# Spring Pendulum with Drag: Newtonian and Lagrangian Approaches

1. May 9, 2017

### StudentOfScience

1. The problem statement, all variables and given/known data
Consider a point mass of mass m suspended from an ideal, massless spring. Let $\theta$ be measured from the vertical. Find the displacement of the mass as a function of time if the spring is initially stretched/compressed a distance $l_0$ and has an initial velocity $v_0$. Do not ignore air drag.

https://www.physicsforums.com/threads/spring-pendulum-with-friction-lagrange.316730/ (the model for friction in this problem is different from the one I have adopted for this problem, thus the equations of motion are different)
https://www.physicsforums.com/threads/spring-pendulum.404865/ (3 dimensions and spherical coordinates, but not including drag considerations)

2. Relevant equations
$$\vec F_{drag} = -\frac{1}{2} \alpha | \vec v |^2\hat v$$ where alpha is dependent on the drag coefficient and other parameters (I am assuming drag force is proportional to velocity squared)
Since the angle is measured from the vertical, the polar base vectors are
$$\hat r = \sin\theta \hat i -\cos\theta \hat j, \hat \theta = \cos\theta \hat i + \sin\theta \hat j$$
The polar acceleration equation is still the same, though (since $\frac{d\hat r}{dt}= \dot \theta \hat \theta, \frac{d\hat \theta}{dt} = -\dot \theta \hat r$)
Here, $l(t)$ is r (the distance from the origin; that is, the length of the spring at time t).

The position of the pendulum bob is given by
$$\vec r(t) = x(t) \hat i+ y(t)\hat j = l(t) \hat r, x(t)=l(t)\sin\theta(t), y(t)=-l(t)\cos\theta(t)$$

$$\vec v(t) = \dot l \hat r + r\dot \theta \hat \theta, \vec a(t) = (\ddot l -l \dot \theta^2) \hat r + (l \ddot \theta + 2\dot l \dot \theta) \hat \theta$$

$L=T-V$

Since drag can't be expressed as a potential all that well, the Euler-Lagrange equations get modified:

$$\frac{d}{dt} \frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j} = Q_j$$
Here, Q is the generalized force from friction (not sure if this is right; I haven't formally learned the Lagrangian yet, but I include the Lagrangian approach since it is many times substantially easier than an F=ma approach). So $Q_j = - \frac{\partial |\vec F_{drag}|}{\partial \dot q_j}$ ( I just looked this up; I'll read the justification later on).

3. The attempt at a solution
I'll start with a Newtonian approach. Gravity resolved in radial and angular components is

$$F_{gr} = mg\cos\theta , F_{g\theta}=-mg\sin\theta$$
For drag, $\hat v$ is somewhat troublesome. Intuitively, I would think $\vec v$ (and thus $\hat v$) is always parallel to the trajectory and thus would only have an angular component. However, the math leads me to believe otherwise when I calculate the components of the drag force:

$$\vec F_{drag} = -\frac{1}{2} \alpha | \vec v |^2\hat v = -\frac{1}{2} \alpha (\dot l^2+l^2\dot \theta^2) \hat v$$

$$\hat v = \frac{\vec v}{|\vec v|} = \frac{\dot l \hat r+l\dot \theta \hat \theta}{\sqrt{\dot l^2+l^2\dot \theta^2}}$$
Where I have used the velocity in polar coordinates.

Intuitively, $\hat v = \pm \hat \theta$ where the plus-minus is there since the bob may change direction and thus the velocity may go in the opposite direction. I say this because at any instant, the velocity vector is tangent to the path, which makes the velocity vector parallel (or anti-parallel, depending on the sign) to $\hat \theta$. I am not sure why there there is a radial component in the previous equation.

For the spring force,

$$\vec F_{sp} = -k(l(t)-l_{eq}) \hat r$$ where $l_{eq}$ is the equilibrium length (which need not equal $l_0=l(t=0)$.

