Calculating the Effective Force (a more general derivation)

There are a lot of nasty derivations of the effective force in mechanics text books. We'd like to make ours simple yet not require too much higher mathematics, so we need to set up the problem as simply as possible.

Newton's laws apply in the inertial frame of reference. Lets work in that frame, but set up a set of unit vectors for an Accelerating frame. For simplicity of notation, lets call those unit vectors \bgroup\color{black}$ \hat{u}_i$\egroup. Lets call the coordinates in the accelerating (rotating) frame \bgroup\color{black}$ x_i$\egroup and the position vector in the Inertial frame \bgroup\color{black}$ \vec{x_I}$\egroup etc. Let the displacement between the origins of the Inertial and Accelerating frame be called \bgroup\color{black}$ \vec{X}$\egroup. We then simply have.

\bgroup\color{black}$\displaystyle \vec{x_I}= \vec{X} +x_j\hat{u}_j $\egroup

We will differentiate this equation twice to get accelerations and hence real and fictitious forces. Note that all the quantities in this equation are defined in the Inertial frame.

\bgroup\color{black}$\displaystyle {d\vec{x_I}\over dt}= {d\vec{X}\over dt} +{dx_j\over dt}\hat{u}_j + x_j {d\hat{u}_j\over dt} $\egroup

The change of \bgroup\color{black}$ \vec{X}$\egroup with time is a relative motion of the origins of the two coordinate systems. The change of the unit vectors with time is due to the rotation of the accelerating coordinate system.

\bgroup\color{black}$\displaystyle {d^2\vec{x_I}\over dt^2}= {d^2\vec{X}\over dt...
...}_j +2{dx_j\over dt}{d\hat{u}_j\over dt} + x_j {d^2\hat{u}_j\over dt^2} $\egroup

These are all the derivatives we need.

We also need the effect of the rotation on the unit vectors \bgroup\color{black}$ \hat{u}_i$\egroup. In an infinitesimal time \bgroup\color{black}$ dt$\egroup the unit vectors rotate through an angle \bgroup\color{black}$ d\vec{\theta}=\vec{\omega}dt$\egroup.

\bgroup\color{black}$\displaystyle d\hat{u}_i= d\vec{\theta}\times\hat{u}_i $\egroup

This is the change of the unit vectors, evaluated in the inertial frame.

\bgroup\color{black}$\displaystyle {d\hat{u}_i\over dt}={d\vec{\theta}\over dt}\times\hat{u}_i=\vec{\omega}\times\hat{u}_i $\egroup

We also need the second derivative.

\bgroup\color{black}$\displaystyle {d^2\hat{u}_i\over dt^2}={d\vec{\omega}\over ...
...\over dt}\times\hat{u}_i +\vec{\omega}\times\vec{\omega}\times\hat{u}_i $\egroup

We can now plug these into the formula for the second derivative of \bgroup\color{black}$ \vec{x_I}$\egroup above.

$\displaystyle {d^2\vec{x_I}\over dt^2}$ $\displaystyle = {d^2\vec{X}\over dt^2} +{d^2x_j\over dt^2}\hat{u}_j +2{dx_j\over dt}{d\hat{u}_j\over dt} + x_j {d^2\hat{u}_j\over dt^2}$    
$\displaystyle {d^2\vec{x_I}\over dt^2}$ $\displaystyle = {d^2\vec{X}\over dt^2} +{d^2x_j\over dt^2}\hat{u}_j +2{dx_j\ove...
...}\over dt}\times\hat{u}_j +\vec{\omega}\times\vec{\omega}\times\hat{u}_j\right)$    
$\displaystyle {d^2\vec{x_I}\over dt^2}$ $\displaystyle = {d^2\vec{X}\over dt^2} +{d^2x_j\over dt^2}\hat{u}_j +2\vec{\ome...
...\over dt}\times x_j\hat{u}_j +\vec{\omega}\times\vec{\omega}\times x_j\hat{u}_j$    

