Calculating Gravity from Inclined plane

[haruspex]

Which means our ratio should be 1 / (2/5 r^2/x^2 + 1) ??

Almost.
I'm afraid I may have confused you a little.
When dealing with bodies that are both rotating and moving linearly, there are two main ways of thinking of it:
- as a linear movement of the mass centre, plus a rotation about the mass centre, or
- as a rotation about the instantaneous centre of rotation.
With the first approach, the total KE is $\frac 12 m(v^2+\frac 25 r^2\omega^2) = \frac m2 (x^2+\frac 25 r^2)\omega^2$, exactly the result the second approach gives.
Hence the fraction of the total KE that is linear is $\frac{x^2}{x^2+\frac 25 r^2}$.
Can you see how to get from there to an expression for v as a function of the height loss?

 

[Fat_Squirrel]

1/[x^2 / (x^2 + 2/5 r^2)] * a * sinθ

so 1/[10.41^2 / (10.41^2 + 2/5 * 15^2) * 0.1646 * (0.061/2)] = 9.866

Answer looks good,how'd I go with this one?

 

[haruspex]

1/[x^2 / (x^2 + 2/5 r^2)] * a * sinθ

so 1/[10.41^2 / (10.41^2 + 2/5 * 15^2) * 0.1646 * (0.061/2)] = 9.866

Answer looks good,how'd I go with this one?

Yes, I believe that works.

 

[Fat_Squirrel]

Yes, I believe that works.

Once again, thanks for your help and patience. Will let you know if we got the answer my teacher expected.

 

[Fat_Squirrel]

Yes, I believe that works.

here's the solution they provided - https://youtu.be/IED3ujslfPY. I got a slightly different answer because I calculated acceleration differently.
I used values associated with t^2 and ran least squares on them to get my value for the slope, he used √d values - i thought they'd give same answer, but slightly different it seems.

Anyway if I use his value for acceleration I get the exact same answer he does. Thankfully my answer was within the margin of error. Thanks again.