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HW9_prob3_dropped particle with rebound

# HW9_prob3_dropped particle with rebound - Dropped Particle...

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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 fi 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 . Assume that we are in the Northern hemisphere, and let the 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 fi 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 denotes an initial value. Falling particle: This problem is solved in the textbook and in the class notes, and the answer is that the de fl
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