We recognize \bgroup\color{black}$ x_j$\egroup as the position in the Accelerating frame (and \bgroup\color{black}$ x_j\hat{u}_j$\egroup as that position written in the inertial frame), \bgroup\color{black}$ {dx_j\over dt}$\egroup as the velocity in the Accelerating frame, and \bgroup\color{black}$ {d^2x_j\over dt^2}$\egroup as the acceleration seen in the Accelerating frame. It is important to note that these are written as vectors in the Inertial system by dotting them into the \bgroup\color{black}$ \hat{u}_j$\egroup, which is why we can do this calculation rather easily. For example, the \bgroup\color{black}$ x_i$\egroup give a position in the Accelerating frame, but when they are dotted into the basis vectors of the Accelerating frame, written in the Inertial frame, \bgroup\color{black}$ \hat{u}_i$\egroup, the vector is now correctly written in the Inertial frame.
\bgroup\color{black}$ \displaystyle \vec{a_I} = \vec{A} + \vec{a_A} +2\vec{\omeg...
...ver dt}\times \vec{x_A} +\vec{\omega}\times\vec{\omega}\times \vec{x_A} $\egroup
In the equation above, \bgroup\color{black}$ \vec{a_I}$\egroup is the acceleration the mass in the Inertial frame, \bgroup\color{black}$ \vec{A}$\egroup is the acceleration of the origin of the Accelerating frame, \bgroup\color{black}$ \vec{a_A}$\egroup is the acceleration of the mass seen in the Accelerating frame, \bgroup\color{black}$ \vec{v_A}$\egroup is the velocity of the mass seen in the Accelerating frame, \bgroup\color{black}$ \vec{x_A}$\egroup is the position of the mass seen in the Accelerating frame, and \bgroup\color{black}$ \vec{\omega}$\egroup is the angular velocity of the rotation of the coordinate axes of the accelerating (rotating) frame. \bgroup\color{black}$ \vec{A}$\egroup and \bgroup\color{black}$ \vec{\omega}$\egroup are properties of the transformation between the Inertial frame and the Accelerating frame.

This equation is written in the inertial frame but at \bgroup\color{black}$ t=0$\egroup the two frames coincide so the equation holds in the accelerating frame too. We can change to the accelerating frame just by removing the \bgroup\color{black}$ \hat{u}_j$\egroup.

We can multiply by the mass \bgroup\color{black}$ m$\egroup to get an equation in the forces, real and fictitious.

\bgroup\color{black}$\displaystyle m\vec{a_I} = m\vec{A} + m\vec{a_A} +2m\vec{\o... dt}\times \vec{x_A} +m\vec{\omega}\times\vec{\omega}\times \vec{x_A} $\egroup

Now solve this equation for the force seen in the Accelerating frame. The general equation for the effective force seen in a frame that is accelerating and rotating is.
\bgroup\color{black}$ \displaystyle \vec{F_A} = \vec{F_I} -m\vec{A} -2m\vec{\ome... dt}\times \vec{x_A} -m\vec{\omega}\times\vec{\omega}\times \vec{x_A} $\egroup

There is a simple fictitious force \bgroup\color{black}$ -m\vec{A}$\egroup if the coordinate system is simply accelerating. To see the effect of a constant rotation, we set \bgroup\color{black}$ \vec{A}=0$\egroup and \bgroup\color{black}$ {d\vec{\omega}\over dt}=0$\egroup

\bgroup\color{black}$ \displaystyle \vec{F_A} = \vec{F_I} -2m\vec{\omega}\times \vec{v_A} -m\vec{\omega}\times\vec{\omega}\times \vec{x_A} $\egroup
The forces are named in the equation below.

\bgroup\color{black}$\displaystyle \vec{F}_{rotating} = \vec{F}_{inertial} + \vec{F}_{Coriolis} +\vec{F}_{centrifugal} $\egroup

The centrifugal and Coriolis forces are fictitious forces that appear in the rotating frame.

The Centrifugal Force grows with the perpendicular distance from the axis of rotation and is in the direction of \bgroup\color{black}$ \vec{r}_\perp$\egroup.

\bgroup\color{black}$\displaystyle \vec{F}_{centrifugal} = -m\vec{\omega}\times\vec{\omega}\times \vec{r} $\egroup

The Coriolis Force is perpendicular to the velocity of the mass and to \bgroup\color{black}$ \vec{\omega}$\egroup. For \bgroup\color{black}$ \vec{\omega}$\egroup pointing up, like in the northern hemisphere, the Coriolis force causes projectiles to deflect to the right. In the southern hemisphere, projectiles deflect to the left.

\bgroup\color{black}$\displaystyle \vec{F}_{Coriolis} = -2m\vec{\omega}\times \vec{v} $\egroup

Its interesting to note that Coriolis's original publication in 1835 was on the energy yield of machines with rotating parts, such as water-wheels.

Jim Branson 2012-10-21