Spring-mass system with a pendulum using Lagrangian dynamics

[MarkTheQuark]

I'm stuck in a problem of a spring mass system with a pendulum attached to it as showed in the figure below:

My goal is to find the movement equation for the mass, using Lagrangian dynamics.

If the spring moves, the wire will move the same amount. Therefore, we can write the x and y position of the mass in terms of this displacement, the length of the wire, and the theta angle.

The problem I'm having is getting to identify the potential energy of the spring, using these terms so that when solving the LaGrange equation it gives a restoring force term of the spring.

Can anyone help me identifying this?

 

[Delta2]

I am not very good in Lagrangian mechanics but if we use $$r(t),\theta(t)$$ for the position of the mass, then the contribution to Lagrangian from the potential energy of the spring should be $$\frac{1}{2}k((r(t)-r(0))^2$$ assuming that at t=0 the spring is at its natural length.

I believe there is more to the Lagrangian from the gravitational potential energy of the mass.

 

[PeroK]

I'm stuck in a problem of a spring mass system with a pendulum attached to it as showed in the figure below:

My goal is to find the movement equation for the mass, using Lagrangian dynamics.

If the spring moves, the wire will move the same amount. Therefore, we can write the x and y position of the mass in terms of this displacement, the length of the wire, and the theta angle.

The problem I'm having is getting to identify the potential energy of the spring, using these terms so that when solving the LaGrange equation it gives a restoring force term of the spring.

Can anyone help me identifying this?View attachment 301707

This looks like homework. You need to show us your best effort. See the guidelines:

https://www.physicsforums.com/help/homeworkhelp/

 

[PhDeezNutz]

I like @Delta2 ‘s suggestion. Treat $r(0)$ as the natural spring/wire length and $r(t)$ as the new length. Then the spring potential energy is exactly what he mentioned. Just add a gravitational term $mg(r(t)-r(0))$ and you've fully formed your Lagrangian.

Edit: You still have to form your kinetic terms.

 

[Delta2]

I think the gravitational potential energy depends on $\theta$ too.

 

[PhDeezNutz]

I think the gravitational potential energy depends on $\theta$ too.

I really ought to start thinking before I start posting. Of course you are right. Tack on a $\cos \theta$.

 

[PeroK]

I like @Delta2 ‘s suggestion. Treat $r(0)$ as the natural spring/wire length and $r(t)$ as the new length. Then the spring potential energy is exactly what he mentioned. Just add a gravitational term $mg(r(t)-r(0))$ and you've fully formed your Lagrangian.

It's not usual to express the quantities in a Lagrangian as a function of time, as these are quantities associated with a configuration space. In fact, in the Lagrangian formalism it's important that they are not the time dependent quantities associated with the system's evolution, but abstract quantities in configuration space.

The gravitational term depends on $r$ and $\theta$.

 

[PhDeezNutz]

@PeroK I definitely agree. IIRC the EL equation (at least the one I know) was derived without an explicit dependence on $t$.

 

[Delta2]

It's not usual to express the quantities in a Lagrangian as a function of time, as these are quantities associated with a configuration space. In fact, in the Lagrangian formalism it's important that they are not the time dependent quantities associated with the system's evolution, but abstract quantities in configuration space.

The gravitational term depends on $r$ and $\theta$.

I don't fully understand this comment, when we going to add the kinetic energy term of Lagrangian we going to use some $\dot r$ and $\dot \theta$ terms with the dot denoting the derivative with respect to time...

 

[PeroK]

I don't fully understand this comment, when we going to add the kinetic energy term of Lagrangian we going to use some $\dot r$ and $\dot \theta$ terms with the dot denoting the derivative with respect to time...

In the Lagrangian formalism $r$, $\dot r$ and $t$ are independent abstract variables.

In particular, $\dot r \not \equiv \frac{ dr} {dt}$.

PS This particular Lagrangian is time-independent.

 

[wrobel]

when we write $\frac{\partial L}{\partial \dot x}$ we consider $\dot x$ as an independent variable;
when we write $\frac{d}{dt}\frac{\partial L}{\partial \dot x}$ we consider both $\dot x,x$ as functions of time

 

[Delta2]

Yes ok I think I get it now, it is like when we have the function$f(x,y)$and we form the function $g(t)=f(x(t),y(t))$then according to the chain rule it will be $$\frac{dg}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$ so though x and y are both functions of time so they can't be independent, we treat them as independent variables when taking the partial derivatives of f with respect to x and y.

 

[PeroK]

when we write $\frac{\partial L}{\partial \dot x}$ we consider $\dot x$ as an independent variable;
when we write $\frac{d}{dt}\frac{\partial L}{\partial \dot x}$ we consider both $\dot x,x$ as functions of time

To add to this, once we have the abstract quantities $\frac{\partial L}{\partial x}$ and $\frac{\partial L}{\partial \dot x}$, we then transform them to time-dependent dynamic quantities and assert that when we evaluate these quantities along a trajectory that represents a valid solution of the system, then the following equation holds:
$$\frac{d}{dt}\frac{\partial L}{\partial \dot x} = \frac{\partial L}{\partial x}$$Which we should all agree is the product of inspirational human thought!

 

[Orodruin]

In the Lagrangian formalism $r$, $\dot r$ and $t$ are independent abstract variables.

In particular, $\dot r \not \equiv \frac{ dr} {dt}$.

PS This particular Lagrangian is time-independent.

More specifically, the Lagrangian is a (possibly also time dependent - although not in this case) function on the tangent bundle of configuration space. The action along a curve $\gamma$ in configuration space is then defined as the time integral of the Lagrangian evaluated along the corresponding curve $(\gamma,\dot\gamma)$ in the tangent bundle, where $\dot\gamma$ is the tangent vector of $\gamma$. The equations if motion that single out a particular time evolution are then given by the principle of stationary action.

 

[Delta2]

More specifically, the Lagrangian is a (possibly also time dependent - although not in this case) function on the tangent bundle of configuration space. The action along a curve $\gamma$ in configuration space is then defined as the time integral of the Lagrangian evaluated along the corresponding curve $(\gamma,\dot\gamma)$ in the tangent bundle, where $\dot\gamma$ is the tangent vector of $\gamma$. The equations if motion that single out a particular time evolution are then given by the principle of stationary action.

Sorry for this as I haven't read any good textbook on Lagrangian Mechanics, if we want to calculate the action for some path in the configuration space, we 'll replace in the Lagrangian $q=q(t)$ and $\dot q=\frac{dq}{dt}$ or we can take $q,\dot q$ as independent and replace them with whatever functions of t we like, I mean not necessarily the one being the derivative/integral of the other?

 

[Orodruin]

Although you could technically compute the action for any path in the tangent bundle, the principle of least action works with paths on configuration space and their corresponding path in the tangent bundle.

If you would try arbitrary paths in the tangent bundle you would just end up with having all partial derivatives of the Lagrangian equal to zero because $\delta q$ and $\delta \dot q$ would be independent variations.