
#1
Aug2712, 06:51 AM

P: 10

I'm a bit unsure about the last couple of bits of this question, and I'm hoping someone might be able to help.
1. The problem statement, all variables and given/known data a) Let a reference frame with origin O & Cartesian axes (x, y, z) be fixed relative to the surface of the rotating earth at colatitude θ (i.e. 0≤θ≤∏, where θ = 0 corresponds to the north pole). Increasing x is east, increasing y is north & increasing z is upwards (opposite direction to gravity g). The earth is assumed to rotate steadily with angular velocity ω. Find the components of ω in this frame of reference. Ignoring the centrifugal force, show that the motion of a particle of mass m under gravity is governed by [itex]\ddot{x} − 2ω\dot{y} cos θ + 2ω\dot{z} sin θ = 0 [/itex] [itex]\ddot{y}+ 2ω\dot{x} cos θ = 0 [/itex] [itex]\ddot{z}− 2ω\dot{x} sin θ = −g [/itex] where ω = ω and g = g. Assuming θ is constant, by integrating the second and third of these equations with respect to time and substituting into the first equation, show that [itex]\ddot{x}+ 4ω^{2}x= 2ω(v_{0}cosθw_{0}sin)+2gtsinθ [/itex] where v_{0} and w_{0} are constants. Hence find the general solution for x. b) If a particle falls from rest at O, find x as a function of t. The particle falls only for a brief time before it hits the ground, so that ωt is small throughout its motion. Use a series xpansion of solution for x in ωt to show, to leading order, [itex]x =\frac{1}{3}gωt^{3} sin θ [/itex] c) Explain briefly how an inertial observer would account for this eastward deflection of the falling particle. 2. Relevant equations [itex]m\textbf{a}=\textbf{F}m\dot{\textbf{ω}}\times\textbf{r}2m\textbf{ω}\times\dot{\textbf{r}}m\textbf{ω}\times(\textbf{ω}\times\textbf{r})m\textbf{A} [/itex] 3. The attempt at a solution For a) I get ω=ωsinθy+ωcosθz, and using the equation above, with the fact that ω is constant and ignoring the centrifugal force, I get the three equations as stated. Integrating then gives [itex]\ddot{x}+ 4ω^{2}x= 2ω(v_{0}cosθw_{0}sin)+2gtsinθ [/itex]. solving this as a 2nd order ODE, I get complementary solution [itex]x=αcos(2ωt)+βsin(2ωt) [/itex] and particular solution [itex]x=\frac{gsinθ}{2ω}t+\frac{v_{0}cosθw_{0}sinθ}{2w^{2}} [/itex] so the general solution for x is these added together. For b), I plugged in the initial values, at t=0, x=0, [itex]\dot{x}[/itex]=0 to get [itex]α=\frac{(v_{0}cosθw_{0})}{(2ω^{2})}[/itex] [itex]β=\frac{gsinθ}{4ω^{2}}[/itex] and using the series expansions for sin and cos, I get for small t [itex]x=\frac{2ω^{2}t^{2}(v_{0}cosθw_{0}sinθ)+2/3gsinθωt^{3}}{2ω^{2}}[/itex] which simplifies to [itex]x=(v_{0}cosθw_{0}sinθ)t^{2}+1/3gsinθt^{3}[/itex] The answer I'm supposed to get here is just the second term, but I'm not entirely sure if I've done this right. Can I just cancel the first term here as t is small? I'm also not entirely sure what answer part c) is looking for. Is it anything to do with Coriolis? I'd be grateful if anyone could shed a bit of light on this. Thanks! 



#2
Aug2712, 09:36 AM

Thanks
P: 5,489

[itex]v_{0}[/itex] and [itex]w_{0}[/itex] are constants of integration. What would they be for the initial conditions given in b)?




#3
Aug2712, 12:11 PM

P: 10

Apologies if I'm being really stupid here, but I don't see how the initial conditions give any information about v_{0} and w_{0}? The two initial conditions give you the constants from the ODE don't they?




#4
Aug2712, 12:21 PM

Thanks
P: 5,489

Inertial/noninertial reference frames
You got them by integrating the equations for y and z. b), however, specifies that the particle is initially at O at rest.




#5
Aug2712, 04:24 PM

P: 10

Ah right of course! I see what you mean now, and that gives the correct answer, thanks.
The only other bit I wasn't sure about was part c), I don't really know what sort of answer they're looking for.. 



#6
Aug2712, 04:42 PM

Thanks
P: 5,489

I think you should take an inertial frame of reference whose origin is coincident with O at t = 0 and see what the motion would look like in that frame (when t is small). I think in should boil down to the fact that in an inertial frame the motion should be the familiar parabolic motion in the XZ plane.




#7
Aug3012, 05:22 PM

P: 10

Hmm.. How do you get parabolic motion? sorry if I'm being slow here!




#8
Aug3012, 05:38 PM

Thanks
P: 5,489

Imagine an inertial frame coincident with the rotating frame but fixed with respect to the center of the planet. If you release at particle at rest in the rotating frame, in the inertial frame it will have a nonzero horizontal velocity. The only force in the inertial frame acting on it will be gravity. The resultant motion is parabolic.




#9
Aug3112, 03:43 AM

P: 10

Oh, right, I see what you mean now... I was thinking that it was released from rest so would have zero velocity in the x direction so wouldn't be parabolic but that's not right if you're in an inertial frame.. Thanks for all the help!



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