Example: Deflection of a Falling Object

If we drop an object from height \bgroup\color{black}$ h$\egroup, the effect of the centrifugal force is already in the local value and local direction of \bgroup\color{black}$ g$\egroup. The Coriolis force is not included and it is proportional to velocity so it will not affect the direction of plumb bobs. The Coriolis force will cause a deflection from a vertical path as the object falls. Lets call the \bgroup\color{black}$ z$\egroup direction the local upward vertical. We need to pick two directions perpendicular to that to form a coordinate system. We can choose the \bgroup\color{black}$ y$\egroup direction be be north and then we must choose east to be in the \bgroup\color{black}$ x$\egroup direction to have a right handed coordinate system.

For a downward velocity, the Coriolis acceleration is to the east,

\bgroup\color{black}$\displaystyle \vec{a}=-2\vec{\omega}\times \vec{v}=-2\omega(\sin\lambda \hat{z}+\cos\lambda\hat{e}_{north})\times v\hat{z} $\egroup

where \bgroup\color{black}$ v$\egroup will be a negative number.

\bgroup\color{black}$\displaystyle \vec{a}=-2\omega v \cos\lambda\hat{e}_{north}\times\hat{z}=-2\omega v \cos\lambda\hat{e}_{east} $\egroup

One can integrate this equation to find the deflection when a cannonball is dropped from a height \bgroup\color{black}$ h$\egroup.

  $\displaystyle v_z=-gt$    
  $\displaystyle a=2\omega gt\cos\lambda \hat{e}_{east}$    
  $\displaystyle v_{east}=2\omega g\cos\lambda\int\limits_0^tt' dt' = \omega g\cos\lambda t^2$    
  $\displaystyle x_{east}=\omega g\cos\lambda \int\limits_0^t t'^2 dt'=\omega g\cos\lambda {t^3\over 3}$    
  $\displaystyle h={1\over 2} gt^2$    
  $\displaystyle t=\sqrt{2h\over g}$    
  $\displaystyle x_{east}={\omega g\cos\lambda\over 3}\sqrt{8h^3\over g^3}={\omega\cos\lambda\over 3}\sqrt{8h^3\over g}$    

For a height of 100 meters, at 32 degrees latitude, the displacement is about 1.9 cm. Explain why the deflection is always to the east.

Jim Branson 2012-10-21