Unformatted text preview: a z = d 2 z dt 2 = − a = ⇒ v z = dz dt = − at and z = − 1 2 at 2 Thus, the circumferential velocity is v θ = 4 3 at Now, we know that v θ = r d θ dt = ⇒ d θ dt = 4 3 at R where we have used the fact that r = R . Integrating over time and imposing the initial condition θ (0) = 0 , we find θ = 2 3 at 2 R Therefore, the circumferential acceleration is a θ = r d 2 θ dt 2 + 2 dr dt d θ dt = 4 3 a Finally, since r = R , necessarily v r = dr/dt = 0 and a r = d 2 r dt 2 − r w d θ dt W 2 = − 16 9 a 2 t 2 R Therefore, the boy’s position, velocity and acceleration vectors are r = R w e r − 1 2 at 2 R k W , v = at w 4 3 e θ − k W , a = − a w 16 9 at 2 R e r − 4 3 e θ + k W...
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 '06
 Shiflett
 Geometry, Velocity, Lighthouse, Cylindrical coordinate system, dt2 dt dt

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