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Homework Help: Displacement Function of an Idealized Circular Loop Coaster

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose we have a roller coaster cart of mass m, which may be approximated as a point mass for our purposes. Ignoring friction and air-drag, what is the displacement of the cart as a function of time? It enters the loop with an initial velocity $$ v_0 \hat i $$ and the radius of the circular loop is R. Also, for the purposes of this problem, assume that once the cart enters the loop, it circles the loop infinitely (we are ignoring losses: friction and air drag). After finding the displacement function, consider the limiting cases (letting mass, radius, initial velocity, time approach infinity, for instance).

    2. Relevant equations
    Acceleration in polar coordinates (I'll define the center of the loop as the origin):
    $$ \vec a= (\ddot r - r(\dot \theta)^2)\hat r + (r \ddot \theta +2 \dot r \dot \theta) \hat \theta $$
    Here, due to how we've defined the origin, $$\dot r = \ddot r = 0 $$ I have previously seen ## \ddot \theta ## being 0 before in certain situations, but I am not certain of the reasoning behind this or even if it applies to this problem; thus, I will leave ## \ddot \theta ## in the equation.

    3. The attempt at a solution
    Ok, let's start with forces (I haven't learned the Lagrangian method yet, so an energy consideration to find the displacement function is off the table) at an arbitrary point along the loop: normal force and gravity. In polar coordinates, we have

    $$ \vec a= \frac{1}{m} ([F_{nr}+F_{gr}]\hat r + [F_{n\theta}+F_{g\theta}]\hat \theta) = - r(\dot \theta)^2 \hat r + r \ddot \theta \hat \theta $$
    where the angular component of the normal force* is simply the angular displacement ( ##\theta (t)\hat \theta ##), and the angular component of the force of gravity on the cart is ## \frac{3\pi}{2} \hat \theta ##, if we measure the angle counterclockwise from the horizontal. I will go ahead and equate the angular components to formulate a differential equation (I am avoiding the radial component since I do not know what the the magnitude of this component of the normal force would be):

    $$ \frac{1}{m}(F_{n\theta}+F_{g\theta}) \rightarrow \frac{1}{m}(\theta (t) +\frac{3\pi}{2})=R\frac{d^2\theta}{dt^2} $$
    *One may also write the angular component of the normal force as ## \theta(t) +\pi ##. This way, one doesn't need to account for the inward pointing vector with a minus sign on the radial component; the ## + \pi ## takes care of that. I would think that they give the same answers if either one was used in the ensuing discussion.

    Now for some math. Define
    $$ \omega(t)=\frac{d\theta}{dt} $$
    Which means that
    $$ \omega \frac{d\omega}{d\theta} = \frac{d^2\theta}{dt^2} $$
    So we have
    $$ R\omega \frac{d\omega}{d\theta}=\frac{1}{m} (\theta+\frac{3\pi}{2})$$
    $$ R\int_{\omega_0}^{\omega} \omega' d\omega' = \int_{\theta_0}^{\theta} \frac{1}{m} (\theta'+\frac{3\pi}{2})d\theta' $$
    where $$\omega_0=\frac{v_0}{R}, \theta_0=\frac{3\pi}{2} $$ (the initial angular velocity is easily derived in terms of the magnitude of the 'total' initial velocity).

    $$ R\frac{1}{2}(\omega^2-\omega_0^2)= \frac{1}{m}(\frac{1}{2}\left(\theta ^2-\theta _0^2\right)+\frac{3\pi }{2}\left(\theta -\theta _0\right)) $$
    Which means

    $$ \omega(\theta) = \pm (\frac{1}{Rm}\left(\left(\theta ^2-\theta _0^2\right)+3\pi \left(\theta -\theta _0\right)\right)+\omega_0^2))^{\frac{1}{2}} $$

    Now, we can say (I think)

    $$ \omega (\theta) = \frac{d\theta}{dt} $$

    Which yields

    $$ \int_{t_0=0}^{t} dt' = \pm \int_{\theta_0}^{\theta} \frac{d\theta'}{(\frac{1}{Rm} [\theta'^2+3\pi\theta'-\theta_0^2-3\pi\theta_0]+\omega_0^2)^{\frac{1}{2}}} $$
    And if we define
    $$ \gamma = -\theta_0^2-3\pi\theta_0+Rm\omega_0^2 $$ to clean up our integral a bit,

    $$ \mp t= \sqrt {mR} \int_{\theta_0}^{\theta} \frac{d\theta'}{(\theta'^2+3\pi\theta'+\gamma)^{\frac{1}{2}}} $$
    Which integrates to

