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For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

http://prola.aps.org/abstract/PR/v99/i2/p587_1

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0$

So that potential gives equations of motion with constant velocity. Not exactly what we expected. So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"

EDIT: Fun Generalization! (Inspired by elfmotat's answer) Take generic Lagrangian $L=\sum_n a_n\dot{q}^n+f(q)$ (Putting all velocity in the first term, generic function of position in the second). Then

$\frac{\partial L}{\partial \dot {q}}=\sum_n na_n \dot{q}^{n-1},\qquad\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}$

So equations of motion are

$\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}=\frac{\partial f(q)}{\partial q}$

So, only when $n=2$ is this "nice". Then $f(q)$ is something like the potential and this last line is Newton's 2nd Law.

For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

Link

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0$

So that potential gives equations of motion with constant velocity. Not exactly what we expected. So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"

EDIT: Fun Generalization! (Inspired by elfmotat's answer) Take generic Lagrangian $L=\sum_n a_n\dot{q}^n+f(q)$ (Putting all velocity in the first term, generic function of position in the second). Then

$\frac{\partial L}{\partial \dot {q}}=\sum_n na_n \dot{q}^{n-1},\qquad\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}$

So equations of motion are

$\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}=\frac{\partial f(q)}{\partial q}$

So, only when $n=2$ is this "nice". Then $f(q)$ is something like the potential and this last line is Newton's 2nd Law.

For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

http://prola.aps.org/abstract/PR/v99/i2/p587_1

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0$

So that potential gives equations of motion with constant velocity. Not exactly what we expected. So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"

For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

http://prola.aps.org/abstract/PR/v99/i2/p587_1

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0$

So that potential gives equations of motion with constant velocity. Not exactly what we expected. So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"

EDIT: Fun Generalization! (Inspired by elfmotat's answer) Take generic Lagrangian $L=\sum_n a_n\dot{q}^n+f(q)$ (Putting all velocity in the first term, generic function of position in the second). Then

$\frac{\partial L}{\partial \dot {q}}=\sum_n na_n \dot{q}^{n-1},\qquad\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}$

So equations of motion are

$\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}=\frac{\partial f(q)}{\partial q}$

So, only when $n=2$ is this "nice". Then $f(q)$ is something like the potential and this last line is Newton's 2nd Law.

For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

http://prola.aps.org/abstract/PR/v99/i2/p587_1

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0\rightarrow \ddot{q}=\frac{1}{m-6\lambda \dot{q}}$

So that potential gives a velocity-dependent acceleration as we might expect; but have you ever seen a force with that form? So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"

For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

http://prola.aps.org/abstract/PR/v99/i2/p587_1

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0$

So that potential gives equations of motion with constant velocity. Not exactly what we expected. So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"