How Much Energy Is There In Your Balance Wheel?

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Energy In A Balance Wheel

How do I calculate the energy in my balance wheel?  Or maybe we should ask why we care how much energy is in the balance wheel? A few weeks ago, I started to look into the details of how I would start to design a watch movement.  As everything in a watch movement goes around in circles, it is essential to define an object’s location, speed, and energy when moving in a circle.  This quest to describe the motion accurately took us into the world of defining arcs drawn on a circle in Radians rather than degrees.  And finally, this gave us a new way to envisage the speed at which a point on a rotating body travels. The question you probably have is, what is this all applicable to?

This week I shall start to delve into this question in detail. This will take us through the hard work of how to calculate the energy in a balance wheel.  Why this is so important will become clear in a few weeks’ time. However, the critical aspects of the first blog will be crucial to every blog in this series. 

Balance Wheel Is A Rotating Body

This time, the goal is to understand how to define a rotating body (a balance wheel), particularly the energy that the rotating body has.  Understanding the energy in a rotating mass will be particularly important when considering the mass used to wind automatic watches, the balance wheel, and how to create size the spring for the balance wheel.

A simple balance wheel with the hairspring. Note that the mass is all at the widest radius.Scharf82, CC BY-SA 4.0, via Wikimedia Commons

To start with, we will consider a simple spinning disc with no forces acting on it and then see how the result can be used in a watch or a balance wheel, to be more exact.  As with the last blog, I will start with what the solution is with linear motion.

Energy In Linear Motion

The kinetic energy of a body moving linearly or without any acceleration has an energy associated with it that can be calculated by the equation.

Where “m” is the mass of the body and “v” is the velocity with which it is moving.

Again, let’s jump back to the previous blog. We can see that the velocity of a point on an arc is moving with a speed that can be calculated as the angle drawn out by the reference point (in radians) in 1 second times the radius to the reference point.

Linear Motion Into Rotational Movement

Here we introduce a new term called the angular velocity, which is the arc, calculated in radians, that is drawn out in a unit of time.  So, for example, a rotating body spinning a full circle in one second will be rotating at 2𝝅 RadS-1   (Radians per second).

If we know the speed at which the disc is rotating (rotational velocity), and we shall define this as “𝛚” (Geek Symbol Omega), the velocity of a rotating point can be represented simply as:

If we place this into the equation for linear kinetic energy, we get to the following equation:

There Is Always A Problem

The problem is that this equation is only valid for each unique bit of mass that is a distance “r” from the center of rotation.  This means that for a rotating disc, the kinetic energy is the summation of all the individual masses throughout the disc and is represented, mathematically, as follows:

Regardless of the mass rotating or the radius at which that mass is rotating, it rotates with the same angular velocity. If the complicated piece, mr2 is now defined as the moment of inertia, or “I” it becomes a lot simpler.

This simplification removes the dependency on the radius of any point in the disc from the equation, so the energy of a spinning disc can be calculated by the equation below where “I” is the moment of inertia around its axis of rotation for the geometry that is in motion, in this case, a disc around its center. 

That Pesky Moment Of Inertia

That does leave one particular issue, how is the moment of inertia for a spinning disc calculated.  This takes us into a bit more mathematics, particularly a bit of integration.  If this strikes the fear of God into you, then please skip forward, but I will include it here for completeness.  That said, through mathematics, the magic of the balance wheel becomes evident, so try and stick with it.

Let us assume that the disc is of uniform density across its radius and a mass “m.”  We can now consider the disc as infinitesimally thin rings of mass spinning at a radius.  We calculate the mass of each of these extremely thin rings by the ratio of the area of the thin ring to the total area of the disc. Suppose we break this down, as in the diagram above. In that case, the thin disc is at radius “r,” it is of thickness “dr” (it is enlarged in the diagram above for ease), and the whole disc has a total radius of “R.” The area of the thin strip is calculated as the circumference at “r” times the width “dr.”

It Is Just A Ratio Of Areas

The mass of the strip is calculated as the ratio of the area of the thin strip and the total area of the disc times the mass.

The moment of inertia of this thin strip is found by substituting this value for the mass into the equation for the moment of inertia.

So the solution is found by the summation of all the infinitesimally small rings for the distance of r =0 until r=R.  We write this as follows:

This equation is solved with a little bit of integration (if you are not sure about this step, don’t worry.  But here it is for those who want to see it all:

Now For The Mathematics

First, reorder the equation to separate the variables and the constants:

The solution for this is :

Simplifying this, we get

This Is The Equation For Balance Wheel Energy

If we put all this together, for a disc of mass “m” with a radius “R” rotating with an angular velocity of 𝛚 the kinetic energy of this disc will be:

the equation to calculate the kinetic energy in a rotating balance wheel
This is the equation to calculate the energy in a rotating balance wheel

That is quite hard going, but what I would want to make sure you leave this blog understanding (and I salute you for making it this far if you have) is that the energy of a rotating object is proportional to the square of the radius of the mass that is spinning.

Sorry, It Is Not That Simple

This equation will be essential but, alas, it is not the final solution.  The problem is in creating this equation we assumed that the disc is spinning with constant velocity and no forces are acting on it and that the disc is made of the same material throughout.  Alas, the first assumption is always wrong and the second assumption is often wrong.  More on that next time.

If you would like more detail on any questions or anything that you would like more detail, please put your questions below, and I will do my best to answer them.   

Next time we will look at how we apply this to have the balance wheel oscillate accurately over time.

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