    $$ \mp \frac{1}{\sqrt {mR}} t = \log \left|\frac{u+\sqrt{1+u^2}}{u_0+\sqrt{1+u_0^2}} \right| $$


    $$ u=\frac{\theta +\frac{3\pi}{2}}{\sqrt{\gamma-\frac{9\pi^2}{4}}}, u_0=\frac{\theta_0 +\frac{3\pi}{2}}{\sqrt{\gamma-\frac{9\pi^2}{4}}} $$

    $$ u=\frac{ (u_0+\sqrt{1+u_0^2})^2\exp(\mp 2\frac{1}{\sqrt {mR}} t)-1}{2((u_0+\sqrt{1+u_0^2}))\exp(\mp \frac{1}{\sqrt {mR}} t)} $$

    Which means,

    $$ \theta(t)= \left (\sqrt{\gamma-\frac{9\pi^2}{4}} \right) \frac{ \left (\frac{\theta_0 +\frac{3\pi}{2}}{\sqrt{\gamma-\frac{9\pi^2}{4}}}+\sqrt{1+(\frac{\theta_0 +\frac{3\pi}{2}}{\sqrt{\gamma-\frac{9\pi^2}{4}}})^2} \right)^2\exp(\mp 2\frac{1}{\sqrt {mR}} t)-1}{2 \left (\left(\left (\frac{\theta_0 +\frac{3\pi}{2}}{\sqrt{\gamma-\frac{9\pi^2}{4}}} \right)+\sqrt{1+(\frac{\theta_0 +\frac{3\pi}{2}}{\sqrt{\gamma-\frac{9\pi^2}{4}}})^2} \right) \right)\exp(\mp \frac{1}{\sqrt {mR}} t)}-\frac{3\pi}{2} $$

    But since $$\theta_0=\frac{3\pi}{2}, \omega_0=\frac{v_0}{R} $$, we have

    $$ u_0= \frac{3\pi}{\sqrt{\frac{v_0^2m}{R}-9\pi^2}} $$

    $$ \theta \left(t\right)=\frac{\frac{3\pi }{\sqrt{\frac{v_0^2m}{R}-9\pi ^2}}+\sqrt{1+\left(\frac{3\pi }{\sqrt{\frac{v_0^2m}{R}-9\pi ^2}}\right)^2}\exp \left(\frac{\mp 2}{\sqrt{mR}}t\right)-1}{2\left(\frac{3\pi }{\sqrt{\frac{v_0^2m}{R}-9\pi ^2}}+\sqrt{1+\left(\frac{3\pi }{\sqrt{\frac{v_0^2m}{R}-9\pi ^2}}\right)^2}\right)\exp \left(\frac{\mp 1}{\sqrt{mR}}t\right)} -\frac{3\pi}{2} $$

    Of course, now with ## \theta(t) ##, we have the displacement function:

    $$ \vec r= R\hat r = R(\cos(\theta(t))\hat i+\sin(\theta(t))) $$

    Now, let's look at some limiting cases. The first concern is all the minus-pluses. I am not sure how to approach this, what should I do with the four cases?

    Limiting cases
    $$ v_0 \rightarrow \infty, \theta(t) \rightarrow \frac{\exp \left(\frac{\mp 2}{\sqrt{mR}}t\right)-1}{2\exp \left(\frac{\mp 1}{\sqrt{mR}}t\right)}-\frac{3\pi}{2} = \frac{1}{2}\exp \left(\frac{\mp 2}{\sqrt{mR}}t\pm\frac{1}{\sqrt{mR}}t\right)-\frac{1}{2\exp \left(\frac{\mp 1}{\sqrt{mR}}t\right)}-\frac{3\pi}{2}$$

    Now for a physical interpretation. Maybe I can 'rule out' some of the plus-minus combinations with some physical intuition (however, this is not rigorous at all). Since we have 3 plus minuses, we have ## 2^3 ## combinations: plus-plus-plus, plus-minus-plus, plus-plus-minus, plus-minus-minus, minus-plus-plus, minus-minus-plus, minus-plus-minus, minus-minus-minus.

    I am also curious what is the physical meaning when ## \frac{v_0^2m}{R} < 9\pi^2 ##.

