## Posts Tagged ‘pendulum’

### Weights on a pendulum

September 23, 2013

A friend of mine is one of the people in charge of winding a clock tower. He was showing my physics class how the tower works, and we were looking at the pendulum.

Here is a photo of the pendulum in question, from this blog. We asked the class:

Why are there weights on the top of the pendulum?

This may sound like a bit of a trick question; one can come up with many possible reasons. But I’ll tell you that the weights are put there by humans deliberately as they maintain the clock. Sometimes they add a few and sometimes they take a few off.

A student then asked why the weights don’t slip off.

For a given pendulum length and swing amplitude, what is the minimum coefficient of friction to keep the weights from sliding off?

### A Brief Illustration of One Forms and Tensors in Mechanics

December 26, 2008

note: Don’t worry too much about dimensions and constants. I drop them like I would a new born child.

other note: I originally identified the eigenvalues incorrectly. Changing it didn’t alter the math significantly. 12/27/08

In this post I will examine the mechanics of a classical particle in the 2-D potential $U(x,y) = \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2$

This is a harmonic oscillator potential, but $\omega$ is different in the $x$ and $y$ directions.

A simple harmonic oscillator in one dimension is canonically the mass on a spring: or a grandfather clock pendulum These correspond to the potential $U(x) = \frac{k x^2}{2}$.

The fastest way to get the equation of motion is Newton’s second law, which yields $\begin{array}{rcl} -\frac{d U}{d x} & = & \frac{d^2 x}{dt^2} = k x \\ x(t) & = & A\sin\left(\omega t + \phi\right)\end{array}$

with $\omega$ some function of $k$.

The easiest way to make the hop to two dimensions is to keep the apparatus the same, but give the mass another degree of freedom. For the mass on a spring, this might correspond to sitting the mass on a table and tacking down one side of the spring, while leaving it free to spin as well as expand/contract. You have to think of this as a spring with no rest length, and collapsible, so that it can pass through itself. Combine the properties of a slinky and a rubber band and you’re pretty much there. The 2-D pendulum is just a pendulum free to swing to and fro in whatever direction it wants.

This scheme would set $\begin{array}{rcl} U(x,y) &=& \left( \frac{ k x^2}{2}\right)+\left(\frac{k y^2}{2}\right) \\ {} & = & \left(\frac{k (x^2 + y^2)}{2}\right) \\ {} & = & \left(\frac{k r^2}{2}\right) \end{array}$

with $r$ the displacement distance from equilibrium. Then $r$ and $\theta$ are decoupled, and we just get simple harmonic motion of $r$, while $\theta$ increases linearly with time. The mass traces out an ellipse in space, with an eccentricity and orientation depending on the initial conditions, and with the same period as before.

It is a bit more interesting to break the symmetry between the two orthogonal directions and treat them distinctly, as you would people with different ethnicities. In the example of the spring, we could set up a situation like this: displacement and force on a mass with springs on either side

Before examining the dynamics of this configuration, let’s first hash out the language we’ll be using by describing the kinematics.

The displacement of the mass from equilibrium is a two-dimensional vector, and the set of all possible displacements constitutes a vector space.

It can be easier to manipulate this vector when we set up a coordinate system, so we’ll set up standard Cartesian coordinates with x the left/right direction and y the up/down direction.

Now we are ready to define a (1,1) tensor that characterizes this system. You might be worried because a (1,1) tensor takes a vector and a one form as its arguments, but we haven’t yet identified any one forms! You’ll see in a moment that we can do quite a bit without knowing exactly what the one forms are, but don’t worry, we’ll return and identify them at the end.

Remember that a (0,1) tensor is a vector. That means that a (1,1) tensor can also be considered a function that takes in a vector and gives a (0,1) tensor – another vector. (1,1) tensors are the set of linear operators on the vector space.

On to the tensor. The spring constant of the system is a (1,1) tensor, $K$, which takes in the displacement of the mass and returns the force on the mass.

Notice that the force does not necessarily point in the same direction as the displacement. If you push the mass straight up/down, the springs will pull it back towards the center, as they will with a pure left/right displacement. But suppose you push the mass at a 45 degree angle. Then the springs push left/right much more than they pull up/down. Let’s quantify this by working out the tensor. We need only work out its action on unit displacements in the x and y directions, and then linearity will take care of the rest.

If we displace the mass a small amount to the right, we pull the left spring out and push the right spring in. They each create a restoring force of 2k . $K(\hat{\mathbf{x}},\_\_) = -2k\hat{\mathbf{x}}$.

