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Coriolis and Centrifugal Acceleration


“Determine the direction of the Coriolis acceleration, $-2\Vec\Omega\times\Vec v_R$.”

“Determine the direction of the centrifugal acceleration, $-\Vec\Omega\times(\Vec\Omega\times\Vec r)$.”


Estimated Time: 10 minutes, including wrap-up

In this SWBQ, students determine the direction of the Coriolis and centrifugal accelerations in various situations. The combination of multiple cross products and overall factors of $-1$ cause some confusion as to which way these accelerations point.

Students will need to know the expression for acceleration in a rotating frame, namely: $${\Vec a}_R = {\Vec a} - 2\,\Vec\Omega\times{\Vec v}_R - \Vec\Omega\times(\Vec\Omega\times\Vec r)$$

Is there a modified version of Newton's Second Law which holds in a rotating reference frame?

Wrap Up

Remind students how to manipulate their right hand in order to determine the direction of the cross product. Use this to determine the orientation of the Coriolis acceleration, relative to the direction of motion (to the right). Then demonstrate that the double cross product in the centrifugal acceleration leads to a reversal of the direction, so that the centrifugal acceleration points radially outward.

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