A bead sliding on a rotating parabola « Reasonable Deviations

A bead sliding on a rotating parabola

By Rod Carvalho

Via jkwiens.com, here’s an interesting problem:

A bead slides along a smooth wire bent in the shape of a parabola, $z = c r^2$ . The bead rotates in a circle, of radius $R$ , when the wire is rotating about its vertical symmetry axis with angular velocity $\omega$ . Find the constant $c$ .

This reminds me of the Classical Mechanics problems I used to enjoy when I was a high-school senior. It can be solved in many ways. I will start with the simplest solution…

If the bead rotates in a circle, then the bead is rotating at constant height and is stationary relative to the reference frame “attached” to the parabola-shaped wire (as illustrated in the image). Let $m$ be the mass of the bead.

If the bead is stationary (relative to the rotating reference frame) and rotating in a circle (relative to the non-rotating inertial reference frame), then the centripetal force being applied on the bead is

$F_c = m R \omega^2$ .

The parabola exerts a “reaction” force on the bead. If that reaction force did not exist, the bead would fly away from the parabola. The reaction force is normal to the parabola-shaped wire. If it were not normal, then the reaction force projected on the parabola would result in a tangential force, which would result in tangential acceleration, and the bead would therefore not be stationary relative to the parabola. The slope of the parabola at point $(R, c R^2)$ is $2 c R$ . If we let

$\tan(\phi) = 2 c R$

be the tangent of the angle defined by the intersection of the tangent line at point $(R, c R^2)$ with the $r$ -axis, then it can easily be seen that the angle that the reaction force exerted on the bead and the $z$ -axis define is $\phi$ . If we project the normal reaction force $\vec{N}$ on the vertical and horizontal axes of the rotating reference frame, we get

$N_r = N \sin(\phi)$

$N_z = N \cos(\phi)$ .

We have two forces being applied on the bead: the gravitational force $\vec{F}_G = m \vec{g}$ , and the normal reaction force $\vec{N}$ . Since the bead is rotating at constant height, then the sum of the vertical components of these two forces must be zero

$N_z = m g$ .

And since $N_z = N \cos(\phi)$ , we get the norm of the reaction force

$N = \displaystyle\frac{m g}{\cos(\phi)}$ .

Since the bead is rotating in a circle of constant radius $R$ , then there’s a normal acceleration and zero tangential acceleration. The acceleration vector will point inwards, towards the rotation axis. This normal acceleration is due to the centripetal force. Note that this centripetal force is nothing but the projection of the reaction force on the horizontal axis:

$N _r = m \omega^2 R$

and consequently

$N \sin(\phi) = m \omega^2 R$ .

We already know that the reaction force’s norm is $N = \displaystyle\frac{m g}{\cos(\phi)}$ , and therefore

$m g \tan(\phi) = m \omega^2 R$ ,

and since $\tan(\phi) = 2 c R$ , we get

$2 g c = \omega^2$

and finally

$c = \displaystyle\frac{\omega^2}{2 g}$ .

Tags: Classical Mechanics

This entry was posted on December 11, 2007 at 9:44 am and is filed under Brain Teasers, Recreational Physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

13 Responses to “A bead sliding on a rotating parabola”

jkwiens.com: Question: A bead sliding on a rotating parabola Says:
December 14, 2007 at 5:33 am | Reply
[...] looking at Rod’s Solution on Reasonable Deviations, I have come to the conclusion that my solution was way to complicated. [...]
ObsessiveMathsFreak Says:
December 14, 2007 at 4:15 pm | Reply
It’s a nice problem. Did you also notice that with the derived constant, the bead will rotate in a horizontal circle no matter how high or low it is placed on the wire. Also, for any other constant, the bead can never rotate in a horizontal circle.

This problem reminds me of the water clock problem, which also involved parabolas. If you consider a vessel in the shape of paraboloid with a hole at the bottom, then the height of water in the vessel will always decrease at a constant rate.
pedja Says:
December 15, 2007 at 6:47 pm | Reply
Isn’t centrifugal force instead of centripetal that is acting on the bead.
rod. Says:
December 15, 2007 at 7:15 pm | Reply
pedja,

Centrifugal force is a fictitious one. For example, let’s say you’re starting your car and you accelerate abruptly… you will feel a force pushing you against the seat, but that is a fictitious force too. Non-fictitious forces are associated with accelerations, while fictitious forces are usually the result of considering non-inertial reference frames.

