I'm a bit unsure about the last couple of bits of this question, and I'm hoping someone might be able to help.(adsbygoogle = window.adsbygoogle || []).push({});

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 co-latitude 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 gravityg). The earth is assumed to rotate steadily with angular velocityω. 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}cosθ-w_{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 getω=ωsinθy+ωcosθz, and using the equation above, with the fact thatωis 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}cosθ-w_{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}cosθ-w_{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}cosθ-w_{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}cosθ-w_{0}sinθ)+2/3gsinθωt^{3}}{2ω^{2}}[/itex]

which simplifies to [itex]x=(v_{0}cosθ-w_{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!

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# Inertial/non-inertial reference frames

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