    In summary:
    • I don't know how to handle all the plus-minuses and what they mean exactly. Should I consider the eight cases mentioned above and see which ones work (as in making sense physically)? Or, is there another, more rigorous way?
    • The physical meaning when ## \frac{v_0^2m}{R} < 9\pi^2 ##. Does this correlate to the cart not making it all the way around the loop?
    • When I graph (yes, it's better to do it by hand; I did it out of convenience though) the displacement function as a pair of parametric equations, it seems that the point that is being traced follows a circular path counter-clockwise (what is expected). But then, for some reason, it starts to trace backwards? And this is when the parameter is increasing values. I am not sure why that is happening. Does it have to do with the periodicity of the sine and cosine functions?

    Thank you all so much for taking the time to read my questions and work, I really appreciate it.
    Last edited: May 3, 2017
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  3. May 4, 2017 #2


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    That doesn't make sense dimensionally. How did you get this?

    I would avoid working with forces and start with energy. Then you only have one integral to do. But be careful that you avoid the normal force becoming negative.
  4. May 4, 2017 #3
    Ok, I decided to use polar coordinates so I equated the angular components of the forces with the angular force: m times the angular acceleration (## \dot r = 0 ##):

    $$ F_{n\theta}+F_{g\theta} = mR\frac{d^2\theta}{dt^2} $$

    Perhaps the m term is off. Since the gravitational force points straight down, we can say the that the angular component is ## \frac{3\pi}{2} ## (in this case, the radial component is positive mg). Also, I figured that the angular component of the normal force is ## \theta(t) ## (if we say the radial component of the normal force is negative: it points inward).

    Hmm, if I did energy, I would need the Lagrangian to get the equations of motion. I haven't learned it yet; I was saving it for a bit later when I felt ready for it. However, I could try to solve the problem using the Lagrangian for the first time (with Morin next to me of course), but I doubt that will go well. Thanks
  5. May 4, 2017 #4


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    I don't understand what you mean by this. How can a component of a force be an angle?
  6. May 4, 2017 #5
    Normally, $$ \vec F_g=-mg \hat j $$

    $$ -mg \hat j = mg \hat r $$
    where the ## \theta## is ## \frac{3\pi}{2} ##, which makes the cosine in the r hat go to 0 ( ##\hat r = \cos\theta \hat i + \sin\theta \hat j ##)
    Yes, I see that I did make a mistake; thanks for pointing that out. So that must mean that we have, after equating angular components of the acceleration and the normal force (I think)

    $$\theta (t)= mR \frac{d^2\theta}{dt^2} $$

    Does this differential equation make more sense?
  7. May 4, 2017 #6
    Solving that differential equation gives

    $$ \theta (t) = \sqrt{mR\gamma} \frac{\left (\frac{\theta_0}{\sqrt{mR\gamma}}+\sqrt{1+\frac{\theta_0^2}{mR\gamma}}\right)^2 \exp \left (\frac{\pm 2t}{\sqrt{mR}}\right )-1}{2 \left (\frac{\theta_0}{\sqrt{mR\gamma}}+\sqrt{1+\frac{\theta_0^2}{mR\gamma}}\right) \exp \left( \frac{\pm t}{\sqrt{mR}} \right)} $$
    Where ## \gamma = \omega_0^2-\frac{\theta_0^2}{mR} ##
    Where again we see the plus-minuses.
  8. May 4, 2017 #7
    Here, I am concerned when ## v_0< R\theta_0 \sqrt{\frac{1}{mR}} ##, which makes the square roots negative. Does this correspond to when the cart doesn't have enough speed to make it around the loop? Also, what should I do with the plus-minuses? Leaving them would mean that there are multiple angular displacement functions. Again, this is all assuming that the differential equation in post #5 makes sense.
  9. May 4, 2017 #8


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    It is still dimensionally wrong. On the left you have an angle, on the right a force. I do not understand how you are turning F into θ. You offer no explanation for that. Nor do I understand how you got rid of the trig functions. Please fill in the gaps.

    It is evident to me that the situation is essentially the same as for a simple pendulum. Since you are not considering the small oscillation case generally used to solve that, you have the full-blown pendulum equation to solve. You might care to look that up.
  10. May 4, 2017 #9


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    Another equation that cannot possibly be right because the dimensions do not match. You would do well to read up on dimensional analysis. It is a very handy way of checking the sanity of equations.
  11. May 4, 2017 #10
    Looking back at the problem statement, it is essentially a pendulum for ## \theta## varying from 0 to 2pi, except that the 'string' (the radius here) doesn't stretch or anything complicated like that. I've looked at the simple pendulum just a bit right now; I'm not sure why tension is not being included in the equation.

    (## \frac{d^2\theta}{dt^2} + \frac{g}{l}\sin\theta =0 ##).

