Inertia force, Coriolis force
Fondamental : From a mathematical point of view, why do we use inertia forces ?
(R) is an inertial frame of reference, (R') is a frame which as a motion with respect to (R).
Let M (masse \(m\)) be a material point, subject to « real » forces, the resultant of which we note \(\vec f\).
Let \(\vec v\) and \(\vec a\) be the velocity and acceleration vectors of M in (R).
Let \(\vec v \;'\) and \(\vec a \;'\) be the velocity and acceleration vectors of M in (R').
In general : (see the lesson "Kinematics : changing the frame of reference")
\(\vec v = \vec v \;' + {\vec v_e}\;\;\;\;\;and\;\;\;\;\;\vec a = \vec a \;' + {\vec a_e} + {\vec a_c}\)
The second law of motion in the inertial frame (R) gives us :
\(m\vec a = \vec f\)
If we compose the acceleration vectors :
\(m(\vec a\;' + {\vec a_e} + {\vec a_c}) = \vec f\;\;\;\;\;so\;\;\;\;\;m\vec a\;' = \vec f\; + ( - m{\vec a_e}) + ( - m{\vec a_c})\)
We note :
\({\vec f_{ie}} = - m{\vec a_e}\) : the inertia force
\({\vec f_{ic}} = - m{\vec a_c}\) : the Coriolis force
Then :
\(m\vec a\;' = \vec f\; + {\vec f_{ie}} + {\vec f_{ic}}\)
We can finally apply the second law of motion in a non-inertial reference frame, but only if we add to the « real » forces the inertia force and the Coriolis force.
Exemple : Elevator and apparent weight :
Here, the frame of reference (R') is in pure translation relative to the frame (R), so there is no Coriolis force.
The equation of the equilibrium in the frame of the elevator is :
\(m\vec a\;' = \vec 0 = \vec R + m\vec g + ( - m{\vec a_e})\)
If we rewrite it using the notation of the « apparent weight » , \(P_{app}\) :
\(\begin{array}{l}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\vec P}_{app}} = - \vec R \\{{\vec P}_{app}} = m(\vec g - {{\vec a}_e}) = - m(g + {a_e})\;{{\vec u}_y} \\\end{array}\)
Note :
\(g\) is an absolute value (\(g > 0\)), whereas \(a_e\) can be positive or negative.
When the elevator goes from the ground floor to the 6th floor for instance, \(a_e\) is positive : the apparent weight is greater than the real weight. That is what we feel when we are in an elevator that goes up.
We can also note that the apparent weight is zero when the elevator is in free fall !
Note :
If (R') is in a rectilinear uniform translation relative to (R), the inertial acceleration is zero.
Then :
\(\vec a\;'(M)=\vec a(M)\)
The second law of motion in (R') is :
\(m\vec a\;'(M)=\vec f\)
The second law of motion is thus still valid in (R').
We finally show that a frame in rectilinear uniform translation relative to an inertial frame of reference is also inertial.
Fondamental : Centrifugal force
The motion of (R') relative to (R) is a rotation around a fixed axis.
The angular velocity \(\omega\) is constant.
The inertia force is :
\({\vec f_{ie}} = - m\;\vec \omega \wedge \left( {\vec \omega \wedge \overrightarrow {OM} } \right)\; = + m{\omega ^2}\overrightarrow {HM} \;\)
It's the well known centrifugal force !
Fondamental : Coriolis force
\(\vec v\;'\) is the velocity vector of a material point in the frame (R').
The Coriolis force at that point is :
\({{\vec f}_{ic}} = - 2m\vec \omega \wedge \vec v\;'\)
A few effects of the Coriolis force in terrestrial mechanics are reminded in the courses on non inertial frames of reference.
A video about Coriolis force :
