- #1
StudentOfScience
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Homework Statement
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.
Other relevant threads:
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)
Homework 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).
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)