A particle of mass \(m\) on the Earth’s surface is confined to move on the quartic curve \(y=ax^4\) where \(y\) is up. Find the Lagrangian of the particle and the equation of motion.
The kinetic energy is
\( K.E. = \dfrac {1}{2} m(\dot x^2 + \dot y^2) \)
The potential energy is
\( P.E. = mgy\)
The Lagrangian is
\( \mathscr L = \dfrac {1}{2} m(\dot x^2 + \dot y^2) - mgy \)
We can eliminate \(y\) with the change of variables
\( y = ax^4 \)
\( \dot y = 4ax^3 \dot x \rightarrow \dot y^2 = 16a^2x^6 \dot x^2 \)
The Lagrangian can be re-written as
\( \mathscr L = \dfrac {1}{2} m\dot x^2(1 + 16a^2x^6) - mgax^4 \)
To find the equation of motion we find the derivatives
\( \dfrac{\partial \mathscr L}{\partial \dot x} = m\dot x(1 + 16a^2x^6) \)
\( \dfrac {d}{dt} \dfrac{\partial \mathscr L}{\partial \dot x} = m\ddot x (1 + 16a^2x^6) + 96ma^2x^5\dot x^2 \)
\( \dfrac{\partial \mathscr L}{\partial x} = 48ma^2x^5\dot x^2 - 4mgax^3 \)
The equation of motion is
\( m\ddot x (1 + 16a^2x^6) + 96ma^2x^5\dot x^2 = 48ma^2x^5\dot x^2 - 4mgax^3 \)
We can rearrange the equation as follows
\( \ddot x (1 + 16a^2x^6) + 48a^2x^5\dot x^2 + 4gax^3 = 0 \)