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## Motion in Non-Inertial Reference Frame, Physics tutorial

Introduction:The non-inertial reference frame is the reference frame which is not tied to state of motion of observer, with property that each physical law portrays itself in same form in every inertial frame. As such, laws of physics in such a frame don't take on their most simple form, as needed by special principle of relativity.:Time Derivatives in Fixed and Rotating FramesLet us consider time derivative of the arbitrary vector A in two reference frames. First reference frame is known as fixed frame and is expressed in terms of Cartesian coordinates r' = (x'; y'; z'). Second reference frame is known as rotating frame and is expressed in terms of Cartesian coordinates r = (x; y; z).

Rotating frame shares similar origin as fixed frame and rotation angular velocity ω of rotating frame (with respect to fixed frame) has components (ω

_{x}, ω_{y}, ω_{z}).As observations are often made in rotating frame of reference, we decompose vector A in terms of components A

_{i }in rotating frame (with unit vectors x^i). Thus, A = A_{i}x^^{i}(using summation rule) and time derivative of A as observed in fixed frame isdA/dt = dA

_{i}/dtx^^{i}+ A_{i}dx^^{i}/dtInterpretation of first term is that of time derivative of A as seen in rotating frame (where unit vector x^

^{i }are constant) while second term involves time dependence of relation between fixed and rotating frames. We now state dxˆ_{i}/ dt as the vector in rotating frame as dx^^{i}/dt = R^{ij}x^_{j}= ε^{ij}kω_{k}x^_{j}where R represents rotation matrix related with rotating frame of reference; this rotation matrix associated is anti-symmetric (R^{ij}= -R^{ji}) and can be written in terms of antisymmetric tensor ε^{ijk}(stated in terms of vector product A x B = A_{i}B_{j}ε^{ijk}x^_{k}for two arbitrary vectors A and B) as R^{ij}= ε^{ijk}ω_{k,}where ω_{k}denotes components of angular velocity ω in rotating frame.This time derivative of arbitrary rotating frame vector A in fixed frame is, thus stated as

(dA/dt)

_{f}= (dA/dt)_{r}+ ω x AWhere (d/dt)

_{f}signifies time derivative as observed in fixed (f) frame while (d/dt)r denotes time derivative as observed in rotating (R) frame. The application of this formular associate to time derivative of rotation angular velocity ω itself. One can easily see that(dw/dt)

_{f }= ω = (dω/dt)_{r}As second term of equation above vanishes for A = ω; time derivative of ω is, thus, same in both frames of reference and is signified ω in what follows.

Accelerations in Rotating Frames:We now consider general case of the rotating frame and fixed frame being associated by translation and rotation. In the figure given above, position of point P according to fixed frame of reference is labeled r' , while position of the same point according to rotating frame of reference is labeled r, and r' = R + r, where R signifies position of origin of the rotating frame according to fixed frame. As velocity of point P involves rate of change of position, now define which time-derivatives operator, (d/dt)

_{f}or (d/dt)_{r}that is to be utilized.Velocities of point P as observed in fixed and rotating frames are stated as V

_{f}= (dr'/dt)_{f }and v_{r}= (dr/dt)_{r }respectively. Using equation relationship between fixed frame and rotating frame velocity is stated asv

_{f }= (dR/dt)_{f }+ (dr/dt)_{f}= V + v_{r}+ ω X r,Where V = (dR/dt)f denotes translation velocity of rotating-frame origin (as observed in fixed frame).

Using above equation, we are now in the position to estimate expressions for acceleration of point P as seen in fixed and rotating frames of reference. Write the expression for acceleration of point P as observed in rotating frame as

a

_{r}= a_{f}- A - ωx(ωxr)-2ω x v_{r}- ω x r, where a_{f}= (dV/dt)_{f }and a_{r}= (dv_{r}/dt)_{f}which represent sum of net inertial acceleration (a

_{f}-A), centrifugal acceleration -ω X (ω X r) and the angular acceleration term -ω X rω that depends explicitly on time dependence of rotation angular velocity ω.Centrifugal acceleration (that is directed outwardly from rotation axis) represents the familiar non-inertial effect in Physics. Less familiar non-inertial effect is Coriolis acceleration.

:Lagrangian Formulation Of Non-Inertia MotionLagrangian for the particle of mass m moving in the non-inertial rotating frame (with its origin coinciding with the fixed-frame origin) in the presence of the potential U(r) is expressed as

L(r, dr) = m/2|dr + ω X r|

^{2}- U(r)Where ω is the angular velocity vector and we use the formula

|dr + ω x r|

^{2}= |r|^{2}+ 2ω.(r X r)+[ω^{2}r^{2}- (ω.r)^{2}]Using Lagrangian we now derive general Euler-Lagrange equation for r. First, deduce expression for canonical momentum

P = ∂L/dr = m(dr + ω X r )

So that Euler-Lagrange equation is

Md

^{2}r/dr = - ∇U(r) - m[ω x dr + 2ωxdr + ωx(ω x r)]Here, potential energy term produces fixed-frame acceleration, - ∇U = ma

_{f}.:Motion relative to earthWe arrange (x, y, z) axis of rotating frame so that z-axis is the continuation of the position vector R of rotating-frame origin, i.e. R

^{→}= Rzˆ in rotating frame (where R=6378km is Earth radius assuming the spherical Earth). When stated in terms of fixed-frame latitude angle λ and azimuthal angle Ψ, the unit vector z^ isZ^ = cos λ(cosΨx^' + sinΨy^') + sinλz^'

Similarly, we select axis to be tangent to a great circle passing through North and South poles, So that

x^ = sinλ(cosΨx^' + sinΨy^') + cosλz^'

Finally, the y axis is selected such that

y^ = z^ z x^ = -sinΨx^' + cosΨy^'

Pure gravitational acceleration is, thus, expressed in rotating frame of Earth as

g

_{0f}= -g_{0r}[((1 + εcosΘ)z^ + εsinΘ(cosΦx^ + sinΦy^))/(1 + 2εcosΘ + ε^{2})^{3/2}]where g

_{or}acceleration because of gravitational pull on earth as observed in rotating frame of earthg

_{or}= GM/R^{2}= 9 .789 m / s^{2}and ε = r/R which is much lesser than 1.Centrifugal acceleration due to R is

d

^{2}R/dR = ωRf = ωx(ωxR) = αg_{or}cosλ(cosλz^ + sinλx^)Each Coriolis drift can be stated as infinite series in power of ω and that all Coriolis effects vanish when ω = 0.

Free-Fall Problem:As the example of significance of Coriolis effects, we consider simple free fall problem, where (x

_{0}, y_{0}, z_{0}) = (0, 0, h) and (dx_{0}, dy_{0}, dz_{0}) = (0, 0, 0) So that constants are C_{x}= 0 = C_{z }and C_{y}= 2ωhcosλSubstituting the constants in equation and keeping only terms up to first order in ω, we find

x(t) = 0,

y(t) = 1/3grt

^{3}ωcosλ,z(t) = h - 1/2grt

^{2}Therefore, a free falling object starting from rest touches ground z(T) = 0 after a time T = √2h/g

_{r}after which object has drifted eastward by the distance ofy(T) = 1/3grT

^{3}ωcosλ = ωcosλ/3√8h^{3}/g_{r}Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

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