Question

The 128 storey Shanghai Centre Tower is the world's second-tallest building: 632 m tall, it is towards China's east coast at a latitude of 31° N. A body is dropped from the top of the building, and falls to the ground beneath. a) Calculate the magnitude and direction of the deflection experienced by the body as a result of the Earth's rotation. Neglect air resistance, wind etc.] Explain your result qualitatively from the point of view of an observer in an inertial frame, not rotating with the Earth, who simply sees the body following Newtonian dynamics subject to the Earth's gravity b) The lifts (elevators) in the Shanghai Tower were, in 2014, the fastest in the world, capable of speeds of 18 ms1 By how much would a passenger be likely to move her/his head to remain upright' during the descent?

An explanation alongside the working would be appreciated!

Answer 1

If is the angular velocity of Earth's rotation, $\omega cos(\lambda )$ will be its component along the tangent of Earth at latitude $\lambda$ . Corriolis acceleration due to this is $2v\omega cos(\lambda )$ on a body falling down at downward velocity v and is directed eastward.

If \omega is the angular velocity of Earth's rotation, \omega cos (\lambda) will be its component along the tangent of Earth at latitude lambda. Corriolis acceleration due to this is 2 v w cos (lambda) on a body falling down at downward velocity v and is directed eastward. v = g t. So, acceleration = 2 g t w cos (lambda). Integrated it once gives velocity, g t^2 w cos (lambda). Integrated again gives displacement in the eastward direction as (1/3)g t^3 w cos (lambda). The height from which the object is falling, h = (1/2) g t^2. Hence required deflection is (1/3)(2h)^{3/2} g^{-1/2}\omega cos (\lambda) = 0.299 m in the eastward direction. b) When a system is viewed from the non-inertial frame like the Earth, the Coriolis force comes to play along with the centripetal force of gravity to explain the motion of the system. For a constant velocity of lift, v = 18 m/s. So acceleration due to

v = gt. So, acceleration = $2gt\omega cos(\lambda )$ . Integrated it once gives velocity, $gt^2\omega cos(\lambda )$ . Integrated again gives displacement in the eastward direction as $(1/3)gt^3\omega cos(\lambda )$ . The height from which the object is falling, h = (1/2)gt² . Hence required deflection is $(1/3)(2h)^{3/2}g^{-1/2}\omega cos(\lambda ) = 0.299m$ in the eastward direction.

b.) When a system is viewed from the non-inertial frame like the Earth, the Coriolis force comes to play along with the centripetal force of gravity to explain the motion of the system.

For a constant velocity of lift, v = 18m/s. So acceleration due to Coriolis is $2v\omega cos(\lambda ) = 0.00225 m/s^2$ at the base of the tower. The angle subtended (or the degree by which we need to move our head) =