    Maybe you could point out the error in my reasoning so that exclusion of tension makes a little more sense (I'll consider the coaster problem for now):

    The net force is simply the normal force + gravity. Nothing special here.

    If we break up the normal force and gravity as follows,

    $$ \vec F_n = F_{nr}\hat r + F_{n\theta} \hat \theta, \vec F_g = F_{gr} \hat r+ F_{g\theta} \hat \theta $$

    and choose the center of the circle as the origin, then

    $$ F_{nr}=?, F_{n\theta} = \theta(t), F_{gr}=mg, F_{g\theta}=0 $$ where the angle between the gravity vector and the positive 'horizontal' axis is 3pi/2. At any point on the circular loop, the angle the normal force vector makes with the positive x-axis is the angular displacement ##\theta (t) ## (assuming that we define ## F_{nr}<0 ## so that it points inward). Is this legal?

    Ok, now I figured that I could equate the angular components to the angular part of the polar-acceleration equation, which is - in this case -

    $$ R\ddot \theta \hat \theta $$ and equating we have the equation that I proposed earlier. However, as you pointed out, it doesn't make sense dimensionally. What is the lapse in my reasoning?

    Thank you
  12. May 4, 2017 #11
    Maybe the misunderstanding is that

    $$\vec F_n = F_{nx} \hat i + F_{ny} \hat j = -|\vec F_n|\hat r = -|\vec F_n|(\cos\theta(t) \hat i + \sin\theta(t) \hat j) $$
    But wouldn't it still be ok to say ## \vec F_n = F_{nr} \hat r + F_{n\theta} \hat \theta ## ?
  13. May 4, 2017 #12


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    Neither does it in the usual simple pendulum analysis.
    That equation comes from considering tangential force and acceleration, so the tension has no component. In your problem, the normal force is exactly the same as the tension in the pendulum over the bottom half of the loop. It is that force required to keep the mass at constant distance from the centre.
    Differences may arise beyond that range. A string cannot have negative tension, so that will be the same as your loop if there is no mechanism to prevent the cart from falling off the track. If there is such a mechanism, a rigid bar pendulum would be the right model.
    This seems to be the root of your dimensional error. How do you justify that equation? A force cannot equal an angle.
  14. May 4, 2017 #13
    Thank you so much! That makes sense now. I think that equation above comes from a mistake that I always make with polar coordinates: I tend to forget about the theta dependence of the polar base vectors, and thus often include an extra theta in there. In other words,
    $$ \vec F_n = F_{nx} \hat i + F_{ny} \hat j = -|\vec F_n|\hat r = -|\vec F_n|(\cos\theta(t) \hat i + \sin\theta(t) \hat j) = F_{nr} \hat r + F_{n\theta} \hat \theta =(F_{nr}\cos\theta(t) - F_{n\theta}\sin\theta(t))\hat i + (F_{nr}\sin\theta(t)+F_{n\theta}\cos\theta)\hat j $$ which means

    $$-| \vec F_n |\cos\theta(t)= F_{nr}\cos\theta(t) - F_{n\theta}\sin\theta(t) $$

    which must mean that ## F_{n\theta} =0 ##, which is apparent after some sketching. Thus,

    $$ -| \vec F_n |\cos\theta(t)= F_{nr}\cos\theta(t) \rightarrow -|\vec F_n|=F_{nr} $$

    It is noteworthy that ## -| \vec F_n |\cos\theta(t) =|\vec F_n| \cos(\pi+\theta(t)) ##

    So after all of this, we have determined that the angular component of acceleration is 0, so ## \ddot \theta ## is 0.

    So, ## \dot \theta ## = A, and ## \theta (t)=At+B ## where A and B are integration constants. Ok, let's see what happens now when we equate radial components of acceleration with radial force:

    $$ F_{nr}+F_{gr}= \ddot r - r(\dot \theta)^2 \rightarrow -|F_n| +mg = -R(\dot \theta)^2 $$ where the +mg is due to the choice of polar coordinates (## mg\hat r= -mg \hat j ##, where the angle its base vector is 3pi/2). So all we need now is the magnitude of the normal force. Since the cart is constrained to the path of the circle, the magnitude of the normal force must be equal to ...? If it is equal to mg, then it cancels with the gravitational force, which gets us nowhere. Input?
  15. May 4, 2017 #14


