Lagrangian mechanics, system of a spring and a pendulum

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  • #1
TimmyD1
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Homework Statement:
System of spring and pendulum
Relevant Equations:
L=T-V
1.png

Hello! I have some problem getting the correct answer for (b).
My FBD:
2.png
For part (a) my lagrangian is
$$L=T-V\iff L=\frac{1}{2}m(b\dot{\theta})^2+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2,\ where\ \boldsymbol{x}=\sqrt{(1.25b-b)^2+(b\sin\theta)^2}-(1.25b-0.25b)$$

Hence my equation of motion become

$$\frac{1}{2}mb^2\ddot{\theta}+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2=0$$

solving this gives me the spring coefficient $$k=111.92\ N/m$$.

for part (b):

when $$\theta=25^{\circ}\ and\ k=100\ N/m$$ then

$$\frac{1}{2}mb^2\dot{\theta}\frac{d\dot{\theta}}{d\theta}+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2=0\iff mb^2\dot{\theta}\frac{d\dot{\theta}}{d\theta}+2mgb(1-\cos\theta)-k\boldsymbol{x}^2=0\iff \dot{\theta} d\dot{\theta} = \Big[\frac{k\boldsymbol{x}^2}{mb^2}-\frac{2g}{b}(1-\cos\theta)\Big]\ d\theta$$

$$\int_0^{\dot{\theta}} \dot{\theta} d\dot{\theta} = \int_0^{\frac{25\pi}{180}}\underbrace{\Big[\frac{k\boldsymbol{x}^2}{mb^2}-\frac{2g}{b}(1-\cos\theta)\Big]}_{-1.7001}\ d\theta\Longrightarrow \frac{\dot{\theta}^2}{2}=0.74384\iff \dot{\theta}=\sqrt{2\cdot 0.74384}=1.2197\ rad/s\Longrightarrow v=r\omega=b\dot{\theta}=0.4\cdot1.2197\approx 0.48788\ m/s$$

The answer should be v = 0.522 m/s but I can't seem to get it right, would really appreciate it if anyone could help me out!
 

Answers and Replies

  • #2
Why do you want to use the Lagrangian formalism here? You don't need to determine the equations of motion, you can just conserve mechanical energy in a single equation.
 
  • #3
TimmyD1
10
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The exercise is to use lagrangian formalism, I have solved it using energy principle but using the lagrangian i ran into problems.
 
  • #4
I don't think your spring potential energy is correct. I assume the ##x## on your diagram labelled next to the curved line segment is the arc length traversed by the mass (in which case I have no idea why you use Pythagorean theorem... it should just be ##x = b\theta##, not ##x=\sqrt{(1.25b-b)^2+(b\sin\theta)^2}##), however this length isn't relevant to the extension of the spring.

(N.B. Also you should not use bold font ##\mathbf{x}## since it is not a vector; just call it ##x##)

The distance you actually want is the magnitude of the vector ##|\overrightarrow{AB}|##,$$|\overrightarrow{AB}| = b\sqrt{\frac{41}{16} - 2.5 \cos{\theta}}$$from this you subtract off ##0.25b## to get the extension ##\delta## of the spring,$$\delta = |\overrightarrow{AB}| - 0.25b = b \left( \sqrt{\frac{41}{16} - 2.5 \cos{\theta}} - 0.25 \right)$$Then the Lagrangian should read$$L=\frac{1}{2}mb^2 \dot{\theta}^2+mgb(1-\cos\theta)-\frac{1}{2}k b^2 \left( \sqrt{\frac{41}{16} - 2.5 \cos{\theta}} - 0.25 \right)^2$$
 
  • #5
TimmyD1
10
1
I ran into the same integral using the lagrangian you defined
 
  • #6
Really? Because you defined some parameter ##x## as$$x =\sqrt{(1.25b-b)^2+(b\sin\theta)^2}-(1.25b-0.25b)$$which doesn't seem to make sense geometrically and doesn't equal my extension ##\delta##. When you plug ##\theta = 25^o## into your ##x##, you get a negative number, which should be a sign that something is off.
 
  • #7
TimmyD1
10
1
Sorry it's a typo that I just realized,

$$x=\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}-0.25b$$

if we can ignore the notation and geometric point of view for now, it just bothers me why I can't get the same answer using lagrangian. I think that using $$v=r\omega$$ relation messes it up, because I can't seem to find the error.
 
  • #8
Sorry it's a typo that I just realized,

$$x=\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}-0.25b$$

That's the correct extension, yes :smile:

But I'll note that what you have labelled ##x## in your diagram is not this length; the ##x## in your diagram equals ##b\theta##, i.e. the length of the curved line segment. This expression you have just written is actually ##|\overrightarrow{AB}| - 0.25b##.

