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Question Number 136109 by physicstutes last updated on 18/Mar/21

$$\mathrm{A}\mathrm{particle}\mathrm{moving}\mathrm{on}\mathrm{the}\mathrm{inside}\mathrm{surface}\mathrm{of}\mathrm{fixed}\mathrm{spherical}\mathrm{bolw}\mathrm{of}$$$$\mathrm{radius}2\mathrm{m}.\mathrm{It}\mathrm{describes}\mathrm{a}\mathrm{horizontal}\mathrm{circle}\mathrm{at}\mathrm{distance}\mathrm{of}\frac{6}{5}\mathrm{m}\mathrm{below}$$$$\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{bowl}.$$$$\left(\mathrm{a}\right)\mathrm{find}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{particle}$$$$\left(\mathrm{b}\right))\mathrm{the}\mathrm{period}\mathrm{of}\mathrm{the}\mathrm{motion}$$

Answered by mr W last updated on 18/Mar/21

Commented by mr W last updated on 18/Mar/21

$$R=2m,h=\frac{6}{5}=1.2m$$$$r=\sqrt{R^2-h^2}=1.6m$$$$\tan \theta =\frac{h}{r}$$$$F=\frac{mg}{\tan \theta }=\frac{{mv}^2}{r}$$$$\Rightarrow v=\sqrt{\frac{gr}{\tan \theta }}=\sqrt{\frac{g(R^2-h^2)}{h}}=\sqrt{\frac{10\times (2^2-1.2^2)}{1.2}}=4.62m/s$$$$period:$$$$T=\frac{2\pi r}{v}=2\pi \sqrt{\frac{h}{g}}=2\pi \sqrt{\frac{1.2}{10}}=2.18s$$

Commented by physicstutes last updated on 19/Mar/21

$$\mathrm{thank}\mathrm{you},\mathrm{just}\mathrm{a}\mathrm{little}\mathrm{doubt},\mathrm{can}\mathrm{these}\mathrm{problem}$$$$\mathrm{be}\mathrm{solved}\mathrm{with}\mathrm{the}\mathrm{angle}\theta \mathrm{being}\mathrm{form}\mathrm{between}$$$$\mathrm{downward}\mathrm{vertical}\mathrm{or}\mathrm{must}\theta \mathrm{be}\mathrm{formed}\mathrm{between}\mathrm{horizontal}?$$

Commented by mr W last updated on 19/Mar/21

$$certainlyyoucanalsousetheangle$$$$tothevertical.$$

Commented by otchereabdullai@gmail.com last updated on 19/Mar/21

$$\mathrm{wow}!$$

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