# Bead on a rotating hoop [closed]

This is problem 10.13 from Fowles and Cassiday, 7e. A bead of constant mass m is constrained to slide along a thin, circular hoop of radius $l$ that rotates with constant angular velocity $\omega$ in a horizontal plane about a point on its rim as shown. I need to figure out the kinetic energy of the bead to write down the Lagrangian.

I want to define a vector $\vec{r}_{0}$ from the origin of the x-y system to the center of the hoop using polar coordinates and a vector $\vec{r}_{1}$ from the center of the hoop to the bead of mass m. Then

$$\vec{r}_{0} = l\hat{e}_{r_{0}}$$

$$\vec{r}_{1} = l\hat{e}_{r_{1}}$$

where $\hat{e}_{r_{1}}$ points radially outward from the center of the hoop and $\hat{e}_{r_{0}}$ points radially outward from the center of the x-y plane. I want to use polar coordinates, with one set emanating from the x-y plane and a second set from the center of the hoop. Then the position of the bead $\vec{r}_{m} = \vec{r}_{0} + \vec{r}_{1}$. I get then that $\dot{\vec{r}}_ {0} = l\omega\hat{e}_{\theta_{0}}$ and $\dot{\vec{r}}_ {1} = l\dot{\theta}\hat{e}_{\theta_{1}}$ where I've defined $\theta_{0} = \omega t$. So

$$\dot{\vec{r}}_{m} \cdot \dot{\vec{r}}_{m} = l^{2}\omega^{2} + l^{2}\dot{\theta}^{2} + 2(l^{2}\omega\dot{\theta}\hat{e}_{r_{1}} \cdot \hat{e}_{r_{0}}) = l^{2}\omega^{2} + l^{2}\dot{\theta}^{2} + 2l^{2}\dot{\theta}\omega cos(\theta)$$

but this isn't right. I suspect that I'm ignoring some consequences of the origin at the center of the hoop rotating with respect to the fixed x-y reference frame, but I'm not sure exactly what consequence. I thought I was taking it into account by adding the rotating vector, but I guess not.

## closed as off-topic by Kyle Kanos, ACuriousMind♦, Jim, BMS, Kyle OmanFeb 21 '15 at 0:03

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• I suggest you that for calculating $\dot {\vec r}_1$ to look in Wikipedia at the article with velocity of a rotating body as seen from another frame. I suppose that it can help. – Sofia Feb 20 '15 at 21:17
• You can see in that article that the velocity of a rotating body, as seen from another frame, contains two terms. So, from the frame rotating around the center of axis, you see the velocity of the bid, as I say, as containing two terms. Just be careful, because in the article, the system indexed by $i$ is at rest, while your system rotating around the origin, and serving in your case as system $i$ is not at rest, is rotating by itself. – Sofia Feb 20 '15 at 21:38
• can I just point out that you haven't asked a question. You stated a problem (cool), told us what you've done (way to be productive), told us that what you did isn't correct (that's a shame), and then said you might not have taken something into account (a fairly reasonable deduction). It's all well and good, but what is it you are asking? (That's kind of important for us to answer you). I'm not being condescending, btw. I'm just a bit of a snarky person – Jim Feb 20 '15 at 21:40
• That's a better structured post than many that do have a question, even though this one doesn't. The implication is there, and the OP seems to be new to the site. – Jiminion Feb 20 '15 at 21:51
• @Jiminion If the implication is "What have I done wrong here?" then the question is still off-topic per our homework policy – Jim Feb 20 '15 at 21:59

The problem is that you are thinking of defining two coordinate systems, and then trying to combine them. This is incredibly dicey and requires extremely precise thinking to make sure all terms are accounted for. I would go so far as to call it a nightmare.

Instead, just think of the position of the particle as the sum of two separate position vectors using the same polar coordinates. Associate your first position vector $\vec{r_0}$ with $\omega t$ and associate your second position vector $\vec{r_1}$ with $\omega t + \theta$. Then in your equations, everywhere there is a $\dot\theta$, it's replaced with $\omega + \dot\theta$. Because your position vectors use the same polar coordinate system the dot product and time derivatives won't pick up any strange extra terms.

A quick test you can use is to test the equation in edge cases where $\theta$ does not vary. If $\dot\theta = 0$, then radius is fixed and the total kinetic energy is easy to calculate. Your more general formula should reduce to the proper expression.

Yeah, it looks like your generalized coordinate is simply $\theta$. If the expressions that you're getting don't seem right, then probably you're just trying to skip too many steps at once. The kinetic energy is going to be $T \propto v_x^2 + v_y^2$; we'll compute it that way alone. Let $\phi = \omega t$. The position vector is then

$\vec r = l [\cos \phi + \cos (\phi + \theta); \sin \phi + \sin (\phi + \theta)].$

Then

$\frac {d\vec r}{dt} = l [-\omega \sin \phi - (\omega + \dot \theta)\sin(\phi + \theta); \omega \cos \phi + (\omega + \dot\theta)\sin (\phi + \theta)].$

After some simplifications on $T \propto (d \vec r / dt)^2$ all of the terms with an explicit $\phi$ (hence explicit $t$-dependence) drop out, but you get terms which are missing from your expression, like a $\omega^2 \cos \theta$ term which acts as a sort of potential well due to the centrifugal force of the ring's rotation about the origin.

The core lesson here is, never use generalized coordinates to derive the Lagrangian. Always use coordinates which you are 100% confident that you understand well, like Cartesian coordinates, and put in the extra "elbow grease" to simplify them.