# Feynman Exercises 19-1: Metal rod framework being pulled in while spinning

## The Attempt at a Solution

The moment of inertia before collapse is for each rod:
BEFORE COLLAPSE:[/B]
Ib = ∫(L2 + x2) dm = m/L ∫(L2 + x2) dx = 4/3 * m * L2
We have 8 of these plus the inertia of the mechanism, giving a total I,
It = 8 * Ib + Ik = (32/3 * m * + 40/3 * M) * L^2
The energy is thus:
Tb = 1/2 * It* ωo2
AFTER COLLAPSE:
Ia = ∫(x2) dm = m/L ∫(x2) dx = 1/3 * m * L2
And the mechanism is the same:
So we have total of:
It_a = 8 * Ia + Ik = (8/3 * m * + 40/3 * M) * L2
The energy is now:
Ta = 1/2 * It_a * ωo2
And energy difference must be:
Ta - Tb = 1/2 * ωo2 * m * (8 / 3 - 32/3) * L2 = -ωo2 * m * 4 * L2

But the solution stated is: ωo2 * M * 6 (note the difference in mass symbol)
What am i doing wrong?

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## Answers and Replies

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kuruman
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Gold Member
Why is the angular speed unchanged through the collapse?

Why is the angular speed unchanged through the collapse?
Because the moment of inertia of the mechanism is unchanged. Since I = L / ω (L is the angular momentum here!).

But maybe I've been to quick in that assumption. All I know is that I is constant for the mechanism - not that L or ω is constant. Am I on the right track?

kuruman
Homework Helper
Gold Member
Because the moment of inertia of the mechanism is unchanged. Since I = L / ω (L is the angular momentum here!).

But maybe I've been to quick in that assumption. All I know is that I is constant for the mechanism - not that L or ω is constant. Am I on the right track?
Sure the moment of inertia of the mechanism is unchanged, but what about the framework? Is that also unchanged? Your expressions for It and It_a are not the same. In fact It_a < It which means that the contraption will speed up much like a spinning skater when she pulls her arms in.,

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Sure the moment of inertia of the mechanism is unchanged, but what about the framework? Is that also unchanged? Your expressions for It and It_a are not the same. In fact It_a < It which means that the contraption will speed up much like a spinning skater when she pulls her arms in.,
Ahh, okay. There's conservation of angular momentum. I'll work it out and return with what i got.

kuruman
Homework Helper
Gold Member
Ahh, okay. There's conservation of angular momentum. I'll work it out and return with what i got.
That's an excellent way to proceed. Note: To get the answer you think is correct, you must set $m=M$.

That's an excellent way to proceed. Note: To get the answer you think is correct, you must set $m=M$.
Ahh yes! Thanks a lot. Quick last question: Do you know if the different symbol for mass is a mistake in the problem? Or am i missing some connection?

To anyone else stuck here:

Just note that angular momemtum is conserved, meaning: I_t * ω_0 = I_t_a * ω_a (ω_a is angular velocity after collapse). After this, it's just algebra.

kuruman