Question

http://techtv.mit.edu/collections/locomotion:1216/videos/8007-dynamic-walking-2010-andy-ruina-cats-astronauts-trucks-bikes-arrows-and-muscle-smarts-stability-translation-and-rotation

at 20 mins.. he starts explaining about angular momentum conservation and about how cat lands on its feet..

but my god.. it gets too complicated..
can someone explain the woman rotating on that hanging thing for me? its at around 30 mins

Answer 1

my goodness this looks like 1000000x more complicated than my physics class

Answer 2

Dustin explains it really well.
http://youtu.be/RtWbpyjJqrU

Answer 3

Insofar as I can tell, this is his explanation - all of this is looking top down


\[ M = \dot H = \frac{ \mathrm{d}H}{\mathrm{d}t} \hspace{25px} \textrm{is like} \hspace{25px} \tau= \frac{ \mathrm{d}L}{\mathrm{d}t}\]
The net torque M is equal to the change in angular momentum, h, in time. But also because of the "viscous friction" in the system, making it act like a "dashpot", some constant c times the angular velocity is proportional to the change in angular momentum in time; aka. torque.
\[ \ c \dot \theta = \dot H \hspace{25px} \textrm{is like} \hspace{25px} c \omega= \frac{ \mathrm{d}I \omega}{\mathrm{d}t}=\tau\]
\[c\frac{ \mathrm{d}\theta}{\mathrm{d}t} = \frac{ \mathrm{d}H}{\mathrm{d}t}\]

c is (apparently) a damping coefficient proportional to the torque per angular velocty
\[c \propto \frac{ \dot H}{\dot \theta}\]

anyways, you integrate that.

\[c\int \mathrm{d}\theta = \int \mathrm{d} H = 0\]

We know the change in angular momentum is zero from all the stuff on the other slide:

--------------------- other slide----------------------
You can show that given there is no torque on the center of mass, the angular momentum if a system of i particles is constant
\[ \vec \tau = \textbf r \times \textbf F = \textbf r \times m \textbf a = \frac{\mathrm{d}}{\mathrm{d}t} \textbf r \times m \textbf v
\\ \
\\ \ \ \textrm{would be for 1 particle. For *many*, sum over i particles, use center of mass}\]
\[ \tau_C=M_C=\sum_i \textbf r_{i/C} \times \textbf F_i=0 \]
\[= \sum_i \textbf r_{i/C} \times (m_i \textbf a_i) = \frac{\mathrm{d}}{\mathrm{d}t} \sum_i \textbf r_{i/C} \times (m_i \textbf v_i) = \frac{\mathrm{d}}{\mathrm{d}t} H = 0 \]
\[\frac{\mathrm{d}}{\mathrm{d}t} H = 0 \hspace{15px} \textrm{implies that H, angular momentum, is constant wrt time}\]
-----------------------------------------------------

Using the viscous friction relationship, since there is no change in her angular momentum H (starts at 0, ends at zero) from start to finish, there can't be any change in the net angular displacement
\[ \Delta \theta = \Delta H = 0\]
**independent** of the magnitude of the damping constant c. If the viscous friction exists, it always effects in the same manner; so she swings back to her initial position.

Except she doesn't...

So he adds that the extra built up angular momentum is from the the torque being built up over time due to some *non-viscous friction that exists in the system. Something vaguely related to this like this

\[F_{friction} \propto M = \dot H\]
\[F_{friction} \propto \int M \ \mathrm{d}t = H \hspace{25px} \textrm{aka} \hspace{25px} \int \tau \ \mathrm{d}t = L\]
So even though she's not increasing her angular displacement, she's building up angular momentum by applying the torque from flailing her body around.

That's what I got out of it :P

Answer 4

ok.. .kinda makes sense.. thanks for all that man.

but what is the meaning of angular momentum being built up!? does it get stored or something like some sort of potential energy.. which then releases and she starts spinning like at the end how she does (after she says she gets tired) ?

Answer 5

I'm not sure it makes sense :P The more I read it, the less it clicks. Anyway.

So he says that you can "do things" with the normal friction, "The friction is a constant times the sign of the direction. Then the accumulated angular momentum in the wrong direction is proportional to the time she's [working]"

It has something to do with this maybe?

\[F_{friction} \propto \int_0^{\textbf t} M \ \mathrm{d}t = H\]

He does say "accumulated," like it's building up. The longer she goes before stopping, the more time gets added to the integral, so the greater the angular momentum. That's the way he says it.

I can't figure out how it would get stored though.

When she stops swinging her legs though, she tightens up as they swing around, so all of the torque that she had been using to try and turn the bar suddenly gets directly translated into angular velocity in the negative theta direction.

I think that if she had just gone limp she would have gone back to her original starting position negative 4 turns from where she stopped. That's just a guess though..... :/

Sorry that got so convoluted! :(