jkwiens.com: Answer: A bead sliding on a rotating parabola

Answer: A bead sliding on a rotating parabola

Thursday, 13 of December , 2007 at 9:33 pm

After looking at Rod’s Solution on Reasonable Deviations, I have come to the conclusion that my solution was way to complicated. Instead of using high school physics, I decided to solve the problem using Hamiltonian Mechanics.

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

Answer: To solve this problem using Hamiltonian Mechanics, we first need to find the Hamiltonian (which is the Kinetic Energy + Potential) of the system.

The system can easily be shown to have the following form:

$L = K + P = \frac{1}{2} m \arrowvert \vec{v} \arrowvert ^{2} - mgz$

If we use cylindrical coordinates and constrain the path of the bead to the parabola, we can simplify the Hamiltonian to a single degree of freedom. Since the velocity $\vec{v}$ in cylindrical coordinates is $\dot{r}\hat{r} + r \dot{\theta}\hat{\theta} + \dot{z}\hat{k}$ and , the Hamiltonian reduces to:

$L = \frac{1}{2} m (\dot{r}\hat{r} + r \dot{\theta}\hat{\theta} + \dot{z}\hat{k})\cdot(\dot{r}\hat{r} + r \dot{\theta}\hat{\theta} + \dot{z}\hat{k}) - mgcr^2$

and since $\dot{\theta}=\omega$ and $\dot{z}=\frac{d}{dt}(cr^2) = 2 c r \dot{r}$ , the equation becomes:

$L = \frac{1}{2} m (\dot{r}^2 + r^2\omega^2 + 4c^2r^2\dot{r}^2) - mgcr^2$

Now that we have the Hamiltonian reduced to its generalized coordinates, we can apply it to the Euler-Lagrange equation. Since our Hamiltonian has only one degree of freedom, the Euler-Lagrange equation will have the following form:

$\frac{\partial L}{\partial r} - \frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{r}}) = 0$

First, lets take the derivatives needed for the differential equation. They can be shown to be:

$\frac{\partial L}{\partial r} = m ( 4c^2r\dot{r}^2 + r \omega ^2) - 2mgcr$

$\frac{\partial L}{\partial \dot{r}} = m (\dot{r} + 4 c^2 r^2 \dot{r})$

$\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{r}}) = m (\ddot{r} + 4 c^2 r^2 \ddot{r} + 8 c^2 r \dot{r}^2)$

When we put these values into the differential equation, we will get:

$m(4c^2r\dot{r}^2 + r \omega^2) - 2mgcr - m(\ddot{r} + 4c^2r^2\ddot{r} + 8c^2r\dot{r}^2) = 0$

$m(1 + 4 c^2 r^2)\ddot{r} + m(4 c^2 r)\dot{r}^2 + m(2gc - \omega ^2)r = 0$

Since the position of the bead is confined to a circle of radius R, we can simplify the differential equation by using the following conditions: $\ddot{r} = 0$ , $\dot{r} = 0$ , and . The differential equation will then reduce to:

$m(2gc - \omega ^2)R = 0$

Therefore, the value of c must be:

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

Category: Answers, Classical Mechanics

4 Comments

Comment by Nikita Nikolaev

Made Friday, 25 of January , 2008 at 5:49 pm

Nice! Nicely written!

Now I think it is the time I write up my solution!

Excellent!

Comment by eldila

Made Friday, 25 of January , 2008 at 6:26 pm

Well, if you do, make sure you don’t copy

Cheers!

Comment by Nikita Nikolaev

Made Monday, 18 of February , 2008 at 12:06 pm

I have posted my solution here.

Enjoy

Comment by eldila

Made Monday, 18 of February , 2008 at 1:24 pm

Very nice.

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