# Angular Momentum

## Table of contents

**Note**: There is an Appendix with more information about rotational motion. This activity assumes basic familiarity with that material.

## Activity

You have a disk attached to a rotary sensor which will measure the angle as a function of time (somewhat like the motion sensor measures position as a function of time). You also have several disks and hoops. If you set the rotary sensor disk spinning and drop one of the other disks/hoops on it, the rotation will abruptly slow down. Your task is to test whether angular momentum and rotational energy are conserved in such collisions. Note that the disk and sensor have all the initial angular momentum and rotational energy. Also note that **unless you drop them perfectly centred, you will need to use the parallel axis theorem** to correct for the misalignment. (See Appendix for details.)

Remember that the best way to show your data is in the form of a graph, not a table of data.

You will need to know that the moment of inertia of a hoop is \(I = \frac{1}{2}m(r_a^2+r_b^2)\) where r_{a} and r_{b} are the inner and outer radii of the hoop. If the inner radius is zero you get a solid disk.

Design an experiment to test both concepts (conservation of angular momentum and conservation of rotational energy). Try to make your uncertainties small. You only need to do one hoop and one disk.

You need to consider:

- All sources of uncertainties, and which is largest, so your final result has an uncertainty.
- How to present your final data (preferably in the form of graphs).
- How to design your experiment to measure what you want and control everything else.
- How to design your experiment to reduce your uncertainties as much as possible.

## Appendix

In linear motion you have the equations \(\vec{a} = \frac{d\vec{v}}{dt}\), \(\vec{v} = \frac{d\vec{x}}{dt}\), and assuming constant acceleration, \(\vec{x}(t) = \vec{x}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2\). You can also define the momentum \(\vec{p} = m \vec{v}\) and kinetic energy \(K = \frac{1}{2} m v^2\), both of which are conserved under certain circumstances.

In rotational motion there are a similar set of equations. We ignore the vector nature of the equations here. In the same order, the equations are \(\alpha = \frac{d\omega}{dt}\), \(\omega = \frac{d\theta}{dt}\), and assuming constant angular acceleration, \(\theta(t) = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2\). We replace the concept of mass with the concept of rotational inertia (or moment of inertia) which we denote with \(I\). The angular momentum is \(L = I \omega\) and the rotational energy is \(K = \frac{1}{2} I \omega^2\), both of which are conserved under certain circumstances.

The moment of inertia is defined by the integral \(I = \int \rho(\vec{r}) r^2 dV\) where \(r\) is the perpendicular distance from the axis of rotation to the infinitessimal volume element \(dV\). The most important result of this theory for this experiment is the parallel axis theorem: \(I = I_{CM} + Md^2\) where \(I\) is the moment of inertia around a point a distance \(d\) away from the centre of mass of the object, and \(I_{CM}\) is the moment of inertia of the object around its centre of mass. You will need this when you drop the hoops/disks onto the platform unless you drop them perfectly centred on the platform (which is very difficult to do).