The equations of motion are, for the radial and angular directions respectively:

$$\ddot l -l \dot \theta^2 = mg\cos\theta - \frac{1}{2} \alpha \dot l \sqrt{ \dot l+l^2\dot \theta^2} - k(l-l_{eq})$$
$$l \ddot \theta + 2\dot l \dot \theta = -mg\sin\theta - \frac{1}{2} \alpha l\dot \theta \sqrt{\dot l+l^2\dot \theta^2}$$

Which I am almost certain must be done numerically; I wouldn't imagine an analytical solution exists. The above equations of motion make use of the drag equation that I am not sure of.
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Now for the Lagrangian approach.

$$L=T-V \rightarrow L= \frac{1}{2} m(\dot l^2+l^2\dot \theta^2)-mgy -U_{sp} \rightarrow L= \frac{1}{2} m(\dot l^2+l^2\dot \theta^2)+mgl\cos\theta -U_{sp}$$

Where
$$U_{sp} = -W = - \int_{l_0}^{l(t)} \vec F_{sp} \cdot d\vec l$$
Carrying out the integration over a straight line (I haven't really dealt with line integrals before; I'm 'improvising' here since I haven't learned the theory of multivariable real-valued functions yet)

$$U_{sp} = \int_{l_0}^{l(t)} k(l(t)-l_{eq})dl = (\frac{1}{2} k l^2(t)-l_{eq}l)-(\frac{1}{2}kl_0^2-l_{eq}l_0)$$
For aesthetic's sake, let $\gamma_0 = -(\frac{1}{2}kl_0^2-l_{eq}l_0)$

So our Lagrangian is

$$L=\frac{1}{2} m(\dot l^2+l^2\dot \theta^2)+mgl\cos\theta-(\frac{1}{2} k l^2(t)-l_{eq}l)-\gamma_0$$

Let the derivatives begin. I will include an image of the computation; if you would like this in LaTeX, let me know.

Which results in the equations of motion:

$$m \ddot l -(m(\dot \theta)^2-k)l-mg\cos\theta-l_{eq}=-\alpha\dot l$$
$$ml^2\ddot \theta +mgl\sin\theta = -\alpha l^2 \dot \theta$$

• Why are the two sets of motion equations different? Shouldn't they give the same result?
• For the Newtonian approach, does the drag force have both a radial and angular component or just an angular one? (See discussion above for more)

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2. May 11, 2017

### StudentOfScience

Bump.

Also note a correction: the Newtonian equations of motion should actually read (I forgot to include the m on the side of the accelerations):

$$m(\ddot l -l \dot \theta^2) = mg\cos\theta - \frac{1}{2} \alpha \dot l \sqrt{ \dot l+l^2\dot \theta^2} - k(l(t)-l_{eq})$$
$$m(l \ddot \theta + 2\dot l \dot \theta) = -mg\sin\theta - \frac{1}{2} \alpha l\dot \theta \sqrt{\dot l+l^2\dot \theta^2}$$

However, these equations are still not equivalent to the ones obtained with the Lagrangian approach. Could you all let me know where I went wrong?

Also, I'v read a little bit of Fitzpatrick's chapter on the chaotic pendulum where he analyzes the simple pendulum with drag. I realize this thread is related to the spring pendulum with drag proportional to v^2, but I'll include how he analyzes drag (where he uses the proportional model rather than v^2 model) since it may shed light on this problem.

See the uploaded image for the equations of motion from Fitzpatrick. It appears he is using v as a drag coefficient. So that we don't confuse this with velocity, I'll redefine it as $\beta$.

$$\vec F_{drag} = -\beta |\vec v| \hat v$$

For the simple pendulum, we have

$$\vec F_{drag} =- \beta R\dot \theta\hat \theta$$

where I have not kept the absolute value on the theta dot after taking the magnitude. This is because if theta dot is positive, then the velocity vector is parallel to $\hat \theta$, thus making drag negative. If theta dot is negative, then drag will be parallel to $\hat \theta$ and thus there is no need for plus-minuses. This finally makes sense to me. Thus, the equations of motion are those obtained in the picture below. However, it does seem that he doesn't include the R (l in the picture) on the drag force. Is that part of the drag coefficient?