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    The angular component of the normal force is zero, but what about gravity?
  16. May 4, 2017 #15
    I'll start with the cartesian form for gravity and the polar form, and then I'll break the polar base vectors into cartesian ones and equate components:
    $$ \vec F_g = -mg \hat j = F_{gr}\hat r + F_{g\theta} \hat \theta \rightarrow -mg \hat j= F_{gr}\cos\theta \hat i + F_{gr}\sin\theta \hat j - F_{g\theta}\sin\theta \hat i + F_{g\theta} \cos\theta \hat j$$ and since ## \theta= \frac{3\pi}{2} ## for gravity,
    $$-mg \hat j= -F_{gr}\hat j+F_{g\theta} \hat i $$
    Wouldn't that imply that the angular component of gravity is 0? Intuitively, I don't agree, but I am not quite sure how to express it mathematically.
  17. May 4, 2017 #16


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    Eh? That is only at one particular angle. You need the expression for any angle.
  18. May 4, 2017 #17
    Even if gravity points in the same direction (with approximately the same magnitude - the same for our purposes) throughout the cart's entire path?
  19. May 4, 2017 #18


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    Of course. Gravity is always down. When the cart is not at the top or bottom, that has a tangential component.
  20. May 5, 2017 #19
    Could you explain? I always thought that if two vectors have the same direction and magnitude, they are equivalent (so it doesn't matter where they are on the euclidean plane - translation doesn't affect the equivalence of the vectors). And here gravity is independent of time, so I am not quite sure how it has a tangential component since it is always the same everywhere on the path of the cart. Also,

    $$ \vec F_g= mg\hat r = mg(\cos(\frac{3\pi}{2})\hat i + \sin(\frac{3\pi}{2}) \hat j) = -mg \hat j $$

    The 3pi/2 is there since it always points down, so where does the tangential component come from?
  21. May 5, 2017 #20


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    You have defined θ(t) as the position of the cart at time t, where θ is measured anticlockwise from the 3o'clock position, yes? (It would have been more convenient to have it as an angle to the vertical, but never mind.)
    Have you drawn a free body diagram for that? If not, do so.
    mg acts straight down. We are interested in tangential forces only, since there is no change in radius.
    We can resolve mg into its radial and tangential components. The radial component has magnitude mg |sin θ| and the tangential component has magnitude mg |cos θ|.
    We do need to be careful with the signs. Because θ is measured counterclockwise, we need the force as positive in that direction. We can arrange that by saying the tangential force is -mg cos θ.

    As noted, the problem is the same as a simple pendulum, so I suggest you look that up any of the usual websites.
  22. May 7, 2017 #21
    Ok, I've come to the realization that when resolving a vector in polar coordinates, the base vectors are drawn with respect to the position of the point of interest (the cart in our case) at a certain instant in time. That is, if the cart is at some point on the loop, ## \hat r, \hat \theta ## are drawn based on the position vector from the origin to the cart at that time. Thus, when resolving ## \vec F_g , \vec F_n ##, this is done with respect to those vectors. (please excuse my ignorance, that just dawned on me)*. As the normal force is anti-parallel to ##\hat r ##, it has no tangential component. For gravity, I got what you got above after working out the geometry from the FBD:

    $$ F_{gr} = -mg \sin \theta(t), F_{g\theta} = -mg\cos\theta(t) $$

    By positive, I assume that you mean that since the tangential and radial components of gravity point in the opposite direction of the basis vectors (and are thus negative), the resultant vector (##\vec F_g ##) must point in the 'positive' direction.

    With that said, our equations of motion are:

    $$ mR(\dot \theta)^2 = -F_{nr}-mg\sin\theta, mR\ddot \theta = -mg\cos\theta $$

    where the second equation is exactly what is obtained from the simple pendulum analysis, except there's a cosine instead of a sine (I assume this comes from us measuring theta from the 3'o clock position, not the vertical). We have two equations in two unknowns: ## F_{nr}, \theta(t) ##. Of course, if we try to solve this, we get an elliptic integral.

    * If we have a system like a ball dropped from rest, and we are ignoring drag, then ##\vec F_g ## would have only a radial component and the angle in ## \hat r ## would be ##\frac {3\pi}{2} ##, which is what I was getting confused with earlier.

    For the bottom half of the circle, this analysis is the same as the simple pendulum. However, for the top half, assuming the angular velocity of the pendulum bob is sufficiently large, the tension points inwards to prevent the bob from 'flying' off of the circular path. So our analysis is really just a simple pendulum for the bottom half and the top half? (note: there is no mechanism in this coaster that keeps the cart on the track; the normal force always points inwards assuming the cart is always on the track)
  23. May 7, 2017 #22


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    Yes, as long as the speed is sufficient to stop it falling off the track.
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