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PS2-5 - a z = d 2 z dt 2 = − a = ⇒ v z = dz dt = − at...

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Problem Set 2: Problem 5. Problem: A lighthouse keeper’s son enjoys sliding down the spiral staircase railing from the light to the ground floor. His path is a spiral of constant radius, R . As a prank, his sister has greased the railing. As the boy descends, because of the reduced friction, he experiences a constant downward vertical acceleration a . The spiral’s pitch is 37 o so that v θ = 4 3 v z . Determine his position, velocity and acceleration vectors as functions of a , R and t . His position at t = 0 is r = R , θ = 0 and z = 0 . Solution: Because the stairway has cylindrical symmetry, we use cylindrical coordinates to describe the motion. We know that the boy’s motion occurs at constant radius r = R , that his vertical acceleration is a z = a and that v θ = 4 3 v z . So, the first thing we do is solve for v z and z by integrating as follows.
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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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