With that cleared up, now let's derive the equations of motion for the Lagrangian:$$L=\frac{1}{2}m(b\dot{\theta})^2+mg(b-b\cos\theta)-\frac{1}{2}k(\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}-0.25b)^2$$We have that$$\frac{\partial L}{\partial \dot{\theta}} = mb^2 \dot{\theta} \implies \frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = mb^2 \ddot{\theta}$$ $$\frac{\partial L}{\partial \theta} = mgb\sin{\theta} - \frac{k(\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}-0.25b)}{2\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}} \left( 2b\sin{\theta} (1.25b - b\cos{\theta}) + 2b^2\sin{\theta} \cos{\theta}\right)$$where I haven't simplified that final expression and hope I haven't made a mistake somewhere. To find equation of motion you need to solve the ODE resulting from $$ \frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = \frac{\partial L}{\partial \theta}$$

I think that using $$v=r\omega$$ relation messes it up, because I can't seem to find the error.

No your kinetic energy term is completely fine. You can get it by working in polars, e.g. ##T = \frac{1}{2}m \vec{v} \cdot \vec{v} = \frac{1}{2}m (r\dot{\theta} \hat{\theta}) \cdot (r\dot{\theta} \hat{\theta}) = \frac{1}{2}mr^2 \dot{\theta}^2##.
 
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  • #9
TimmyD1
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So the acceleration would simply be

$$\ddot{\theta}=\frac{\dfrac{\partial L}{\partial \theta}}{mb^2}$$

now I would assume to integrate the expression to get velocity but not sure what the bounds would be

$$\int_0^{\dot{\theta}}\ddot{\theta}d\ddot{\theta}=\int_0^{\frac{25\pi}{180}}\frac{\dfrac{\partial L}{\partial \theta}}{mb^2}\ d\theta\ ?$$
 
  • #10
$$\int_0^{\dot{\theta}}\ddot{\theta}d\ddot{\theta}=\int_0^{\frac{25\pi}{180}}\frac{\dfrac{\partial L}{\partial \theta}}{mb^2}\ d\theta\ ?$$

If you have the equation$$\ddot{\theta} = \frac{1}{mb^2} \frac{\partial L}{\partial \theta}$$then both sides must be integrated with respect to the same variable*; but you integrated the LHS w.r.t. ##\ddot{\theta}## and the RHS w.r.t. ##\theta##, which is an incorrect step.

*(As an example, remember that if ##f=g##, then ##\Gamma(f) = \Gamma(g)##. You can define ##\Gamma(f) := \int_{x_1}^{x_2} f(x) dx##. It follows that you can integrate both sides w.r.t. the same variable)

What you might think of is to try$$c+\dot{\theta} = \int \ddot{\theta} dt = \frac{1}{mb^2} \int \frac{\partial L}{\partial \theta} dt$$however you can't do this because the RHS depends on ##\theta##. This way won't work.
 
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  • #11
TimmyD1
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So with the help of MATLAB, I've found that

$$\frac{1}{mb^2}\int\frac{\partial L}{\partial \theta}dt\approx 0.83719t+C$$

What next!?
 
  • #12
No you can't do it, because ##\frac{\partial L}{\partial \theta}## is a function of ##\theta##. How are you going to integrate with respect to time? You will require another approach.
 
  • #13
To reiterate, you want to solve the (unsimplified) differential equation in ##\theta(t)##,$$mb^2 \frac{d^2 \theta}{dt^2} = mgb\sin{\theta} - \frac{k(\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}-0.25b)}{2\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}} \left( 2b\sin{\theta} (1.25b - b\cos{\theta}) + 2b^2\sin{\theta} \cos{\theta}\right)$$
 
  • #14
TimmyD1
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Alright! Before I tackle this, do we have any initial conditions?
 
  • #15
You will need to specify a finite ##\theta(0) = \theta_0##. Because if ##\theta(0) = 0## then the system will remain at rest indefinitely. You can also set ##\dot{\theta}(0) = 0##.

Once you've done that, and checked that I didn't make a mistake in ##\partial L/\partial \theta##, then Mathematica's probably the way to go...
 
  • #16
TSny
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So the acceleration would simply be
$$\ddot{\theta}=\frac{\dfrac{\partial L}{\partial \theta}}{mb^2}$$
A trick is to multiply both sides by ##\dot \theta##.
Each side can then be expressed as the time derivative of some expression (a different expression for each side).
 
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  • #17
TSny
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A trick is to multiply both sides by ##\dot \theta##.
It helps to note that for your particular Lagrangian, ##\frac{\partial L}{\partial \theta} = -\frac{\partial V}{\partial \theta}##.

