B How to find the Moment of Inertia for a Sphero robot?

[GopherTv]

TL;DR Summary

A Sphero is a programable robot that has a Moment of inertia somewhere between 2/3MR^2 (Thin spherical shell) and 2/5MR^2 (Solid Sphere)

What kind of experiment can I design to determine the actual value of the moment of inertia. What should I instruct the sphero to do and what data should I collect?

 

[kuruman]

Get it to roll without slipping down an incline starting from rest and have it report its speed at the bottom of the incline. Energy conservation says that for a vertical drop $h$ the potential energy is converted into kinetic energy of the center of mass and rotational energy about the center of mass,$$mgh=\frac{1}{2}mV_{\text{cm}}^2+\frac{1}{2}I_{\text{cm}}\omega^2.$$As you noted, you can write the moment of inertia about the CoM as $I_{\text{cm}}=qmR^2$ where $q$ is the constant fraction that you want to determine. If this thing rolls without slipping, $\omega =\dfrac{V_{\text{cm}}}{R}$ in which case the energy conservation equation becomes $$mgh=\frac{1}{2}mV_{\text{cm}}^2+\frac{1}{2}(qmR^2)\left( \frac{V_{\text{cm}}}{R}\right)^2.$$After the obvious cancellations you get $$gh=\frac{1}{2}V_{\text{cm}}^2+\frac{1}{2}qV_{\text{cm}}^2\implies q=\frac{2gh}{V_{\text{cm}}^2}-1.$$So all you have to do is measure the vertical height by which it drops and have it tell you how fast it is moving after it drops by that height and plug in. Its mass or radius don't matter with this technique. This is counterintuitive when what you are trying to determine is its moment of inertia but there you have it. Just make sure there is no slipping.

 

[Dale]

Just make sure there is no slipping.

And that it rolls down without power

 

[GopherTv]

Ok, this makes sense, i appreciate the help.

Thanks!

 

[erobz]

Get it to roll without slipping down an incline starting from rest and have it report its speed at the bottom of the incline. Energy conservation says that for a vertical drop $h$ the potential energy is converted into kinetic energy of the center of mass and rotational energy about the center of mass,$$mgh=\frac{1}{2}mV_{\text{cm}}^2+\frac{1}{2}I_{\text{cm}}\omega^2.$$As you noted, you can write the moment of inertia about the CoM as $I_{\text{cm}}=qmR^2$ where $q$ is the constant fraction that you want to determine. If this thing rolls without slipping, $\omega =\dfrac{V_{\text{cm}}}{R}$ in which case the energy conservation equation becomes $$mgh=\frac{1}{2}mV_{\text{cm}}^2+\frac{1}{2}(qmR^2)\left( \frac{V_{\text{cm}}}{R}\right)^2.$$After the obvious cancellations you get $$gh=\frac{1}{2}V_{\text{cm}}^2+\frac{1}{2}qV_{\text{cm}}^2\implies q=\frac{2gh}{V_{\text{cm}}^2}-1.$$So all you have to do is measure the vertical height by which it drops and have it tell you how fast it is moving after it drops by that height and plug in. Its mass or radius don't matter with this technique. This is counterintuitive when what you are trying to determine is its moment of inertia but there you have it. Just make sure there is no slipping.

The CM might not be in the center of the sphere ( its axis of rotation) as it is in a solid sphere or shell with uniform density which I think adds another parameter to finding the MOI, thus making this inconclusive?

Does the robot itself rotate, or does it remain "level" while the shell rotates?

EDIT:
From what I can tell, the shell rotates, the "bot" driving it tries not to. So, if its rolling down a hill ( and it its on ) it is actively applying forces to the shell through it rollers to maintain it orientation. These forces ( and other internal rotating components ) are probably the reason the MOI appears to be between the shell and the solid sphere.