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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
A hemispherical bowl of radius R = 0.1 m is rotating about its own axis (which is vertical) with an angular velocity ω. A particle of mass 10^(−2) kg on the frictionless inner surface of the bowl is also rotating with the same ω. The particle is at a height h from the bottom of the bowl. It is desired to measure g (acceleration due to gravity) using the set up by measuring h accurately. Assuming that R and ω are known precisely and that the least count in the measurement of h is 10^(−4) m, what is the minimum possible error Δg in the measured value of g?
$$\mathrm{A}\:\mathrm{hemispherical}\:\mathrm{bowl}\:\mathrm{of}\:\mathrm{radius}\:{R}\:=\:\mathrm{0}.\mathrm{1} \\ $$$$\mathrm{m}\:\mathrm{is}\:\mathrm{rotating}\:\mathrm{about}\:\mathrm{its}\:\mathrm{own}\:\mathrm{axis}\:\left(\mathrm{which}\right. \\ $$$$\left.\mathrm{is}\:\mathrm{vertical}\right)\:\mathrm{with}\:\mathrm{an}\:\mathrm{angular}\:\mathrm{velocity}\:\omega. \\ $$$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{10}^{−\mathrm{2}} \:\mathrm{kg}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{frictionless}\:\mathrm{inner}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bowl}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{rotating}\:\mathrm{with}\:\mathrm{the}\:\mathrm{same}\:\omega.\:\mathrm{The} \\ $$$$\mathrm{particle}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:{h}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bottom} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{bowl}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{desired}\:\mathrm{to}\:\mathrm{measure}\:{g}\:\left(\mathrm{acceleration}\right. \\ $$$$\left.\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\right)\:\mathrm{using}\:\mathrm{the}\:\mathrm{set}\:\mathrm{up}\:\mathrm{by} \\ $$$$\mathrm{measuring}\:{h}\:\mathrm{accurately}.\:\mathrm{Assuming} \\ $$$$\mathrm{that}\:{R}\:\mathrm{and}\:\omega\:\mathrm{are}\:\mathrm{known}\:\mathrm{precisely}\:\mathrm{and} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{least}\:\mathrm{count}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measurement} \\ $$$$\mathrm{of}\:{h}\:\mathrm{is}\:\mathrm{10}^{−\mathrm{4}} \:\mathrm{m},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{possible}\:\mathrm{error}\:\Delta{g}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measured}\:\mathrm{value} \\ $$$$\mathrm{of}\:{g}? \\ $$
Commented by ajfour last updated on 11/Nov/17
Commented by Tinkutara last updated on 12/Nov/17
Thank you very much Sir! But why − in ω^2 Δh? Errors are always + and added.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$$$\mathrm{But}\:\mathrm{why}\:−\:\mathrm{in}\:\omega^{\mathrm{2}} \Delta{h}?\:\mathrm{Errors}\:\mathrm{are}\:\mathrm{always} \\ $$$$+\:\mathrm{and}\:\mathrm{added}. \\ $$
Commented by ajfour last updated on 11/Nov/17
Ncos θ=mg Nsin θ =mω^2 Rsin θ ⇒ ω^2 Rcos θ = g ....(i) and h=R(1−cos θ) ⇒ g=ω^2 (R−h) △g =−ω^2 △h ⇒ △g = −((g/(Rcos θ)))△h ∣(△g)_(min) ∣ = (((9.8)/(0.1)))×10^(−4) m/s^2 =9.8×10^(−3) m/s^2 .
$${N}\mathrm{cos}\:\theta={mg} \\ $$$${N}\mathrm{sin}\:\theta\:={m}\omega^{\mathrm{2}} {R}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\omega^{\mathrm{2}} {R}\mathrm{cos}\:\theta\:=\:{g}\:\:\:\:\:….\left({i}\right) \\ $$$${and}\:\:\:\:{h}={R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\:\:\:{g}=\omega^{\mathrm{2}} \left({R}−{h}\right) \\ $$$$\:\:\:\:\:\:\:\:\bigtriangleup{g}\:=−\omega^{\mathrm{2}} \bigtriangleup{h} \\ $$$$\Rightarrow\:\:\bigtriangleup{g}\:=\:−\left(\frac{{g}}{{R}\mathrm{cos}\:\theta}\right)\bigtriangleup{h} \\ $$$$\:\mid\left(\bigtriangleup{g}\right)_{{min}} \mid\:=\:\left(\frac{\mathrm{9}.\mathrm{8}}{\mathrm{0}.\mathrm{1}}\right)×\mathrm{10}^{−\mathrm{4}} \:{m}/{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}.\mathrm{8}×\mathrm{10}^{−\mathrm{3}} \:{m}/{s}^{\mathrm{2}} \:. \\ $$
Commented by NECx last updated on 12/Nov/17
when you draw with this app do you use your hand or a stylus pen? i′m asking because i try to use it with my hands but my drawings are not as nice as the ones i see you post here.
$${when}\:{you}\:{draw}\:{with}\:{this}\:{app}\:{do}\:{you} \\ $$$${use}\:{your}\:{hand}\:{or}\:{a}\:{stylus}\:{pen}? \\ $$$$ \\ $$$${i}'{m}\:{asking}\:{because}\:{i}\:{try}\:{to}\:{use}\:{it} \\ $$$${with}\:{my}\:{hands}\:{but}\:{my}\:{drawings}\:{are} \\ $$$${not}\:{as}\:{nice}\:{as}\:{the}\:{ones}\:{i}\:{see}\:{you} \\ $$$${post}\:{here}. \\ $$
Commented by ajfour last updated on 12/Nov/17
dont have stylus , i draw with finger after necessary zoom using Handraw.
$${dont}\:{have}\:{stylus}\:,\:{i}\:{draw}\:{with} \\ $$$${finger}\:{after}\:{necessary}\:{zoom} \\ $$$${using}\:{Handraw}. \\ $$
Commented by ajfour last updated on 12/Nov/17
this implies that if measured h is little greater than true value, then calculated g will be less than the true value..
$${this}\:{implies}\:{that}\:{if}\:{measured} \\ $$$${h}\:{is}\:{little}\:{greater}\:{than}\:{true} \\ $$$${value},\:{then}\:{calculated}\:{g}\:{will}\:{be} \\ $$$${less}\:{than}\:{the}\:{true}\:{value}.. \\ $$
Commented by NECx last updated on 12/Nov/17
ok boss
$${ok}\:{boss} \\ $$

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