This sounds a bit sketchy, but there’s nothing like drawing a free-body diagram, and see all the forces and reactive forces being applied. It’s kind of fun, and also instructive.
Nikita Nikolaev Says:
January 25, 2008 at 8:52 pm | Reply
There is a quite nicer variation on this problem, which explains a lot more. For example, the solution does not tell you anything about why the constant c is exactly what it is. Although we can guess why, simply from our physical intuition, that isn’t a great answer to my mind (because, in such cases, I always say: “well, suppose we have never seen a wire with bed on it that rotates and we don’t know what it is to be in a ‘classical motion’. How then will you be able to predict that?”

Consider the mentioned variation on the problem. The wire is still bent in a shape $z = cr^2$ and oriented upwards, and the bed of mass $m$ is still there; but now, suppose, the wire rotates at a constant angular velocity $\omega$ . Now, our task is to analyse the system for stability.

We ought to find the Lagrangian and hence the equation of motion (there’s only one, since the problem really is one-dimensional). The obtained equation is (highly) nonlinear and I do not recommend solving it explicitly. Rather, what we do is we find the fixed points by setting the velocity and acceleration to zero (that’s the definition of a fixed point).

Anyway, after some gibberish, what we finally obtain is a second order linear differential equation, which really much reminds of the simple harmonic oscillator. The coefficient there is $2gc - \omega^2$ . Now, everything is solely up to this coefficient. if it is positive, then the origin is an equilibrium point. If it is negative, however, then there are no equilibrium points and a point will move away to infinity. If, however, the coefficient is zero, which is what you have found $c = \frac{\omega^2}{2g}$ , then we see that any point on the curve is an equilibrium point.

So this is where that perhaps strange value of the coefficient comes from.

P.S.: It’s the first time I’m trying out LaTeX in comments, so it might not be really nice. :P
Nikita Nikolaev Says:
January 25, 2008 at 8:52 pm | Reply
Nope, it worked really well, apparently :)
eldila Says:
January 25, 2008 at 10:21 pm | Reply
If you check out my solution, it basically how I solved the problem. I didn’t use the same wording, but it is basically the same approach that you mentioned.

http://www.jkwiens.com/2007/12/13/question-a-bead-sliding-on-a-rotating-parabola-2/
Nikita Nikolaev Says:
January 26, 2008 at 1:50 am | Reply
Eldila,

the solution is very nicely written indeed, as I mention in the comment over there.
Great!
A Bead Sliding on a Parabola-Shaped Smooth Wire « Where Everything Equals to 0 Says:
February 18, 2008 at 8:03 pm | Reply
[...] second solution to the Lagrange equation for fixed points (namely, .) Such arguments are presented here. Indeed, we can see that we here have arrived at the fact that the bead will rotate in a circle of [...]
Nikita Nikolaev Says:
February 18, 2008 at 8:05 pm | Reply
Update: I have posted my solution here.

Enjoy :)
rod. Says:
February 18, 2008 at 8:19 pm | Reply
@ Nikita

Thanks for your comments. I agree with what you wrote on January 25. The solution I have presented here is sketchy at best, and ugly at worst. I have worked on more sophisticated solutions, using Lagrangian mechanics (as you suggested) and also vector mechanics. I got stuck at some point, and now I can’t find my handwritten notes on that problem.

The way I would like to solve the problem is to start from the beginning: the wire is not rotating, the bead is placed somewhere in the wire, and then we apply some torque on the wire, until the motor torque equals the frictional torque and the bead is rotating at constant height and constant radius. This should be easy, but it gives quite long expressions, and I kept making small errors. I will re-check my derivations one of these days.
Nikita Nikolaev Says:
February 18, 2008 at 8:28 pm | Reply
Rod.,

when you will, please make sure you post at least some of it, as I grew interested! :)
rod. Says:
February 18, 2008 at 8:45 pm | Reply
Will do! :-)

Reasonable Deviations

A bead sliding on a rotating parabola

13 Responses to “A bead sliding on a rotating parabola”

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