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

[aa_o]

Homework Statement

Homework Equations

The Attempt at a Solution

The moment of inertia before collapse is for each rod:
BEFORE COLLAPSE:[/B]
I_b = ∫(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,
I_t = 8 * I_b + I_k = (32/3 * m * + 40/3 * M) * L^2
The energy is thus:
T_b = 1/2 * I_t* ω_o²
AFTER COLLAPSE:
I_a = ∫(x²) dm = m/L ∫(x²) dx = 1/3 * m * L²
And the mechanism is the same:
So we have total of:
I_{t_a} = 8 * I_a + I_k = (8/3 * m * + 40/3 * M) * L²
The energy is now:
T_a = 1/2 * I_{t_a} * ω_o²
And energy difference must be:
T_a - T_b = 1/2 * ω_o² * m * (8 / 3 - 32/3) * L² = -ω_o² * m * 4 * L²

But the solution stated is: ω_o² * M * 6 (note the difference in mass symbol)
What am i doing wrong?

 

[kuruman]

Why is the angular speed unchanged through the collapse?

 

[aa_o]

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]

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 I_t and I_{t_a} are not the same. In fact I_{t_a} < I_t which means that the contraption will speed up much like a spinning skater when she pulls her arms in.,

 

[aa_o]

Sure the moment of inertia of the mechanism is unchanged, but what about the framework? Is that also unchanged? Your expressions for I_t and I_{t_a} are not the same. In fact I_{t_a} < I_t 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]

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 $m=M$.

 

[aa_o]

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 momentum 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]

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?

I have no idea. Probably a typo by whoever transcribed the problem or by somebody who didn't pay attention to the difference between upper and lower case characters. It looks like the question was typed using a typewriter, a technology that faded about 30 years ago.