Experimental determination of the moment inertia of a sphere

[Dimitris Papadim]

Hello, I was recently given the task to find experimentally the moment inertia of a sphere. I thought of rolling the sphere down an inclined plane and applying conservation of energy to the sphere. The equations i came up with are: mgh = 1/2mv² + 1/2Iω² solving for v^2 we come up with the equation:
v² = (2mgR²h)/(mR²+ I) now if we plot v²(h) we come up with a straight line through the origin and (2mgR^2)/(mR^2+ I)should be its slope. solving for I we come up with I = mR²/k(2g-k) where k is the gradient. Now if we equate this with 2/5 mR² which is the mathematical formula for the moment inertia of the sphere we should come up with the same result, but the mass and radius cancel. This makes no sense. Please help :(

 

[andrewkirk]

we come up with I = mR2/k(2g-k)

I don't get that result. I get $\frac{mR^2(2g-km)}k$, which is $mR^2(2g/k-m)$.

Now if we equate this with 2/5 mR2 which is the mathematical formula for the moment inertia of the sphere we should come up with the same result, but the mass and radius cancel.

That doesn't matter. Our task is not to equate our empirical formula to the theoretical formula and solve for something. It is to calculate the value given by the empirical formula and compare it to that given by the theoretical formula. So let's not cancel anything at that stage, instead do the following:

You will have observed values of m, R, g and k, so you can calculate the value of $mR^2(2g/k-m)$, which is an estimate of I. If the sphere is solid with uniform density and the experiment was accurate, that should give a number that is approximately $\frac25 mR^2$. In other words, $2g/k-m$ should be approximately 2/5.

 

[Dimitris Papadim]

Thank you very much!