I have used the underscore in the second argument of $K$ to indicate this is still an open slot. $K$ must still take on a one form before it can spit out a scalar. Right now, it’s a vector – a (0,1) tensor.

If we push the mass up a small amount in the y direction, we don’t make the springs any longer to first order, but the forces they exert on the mass are no longer balanced. Each spring creates a restoring force that is the equilibrium tension in the spring times the sine of the angle the springs make with the horizontal. $K(\hat{\mathbf{y}},\_\_) = -2\frac{\textrm{T}_0}{L}\hat{\mathbf{y}}$

with $\textrm{T}_0$ the equilibrium tension of the springs and $L$ their length. If the springs had zero rest length, this would evaluate to the same thing as in the x direction, but let’s just pretend these springs don’t have that.

This defines the tensor spring constant, and allows us to write the force on the mass by the equation $\mathbf{F} = K(\mathbf{x},\_\_)$.

We can again use Newton’s second law to find the equations of motion. $K(\mathbf{x},\_\_) = m \frac{d^2 \mathbf{x}}{d t^2}$.

Note that this is a vector equation, not a coordinate equation. Before we solve this equation, we’ll return to the discussion of kinematics left incomplete earlier.

We have a vector space of displacements, so we know there exists a dual space of one forms. This is the space of linear, scalar-valued functions of the displacement. It also has dimension two. We can find a basis for this space by finding two independent one-forms. Call them $p_x$ and $p_y$, and define them by $p_x(\hat{\mathbf{x}}) = 1$ $p_x(\hat{\mathbf{y}}) = 0$ $p_y(\hat{\mathbf{x}}) = 0$ $p_y(\hat{\mathbf{y}}) = 1$

A general one form can be expressed in this basis as follows: $\begin{array}{rcl} p(\mathbf{x} &=& p(x\hat{\mathbf{x}} + y\hat{\mathbf{y}}) \\ {} &=& x*p(\hat{\mathbf{x}}) + y*p(\hat{\mathbf{y}}) \\ {} &=& x*\alpha + y*\beta \\ {} &=& \alpha*p_x(\mathbf{x}) + \beta*p_y(\mathbf{x}) \\ p &=& \alpha p_x + \beta p_y \end{array}$

You may have noticed that the basis one forms have a simple interpretation as projection operators. From the above discussion, we see that $\left[\alpha p_x + \beta p_y\right](\mathbf{x}) = \alpha*x + \beta*y$

So that the one forms are acting in a way similar to the dot product in elementary physics. However, since I haven’t defined the dot product in this little series of posts, I won’t pursue that notion.

Now we can return to the equations of motion, and put each of the basis one forms into the vector equations. $\mathbf{F}(p_x) = K(\mathbf{x},p_x) = -2 k x$, $\mathbf{F}(p_y) = K(\mathbf{x},p_y) = -2 \frac{T_0}{L} y$.

You might verify that these forces are consistent with the potential I quoted at the beginning of the post. You probably won’t, though. Nobody ever does shit like that. Anyway, the solutions are of the form $\mathbf{x}(t) = A\sin(\omega_1 t)\hat{\mathbf{x}} + B\sin\left(\omega_2 t+\phi\right)\hat{\mathbf{y}}$

where $\omega_1$ is a function of $k$ and $\omega_2$ is a function of $T_0$ and $L$. In general they are different, and their ratio will determine the orbit of the mass in 2-D space. The motion of the mass traces out a Lissajous figure, one example of which is this: A Lissajous figure with frequency ratio 1.8

It might not seem like the tensor accomplished anything, since we could easily have found the same system of equations without knowing anything about tensors. The payoff here is theoretical. The equation $\mathbf{F} = K(\mathbf{x},\_\_)$

involves no coordinates. It is simply a statement about vectors and linear functions of vectors. The coordinate system was an easy way to do some calculations, but in writing the tensor equation we’ve mathematically depicted the system geometrically instead of algebraically.

In the x-y basis, $K$ can be written as a diagonal matrix with eigenvalues $-2k$, $-2\frac{T_0}{L}$. If we were to change coordinate systems, for example rotating x and y by some angle, a coordinate-based approach would have to recalculate all the forces. On the other hand, our tensor equation would still hold, and we could evaluate it just as before – by plugging in the one forms corresponding to projection operator on the axis of the springs and orthogonal to that axis. The only work would be in finding new coordinate representation of the one forms.