# Calculating Gravity from Inclined plane

haruspex
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What are we working towards here, total Kinetic energy ? Sorry, it's just not clicking.

What's our end goal here? If I know what we're working towards that might help me put the pieces into place.
We have a value for x, are we trying to work out ω, or looking for something to substite in place of v? Sorry, i've looked at so many formulas its making less and less sense.

Was the other person here posting about effective mass on point, or was that a red herring?
We are now in a position to work out what fraction of the KE goes into linear motion and what into rotational. That will give us the ralationship between distance moved down the ramp and linear velocity.

We are now in a position to work out what fraction of the KE goes into linear motion and what into rotational. That will give us the ralationship between distance moved down the ramp and linear velocity.

K_linear = 1/2mv^2
K_rotational = 2/5mr^2 + mx^2 ?

Total KE = 1/2mv^2 + 2/5mr^2 + mx^2 ??

So rotional/linear = (1/2 v^2) / (2/5r^2 + x^2) ??

???

haruspex
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K_rotational = 2/5mr^2 + mx^2 ?
That's the moment of inertia about the instantaneous axis. What do you have to do with that to get the KE?

That's the moment of inertia about the instantaneous axis. What do you have to do with that to get the KE?

KE=1/2Iω^2

KE=1/2I(v^2/x^2) ??

KE=1/2Iω^2

KE=1/2*I*(v^2/x^2) ??

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

haruspex
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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?

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
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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.

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.

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.