Question and Answers Forum

All Questions      Topic List

Probability and Statistics Questions

Previous in All Question      Next in All Question      

Previous in Probability and Statistics      Next in Probability and Statistics      

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!

$$\mathrm{wow}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com