Say I have a pulley and its attached to a string which hangs onto a mass of $1\,kg$ at its end. On the surface of this pulley I place a sphere with radius $0.1\, m$. and mass of 1 kg. If the mass of this string is $0.1\, kg$ and the pulley radius is $0.05655\, m$. and the angular acceleration measured is $12.83\, rad/s^2$, what kind of sphere is positioned on top of the pulley? Is it a hollow sphere or a solid sphere?
Here's an image of the set up: http://spiff.rit.edu/classes/phys312/workshops/w2a/w2a_longer.html
Just imagine that instead of the disk on the first image there is a sphere in its place. How would you go about solving this problem?
After a more thorough look at this I will need to make a couple assumptions along the way:
The moment of inertia of the solid sphere $ I_{\text{solid}} =\frac{2}{5}MR^2 =\frac{1}{250}kg\cdot m^2 $
The moment of inertia of the spherical shell $ I_{\text{shell}} =\frac{2}{3}MR^2 = \frac{1}{150}kg\cdot m^2 $
The rotational kinetic energy of the sphere will be $\frac{1}{2}I\omega^2$. There is a typo in my comment.
I assume the $1kg$ mass starts from rest and we can ignore the mass of the clamp on super pulley. There is negligible friction in the connection of the RMS and 3 stage pulley and the string doesn't stretch or slip.
Now the change in height of the mass will give the total change in potential energy of the mass at the end of the string:
$\Delta U = mg\Delta h$
And this must be equal to the sum of the change in rotational kinetic energy of the sphere and the kinetic energy of the mass on the end of the string:(I am ignoring the mass of the string as well)
$\Delta U = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $
Since $\frac{d\omega}{dt} = \alpha$ and $\omega(0)=0$, we have $\omega(t) = \alpha t$ giving:
$\Delta U = \frac{1}{2}mv^2 + \frac{1}{2}I\alpha^2t^2 $
As a first approximation, in your experiment if you have a meter stick and a stop watch then you have enough information to evaluate:
$\sqrt{\frac{2\Delta U}{\left(mr^2+I\right)t^2}} = \alpha $
where $m$ is the mass on the end of the string and $r$ is the radius of the pulley. Then check which $I$ most closely approximates the equality. So it looks like we can make significant progress with energy alone. We also have available to us $|\vec\tau| = |\vec r\times\vec F| = rF\sin\theta = rF$ where $F$ is the tension in the string and $r$ is the radius of the pulley and $\theta$ is the angle between them which is $\frac{\pi}{2}$. and Newton's second law on the hanging mass which gives $F-mg=ma$. Finally, the acceleration of the hanging mass, $a$ is related to the agular acceleration of the sphere through $a=r\alpha$ and $t$ could also be eliminated in favor of $\alpha$ which means you would only need to measure the change in height. Ignoring the string's mass is probably a stretch.