A-hemispherical-bowl-of-radius-R-0-1-m-is-rotating-about-its-own-axis-which-is-vertical-with-an-angular-velocity-A-particle-of-mass-10-2-kg-on-the-frictionless-inner-surface-of-the-bowl-is-

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Question Number 24032 by Tinkutara last updated on 11/Nov/17

$$\mathrm{A}\mathrm{hemispherical}\mathrm{bowl}\mathrm{of}\mathrm{radius}R=0.1$$$$\mathrm{m}\mathrm{is}\mathrm{rotating}\mathrm{about}\mathrm{its}\mathrm{own}\mathrm{axis}(\mathrm{which}$$$$\mathrm{is}\mathrm{vertical})\mathrm{with}\mathrm{an}\mathrm{angular}\mathrm{velocity}\omega .$$$$\mathrm{A}\mathrm{particle}\mathrm{of}\mathrm{mass}{10}^{-2}\mathrm{kg}\mathrm{on}\mathrm{the}$$$$\mathrm{frictionless}\mathrm{inner}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{bowl}\mathrm{is}$$$$\mathrm{also}\mathrm{rotating}\mathrm{with}\mathrm{the}\mathrm{same}\omega .\mathrm{The}$$$$\mathrm{particle}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{height}h\mathrm{from}\mathrm{the}\mathrm{bottom}$$$$\mathrm{of}\mathrm{the}\mathrm{bowl}.$$$$\mathrm{It}\mathrm{is}\mathrm{desired}\mathrm{to}\mathrm{measure}g(\mathrm{acceleration}$$$$\mathrm{due}\mathrm{to}\mathrm{gravity})\mathrm{using}\mathrm{the}\mathrm{set}\mathrm{up}\mathrm{by}$$$$\mathrm{measuring}h\mathrm{accurately}.\mathrm{Assuming}$$$$\mathrm{that}R\mathrm{and}\omega \mathrm{are}\mathrm{known}\mathrm{precisely}\mathrm{and}$$$$\mathrm{that}\mathrm{the}\mathrm{least}\mathrm{count}\mathrm{in}\mathrm{the}\mathrm{measurement}$$$$\mathrm{of}h\mathrm{is}{10}^{-4}\mathrm{m},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{minimum}$$$$\mathrm{possible}\mathrm{error}\mathrm{\Delta }g\mathrm{in}\mathrm{the}\mathrm{measured}\mathrm{value}$$$$\mathrm{of}g?$$

Commented by ajfour last updated on 11/Nov/17

Commented by Tinkutara last updated on 12/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$$$\mathrm{But}\mathrm{why}-\mathrm{in}{\omega }^2\mathrm{\Delta }h?\mathrm{Errors}\mathrm{are}\mathrm{always}$$$$+\mathrm{and}\mathrm{added}.$$

Commented by ajfour last updated on 11/Nov/17

$$N\cos \theta =mg$$$$N\sin \theta =m{\omega }^{2}R\sin \theta$$$$\Rightarrow {\omega }^{2}R\cos \theta =g\dots .\left(i\right)$$$$andh=R(1-\cos \theta )$$$$\Rightarrow g={\omega }^2(R-h)$$$$△g=-{\omega }^2△h$$$$\Rightarrow △g=-\left(\frac{g}{R\cos \theta }\right)△h$$$$\mid {(△g)}_{min}\mid =\left(\frac{9.8}{0.1}\right)\times {10}^{-4}m/s^2$$$$=9.8\times {10}^{-3}m/s^2.$$

Commented by NECx last updated on 12/Nov/17

$$whenyoudrawwiththisappdoyou$$$$useyourhandorastyluspen?$$$$$$$$i^{\prime }maskingbecauseitrytouseit$$$$withmyhandsbutmydrawingsare$$$$notasniceastheonesiseeyou$$$$posthere.$$

Commented by ajfour last updated on 12/Nov/17

$$donthavestylus,idrawwith$$$$fingerafternecessaryzoom$$$$usingHandraw.$$

Commented by ajfour last updated on 12/Nov/17

$$thisimpliesthatifmeasured$$$$hislittlegreaterthantrue$$$$value,thencalculatedgwillbe$$$$lessthanthetruevalue..$$

Commented by NECx last updated on 12/Nov/17

$$okboss$$

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