\cos \theta [\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]$$$$V_x=[(1+e){\sin }^2\phi -1]u\cos \theta -(1+e)\sin \phi \cos \phi u(\sin \theta -\frac{2\lambda \cos \phi }{\cos \theta })$$$$=u\cos \theta \{(1+e){\sin }^2\phi -1-(1+e)\sin \phi \cos \phi [\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]\}$$$$V_y=(1+e)\sin \phi \cos \phi u\cos \theta -[(1+e){\cos }^2\phi -1]u\cos \theta [\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]$$$$=u\cos \theta \{(1+e)\sin \phi \cos \phi -[(1+e){\cos }^2\phi -1][\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]\}$$$$fromAtoB:$$$$t=\frac{\xi R}{u}$$$$\xi =\frac{(1+\cos \phi )}{\cos \theta \{(1+e){\sin }^2\phi -1-(1+e)\sin \phi \cos \phi [\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]\}}$$$$R\sin \phi =u\cos \theta \{(1+e)\sin \phi \cos \phi -(e{\cos }^2\phi -{\sin }^2\phi )[\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]\}t-\frac{{gt}^2}{2}$$$$\sin \phi =\cos \theta \{(1+e)\sin \phi \cos \phi -[(1+e){\cos }^2\phi -1][\tan \theta -2\lambda \cos \phi (1+{\tan }^2\theta )]\}\xi -\lambda {\xi }^2$$$$foragivenewefinda{\lambda }_{max}suchthat$$$$thisequationhasoneandonlyone$$$$rootfor0<\phi <\frac{\pi }{2}.$$$$u_{min}=\sqrt{\frac{gR}{2{\lambda }_{max}}}$$$$fore=1:u_{min}\approx \frac{\sqrt{gR}}{1.274}$$

Commented by mr W last updated on 22/Aug/23

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