However, for this problem, do I have to consider the radial components of the drag force since the length of the spring changes (see questions in the previous post, also)?

Thank you all

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3. May 11, 2017

### haruspex

Yes, but that does not make it parallel to $\hat\theta$. Consider e.g. just bobbing up and down vertically.

4. May 11, 2017

### zwierz

I did not check the text; just a single remark
$$Q_j=\Big(\boldsymbol F_{drag},\frac{\partial\boldsymbol v}{\partial \dot q_j}\Big)$$

5. May 11, 2017

### StudentOfScience

Ah yes, thank you for pointing that out. This makes more sense now; velocity will only be parallel/anti-parallel to $\hat \theta$ if the position vector is normal to the trajectory. I suppose some of my sketching led me astray. Of course it may happen that at some points along the trajectory the position vector may be normal or tangent to the trajectory (resulting in the velocity just having an angular or radial component, respectively), but we don't know this until we analyze the equations of motion to see when $\dot l$ or $\dot \theta$ are 0.

Nevertheless, I think - but I am not sure - that my mistake lies with my solution using the Lagrangian approach as I am now fairly confident in the Newtonian equations of motion (one can never be too confident I suppose, as it is very easy to lose track of a minus sign, an all-too common occurrence).

I used $Q_j = - \frac{\partial |\vec F_{drag}|}{\partial \dot q_j}$ where I used the drag equation that was mentioned earlier (and used in the Newtonian approach):
$$\vec F_{drag} = - \frac{1}{2} \alpha |\vec v|^2 \hat v= -\frac{1}{2} \alpha (\dot l^2+l^2\dot \theta^2) \frac{\dot l \hat r+l\dot \theta \hat \theta}{\sqrt{\dot l^2+l^2\dot \theta^2}} = -\frac{1}{2} \alpha \sqrt{\dot l^2+l^2\dot \theta^2}(\dot l \hat r+l\dot \theta \hat \theta)$$

Any input? Thank you

6. May 12, 2017

### zwierz

that is wrong

7. May 13, 2017

### zwierz

Unfortunately in courses of mechanics it is paid very few attention to generalized forces. So students commit many errors while calculating the generalized forces in different problems.
Especially this concerns to problems with rigid bodies. For the case of particles formulas for calculating of generalized forces are well known and mentioned in many places.
For example, the following formula is little known. Assume the system consists of a rigid body and let $A$ be a point of the rigid body. The velocity of this point is $\boldsymbol v_A$; and $\boldsymbol M_A$ is a net torque applied to the rigid body about the point $A$ ; $\boldsymbol F$ is a net force applied to the rigid body. Both vectors are calculated excluding constraint forces. Then
$$Q_j=\Big(\boldsymbol F,\frac{\partial \boldsymbol v_A}{\partial \dot q_j}\Big)+\Big(\boldsymbol M_A,\frac{\partial \boldsymbol \omega}{\partial \dot q_j}\Big)$$
For example if we consider a spinning top which position is given by Euler angles, and this top experiences a drag torque then we can easily convert this drag torque into generalized forces by using written above formula.

8. May 14, 2017

### StudentOfScience

Thank you for pointing that out! Currently, I have not formally studied Euler angles and tops (and all the other complexities in rigid body motion like the inertia tensor). If you are wondering, I am using Morin and Kleppner as a primary texts (more-so Morin) for self-study. I'm not too far into the texts (obviously, as indicated by my limited knowledge). Would supplementing Morin with some sections in Goldstein/Landau vol 1 give me more information on generalized forces and the derivations of such rare formulas? As it seems, I probably will have to postpone the Lagrangian consideration of the problem until my understanding of generalized forces is ironed out. Again, thank you.