Dropped Particle with Rebound Problem (a) If we neglect the very small “centrifugal force” term in GPS Eq.(4.91), the equation of motion for a particle of mass m in the rotating frame of the earth is m d v dt = m g +2 m v × ω , (1) where v is the particle velocity, ω is the earth’s angular velocity vector, and g is the acceleration of gravity at the earth’s surface, g = − g e z . All the vectors and derivatives are measured in a body f xed frame in which the z axis is normal to the earth’s surface and the x-y plane is tangent to the earth’s surface and is located at z =0 .AssumethatweareintheNorthernhem isphere ,andletthe y axis point north and the x axis point east. As we saw in class, the solutions to Eq.(1) for v and r to f rst order in ω are v= v 0 + g t+2( v 0 × ω ) t +( g × ω ) t 2 , (2) r= r 0+ v 0 t + 1 2 g t 2+( v 0 × ω ) t 2 + 1 3 ( g × ω ) t 3 ,(3) where the subscript 0
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This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.