# How to calculate the inertia force  Now let's consider an example of calculating the moment of inertia. First, the moment of inertia of a cylinder should be calculated:

When calculating a moment of inertia, it is important to consider symmetry. If you choose the coordinates cleverly and define small volume elements in such a way that you can easily calculate their content and then total over all elements, you save a lot of time and work. So let's look for 'clever' coordinates:

The first attempt is at the known Cartesian coordinates. According to the formula for the moment of inertia we can with the condition of constant density write.

The cylinder has the thickness or length and the radius R. Now the integral must be over the volume

dxdydz are calculated, so = This approach looks complicated. Therefore we are trying a new approach with cylindrical coordinates. For this we consider an infinitely thin hollow cylinder with the radius r and the thickness dr. Now the volume can be specified directly as the product of the base area r2and height of the cylinder. This applies

volume

dV = 2 rdr

With dm = 2 rdr

Out I.    With V = R2 .

This is the formula for the moment of inertia of a cylinder of radius R and volume V.

If you let two cylinders of the same mass roll down an inclined plane, they would have to perform the same movement if they have the same volume and the same radius. We consider this result in an experiment:

In this experiment, two cylinders of the same mass and dimensions are allowed to roll down an inclined plane. Contrary to expectations, one cylinder rolls faster than the other. If we again assume that the calculated principles were not wrong, then, analogous to the wobbly wood, a quantity does not have to be what we suspect. The result is found when the body is opened: a cylinder is hollow. Let's now calculate which cylinder is the faster:

For this we need the moment of inertia. With an analogous calculation it then follows for the moment of inertia of the hollow cylinder:

More accurate

I.H = (Ra2 + ri2)

Moment of inertia of the full cylinder:

I.V = According to Chapter VI.2, the following applies to the forces acting on rolling cylinders:

Power to accelerate

ma = mg sin - f

With

With and    In

M = R f = IZ with I.Z = Moment of inertia about axis of symmetry

The comparison with leads to the result Inserting the moments of inertia for solid cylinders and hollow cylinders results in:

Full cylinder: a = g sin the force is used on rotation.
Hollow cylinder: a = g sin the force is used on rotation.

The full cylinder has the greater acceleration and therefore rolls faster.

Another example for calculating a rotational movement with the help of the moment of inertia is Maxwell's disk:

The Maxwell disc is constructed like a yo-yo with an inner disc on which a thread is rolled up and an outer disc. If you drop the yo-yo while holding the thread, gravity and thread tension act. The center of gravity of the yoyo, on which the gravitational pull of the earth acts, lies in the center of the two concentric discs. The point at which the force acts via the thread, however, lies on a point on the inner disk. The two forces can be broken down again into a residual force and a force couple. The yo-yo rotates and runs downwards. However, if the thread is completely unwound, the yo-yo runs back up to the starting point. The potential energy created by lifting the yo-yo is converted into rotational energy during the process. Obviously, the conservation of energy also works when converting potential energy into rotational energy. At the lowest point, the conservation of angular momentum must also cause the thread to 'wind up' on the way up. We want to summarize these principles mathematically in order to calculate the acceleration with which the disk moves downwards. The thread force attacks vertically upwards at a point that is the small radius away from the center of gravity.

The force of gravity attacks vertically downwards in the center of gravity.

The resultant of the two forces is a force that causes acceleration in the Newtonian sense. So the following applies:  .

The thread force is unfortunately not directly accessible to us via a known formula, but we do know that the thread force exerts a torque on the disc.

Because the connecting lines perpendicular to amounts can be expected. Instead of we can do that

Torque to use.

The same applies .

Off and    If we substitute this expression for the thread force in the amount of, it follows    If we insert the moment of inertia of a disk or a cylinder into this formula for the acceleration of a yoyo, it follows

with I =  Is R2/ r2 so much larger than 1, the disk rolls off extremely slowly.

In this experiment, a Maxwell disk is hung on a pan of a balance.

1) When the yo-yo is rolled up, the balance is adjusted by placing weights in the other weighing pan. The yo-yo is chosen so that the outer radius is considerably larger than the inner one.

2) If you let the yo-yo run down, the scale with the yo-yo goes up; it shows a lower weight than before. This can be justified as follows:

In state 1) only the gravitational pull of the earth G = Mg acts on the shell.

If the yo-yo is running, the forces listed above act . The comparison of the forces shows that the force in state 2) is smaller than the first. The balance shows this difference.

3) After the yo-yo is turned over, the scales show a lower weight again when walking up. Here, too, both forces must be viewed from above. This time the acceleration a acts as an upward deceleration, but it has a negative sign. The resultant then acts downwards again and the acting force is reduced.