As @etotheipi noted, you obtain a first-order differential equation in ##\dot \theta## and ##\theta##.
It expresses conservation of _____________ (fill in the blank :oldsmile:).
 
  • #18
TimmyD1
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I could not solve the differential equation and I think finding the acceleration of the system will be enough for me!

Amazing guidance by the way!
 
  • #19
TimmyD1
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It helps to note that for your particular Lagrangian, ##\frac{\partial L}{\partial \theta} = -\frac{\partial V}{\partial \theta}##.

As @etotheipi noted, you obtain a first-order differential equation in ##\dot \theta## and ##\theta##.
It expresses conservation of _____________ (fill in the blank :oldsmile:).

So I thought more about this

$$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}=mb^2\ddot{\theta}=\dfrac{\partial L}{\partial\theta}$$ Multiply both sides with $$\dot{\theta}$$

gives

$$\frac{1}{2}mb^2\frac{d\dot{\theta}^2}{dt}=\dfrac{\partial L}{\partial \theta}\dot{\theta}=-\frac{dV}{d\theta}\dot{\theta}=-\frac{dV}{dt}$$ such that $$\dfrac{d}{dt}\Big(\frac{1}{2}mb^2\dot{\theta}+V\Big)=0$$ which gives us a constant.
 
  • #20
$$\dfrac{\partial L}{\partial \theta}\dot{\theta}=-\frac{dV}{d\theta}\dot{\theta}=-\frac{dV}{dt}$$
I think you should have a partial here instead,$$\dfrac{\partial L}{\partial \theta}\dot{\theta}=-\frac{\partial V}{\partial \theta}\dot{\theta}=-\frac{dV}{dt}$$but that's not such a big deal :wink:
$$\dfrac{d}{dt}\Big(\frac{1}{2}mb^2\dot{\theta}+V\Big)=0$$
You miss a square (probably just typo, since you had it in previous line)$$\dfrac{d}{dt} \left(\frac{1}{2}mb^2\dot{\theta}^2+V \right)=0$$And ##\frac{1}{2}mb^2\dot{\theta}^2+V## is a special constant, it is the energy ##E##!
 
  • #21
TimmyD1
10
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Another way of looking at the solution is to study what is usually called the energy function

$$h(q_1...\dot{q}_n,\dot{q}_1...\dot{q}_n,t)=\sum_j\dot{q}_j\frac{\partial L}{\partial \dot{q}_j}-L$$

The energy function is derived as one term from the expression for

$$\frac{dL}{dt}$$


when replacing (from Euler Lagrange :)

$$\frac{\partial L}{\partial q_j}=\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_j}$$

The energy function is identical with the Hamiltonian (albeit in disguised form), cf.


$$H(q_j,p_j,t)$$

If L does not depend explicitly on time, but only implicitly, ie the second term ##\dfrac{\partial L}{\partial t}=0## in the derivation, ##h## is conserved. We can set it to an arbitrary value (the potential is arbitrarily determined), e.g. 0.

For our specific case, the following applies:

$$h(\theta,\dot{\theta})=\dot{\theta}\frac{\partial L}{\partial \dot{\theta}}-L=0$$

With inserted expressions
$$T=\frac{mb^2\dot{\theta}^2}{2},\ V=\frac{kx^2}{2}-mgb(1-cos\theta)\leadsto mb^2\dot{\theta}^2-(\frac{1}{2}mb^2\dot{\theta}^2-V)=0$$

one can add that the last equation is the same as ##T + V = 0##, which means that we have shown that the energy of the system, the sum of the kinetic and the potential energy, must be constant. A constant we can choose arbitrarily through the potential.
 
  • #22
Fred Wright
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Two observations about your problem:
1. In the diagram below
springpend.jpg

##d^2=(1.25b)^2+ b^2 - 2(1.25b)b\cos(\theta)##

##d(\theta)=b\sqrt (2.5625-2.5\cos(\theta))##

The spring extension is ##d(\theta)-.25b## and therefore the spring potential energy is ##\frac{1}{2}k(b\sqrt (2.5625-2.5\cos(\theta)) - .25b)^2##



2. You can show, using the chain rule,
##\ddot{\theta}=\dot{\theta}\frac{d\dot{\theta}}{d\theta}##

From this if you have,$$\ddot{\theta}=f(\theta)$$
you can integrate to get$$\int_{\dot{\theta_{init}}}^{\dot{\theta_{final}}}\dot{\theta}d\dot{\theta}=\int_{\theta_{init}}^{\theta_{final}}f(\theta)d\theta$$
 
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