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A-particle-moving-on-the-inside-surface-of-fixed-spherical-bolw-of-radius-2-m-It-describes-a-horizontal-circle-at-distance-of-6-5-m-below-the-centre-of-the-bowl-a-find-the-speed-of-the-partic

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Question Number 136109 by physicstutes last updated on 18/Mar/21
A particle moving on the inside surface of fixed spherical bolw of radius 2 m. It describes a horizontal circle at distance of (6/5) m below the centre of the bowl. (a) find the speed of the particle (b))the period of the motion
$$\mathrm{A}\:\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inside}\:\mathrm{surface}\:\mathrm{of}\:\:\mathrm{fixed}\:\mathrm{spherical}\:\mathrm{bolw}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{2}\:\mathrm{m}.\:\mathrm{It}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{at}\:\:\mathrm{distance}\:\mathrm{of}\:\frac{\mathrm{6}}{\mathrm{5}}\:\mathrm{m}\:\mathrm{below} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bowl}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\left.\left(\mathrm{b}\right)\right)\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{the}\:\mathrm{motion} \\ $$
Answered by mr W last updated on 18/Mar/21
Commented by mr W last updated on 18/Mar/21
R=2 m, h=(6/5)=1.2 m r=(√(R^2 −h^2 ))=1.6 m tan θ=(h/r) F=((mg)/(tan θ))=((mv^2 )/r) ⇒v=(√((gr)/(tan θ)))=(√((g(R^2 −h^2 ))/h))=(√((10×(2^2 −1.2^2 ))/(1.2)))=4.62 m/s period: T=((2πr)/v)=2π(√(h/g))=2π(√((1.2)/(10)))=2.18 s
$${R}=\mathrm{2}\:{m},\:{h}=\frac{\mathrm{6}}{\mathrm{5}}=\mathrm{1}.\mathrm{2}\:{m} \\ $$$${r}=\sqrt{{R}^{\mathrm{2}} −{h}^{\mathrm{2}} }=\mathrm{1}.\mathrm{6}\:{m} \\ $$$$\mathrm{tan}\:\theta=\frac{{h}}{{r}} \\ $$$${F}=\frac{{mg}}{\mathrm{tan}\:\theta}=\frac{{mv}^{\mathrm{2}} }{{r}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{gr}}{\mathrm{tan}\:\theta}}=\sqrt{\frac{{g}\left({R}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)}{{h}}}=\sqrt{\frac{\mathrm{10}×\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}.\mathrm{2}^{\mathrm{2}} \right)}{\mathrm{1}.\mathrm{2}}}=\mathrm{4}.\mathrm{62}\:{m}/{s} \\ $$$${period}: \\ $$$${T}=\frac{\mathrm{2}\pi{r}}{{v}}=\mathrm{2}\pi\sqrt{\frac{{h}}{{g}}}=\mathrm{2}\pi\sqrt{\frac{\mathrm{1}.\mathrm{2}}{\mathrm{10}}}=\mathrm{2}.\mathrm{18}\:{s} \\ $$
Commented by physicstutes last updated on 19/Mar/21
thank you,just a little doubt, can these problem be solved with the angle θ being form between downward vertical or must θ be formed between horizontal?
$$\mathrm{thank}\:\mathrm{you},\mathrm{just}\:\mathrm{a}\:\mathrm{little}\:\mathrm{doubt},\:\mathrm{can}\:\mathrm{these}\:\mathrm{problem} \\ $$$$\mathrm{be}\:\mathrm{solved}\:\mathrm{with}\:\mathrm{the}\:\mathrm{angle}\:\theta\:\mathrm{being}\:\mathrm{form}\:\mathrm{between} \\ $$$$\mathrm{downward}\:\mathrm{vertical}\:\mathrm{or}\:\mathrm{must}\:\theta\:\mathrm{be}\:\mathrm{formed}\:\mathrm{between}\:\mathrm{horizontal}? \\ $$
Commented by mr W last updated on 19/Mar/21
certainly you can also use the angle to the vertical.
$${certainly}\:{you}\:{can}\:{also}\:{use}\:{the}\:{angle} \\ $$$${to}\:{the}\:{vertical}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 19/Mar/21
wow!
$$\mathrm{wow}! \\ $$

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