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Suppose-the-radius-of-the-earth-is-6400km-the-acceleration-of-a-person-at-latitude-60-due-to-the-earth-rotation-is-a-0-034m-s-2-b-232-7m-s-2-c-0-0169m-s-2-d-465-4m-s-2-Please-help-

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Question Number 5350 by sanusihammed last updated on 09/May/16
Suppose the radius of the earth is 6400km. the acceleration of a person at latitude 60° due to the earth rotation is ? (a) 0.034m/s^2 (b) 232.7m/s^2 (c) 0.0169m/s^2 (d) 465.4m/s^2 Please help. thanks in advance.
$${Suppose}\:{the}\:{radius}\:{of}\:{the}\:{earth}\:{is}\:\mathrm{6400}{km}.\:{the}\:{acceleration}\: \\ $$$${of}\:{a}\:{person}\:{at}\:{latitude}\:\mathrm{60}°\:{due}\:{to}\:{the}\:{earth}\:{rotation}\:{is}\:? \\ $$$$ \\ $$$$\left({a}\right)\:\mathrm{0}.\mathrm{034}{m}/{s}^{\mathrm{2}} \\ $$$$\left({b}\right)\:\mathrm{232}.\mathrm{7}{m}/{s}^{\mathrm{2}} \\ $$$$\left({c}\right)\:\mathrm{0}.\mathrm{0169}{m}/{s}^{\mathrm{2}} \\ $$$$\left({d}\right)\:\mathrm{465}.\mathrm{4}{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$$${Please}\:{help}.\:{thanks}\:{in}\:{advance}. \\ $$
Commented by Yozzii last updated on 11/May/16
(b) is right, assuming the earth is spherical. with respect to the axis of rotation of the earth, the person moves in a circle corresponding to the latitude θ=60°. Its radius is hence r_θ =r_E cosθ. If the Earth is assumed uniformly dense then all particles that constitute it and lie on it rotate about the axis of rotation of the earth with period T≈24×3600 s=86400 s. So, the angular velocity of all particles of the earth is ω=((2π)/T). If we consider the person′s acceleration towards the axis, at any θ, then a_θ =r_θ ω^2 =((4π^2 r_E cosθ)/T^2 ). Putting in θ=60°,r_E =6400×10^3 m and T≈86400 s gives the answer (b).
$$\left({b}\right)\:{is}\:{right},\:{assuming}\:{the}\:{earth}\:{is}\:{spherical}. \\ $$$${with}\:{respect}\:{to}\:{the}\:{axis}\:{of}\:{rotation}\:{of}\:{the}\:{earth},\:{the} \\ $$$${person}\:{moves}\:{in}\:{a}\:{circle}\:{corresponding} \\ $$$${to}\:{the}\:{latitude}\:\theta=\mathrm{60}°.\:{Its}\:{radius}\:{is}\:{hence} \\ $$$${r}_{\theta} ={r}_{{E}} {cos}\theta.\:{If}\:{the}\:{Earth}\:{is}\:{assumed} \\ $$$${uniformly}\:{dense}\:{then}\:{all}\:{particles}\:{that} \\ $$$${constitute}\:{it}\:{and}\:{lie}\:{on}\:{it}\:{rotate}\:{about} \\ $$$${the}\:{axis}\:{of}\:{rotation}\:{of}\:{the}\:{earth}\:{with} \\ $$$${period}\:{T}\approx\mathrm{24}×\mathrm{3600}\:{s}=\mathrm{86400}\:{s}. \\ $$$${So},\:{the}\:{angular}\:{velocity}\:{of}\:{all}\:{particles} \\ $$$${of}\:{the}\:{earth}\:{is}\:\omega=\frac{\mathrm{2}\pi}{{T}}. \\ $$$${If}\:{we}\:{consider}\:{the}\:{person}'{s}\:{acceleration} \\ $$$${towards}\:{the}\:{axis},\:{at}\:{any}\:\theta,\:{then}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\theta} ={r}_{\theta} \omega^{\mathrm{2}} =\frac{\mathrm{4}\pi^{\mathrm{2}} {r}_{{E}} {cos}\theta}{{T}^{\mathrm{2}} }. \\ $$$${Putting}\:{in}\:\theta=\mathrm{60}°,{r}_{{E}} =\mathrm{6400}×\mathrm{10}^{\mathrm{3}} \:{m}\:{and}\:{T}\approx\mathrm{86400}\:{s} \\ $$$${gives}\:{the}\:{answer}\:\left({b}\right). \\ $$$$ \\ $$$$ \\ $$
Commented by sanusihammed last updated on 12/May/16
thanks so much
$${thanks}\:{so}\:{much} \\ $$

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