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Photo courtesy NASA images.

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Excerpt
hiddentrue

The dangers of angular momentum in outer space.


A common problem from the conservation of linear momentum is that of the stranded astronaut. An astronaut is stuck a few meters from their spacecraft and decides to throw a tool away from the craft so that they recoil toward the craft. In this problem, we consider why it is important to take angular momentum into account when deciding how to move around in space.

Suppose an astronaut who has a mass of 95 kg when gear is included is at rest 5.0 m from their ship and decides to test Newton's 3rd Law by throwing a 1.5 kg tool directly away from the ship at a speed of 3.5 m/s. Suppose the astronaut throws the tool "overarm". The tool is released at the same level as the top of the astronaut's head. Model the astronaut as a uniform thin rod 1.75 m in height and model the tool as a point mass. By the time the astronaut has reached the ship, how many head-over-heels revolutions has the astronaut undergone? (For this very crude estimate, ignore any asymmetries – such as right/left handedness – which would cause the astronaut to spin about other axes as well.)

Solution

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System:
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Astronaut as plus tool as .

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Interactions:
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There are no external influences on the system.

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Model:
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Constant Linear Momentum and plus One-Dimensional Motion with Constant Velocity.

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Approach:

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Diagrammatic Representation

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It is important to sketch the situation and to define linear and rotational coordinate axes.

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Mathematical Representation

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There are effectively no important external forces (gravity can be neglected since the astronaut and tool are in freefall), so linear momentum conservation will give us the astronaut's linear speed toward the ship. The relationship is straightforward:

Latex
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|!spacewalk.jpg!|
|Photo courtesy NASA images.|
A common problem from the conservation of linear momentum is that of the stranded astronaut.  An astronaut is stuck a few meters from their spacecraft and decides to throw a tool away from the craft so that they recoil toward the craft.  In this problem, we consider why it is important to take angular momentum into account when deciding how to move around in space.

Suppose an astronaut who has a mass of 95 kg when gear is included is at rest 5.0 m from their ship and decides to test Newton's 3rd Law by throwing a 1.5 kg tool directly away from the ship at a speed of 3.5 m/s.  Suppose the astronaut throws the tool "overarm".  The tool is released at the same level as the top of the astronaut's head.  Model the astronaut as a uniform thin rod 1.75 m in height and model the tool as a point mass.  By the time the astronaut has reached the ship, how many head-over-heels revolutions has the astronaut undergone?  (For this very crude estimate, ignore any asymmetries that might cause the astronaut to spin about other axes as well.)

System:  Astronaut as [rigid body] plus tool as [point particle].

Model:  [Constant Linear Momentum|Constant Momentum] and [Constant Angular Momentum] plus [One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)].

Approach:  It is important to sketch the situation and to define linear and rotational coordinate axes.

!spacewalkcoord.png!

There are effectively no important external forces (gravity can be neglected since the astronaut and tool are in freefall), so linear momentum conservation will give us the astronaut's linear speed toward the ship.  The relationship is straightforward:

{latex}\begin{large}\[ m_{a}v_{a,{\rm f}} - m_{t}v_{t,{\rm f}} = 0 \]\end{large}{latex}

This

...

gives:

{
Latex
}\begin{large}\[ v_{a,{\rm f}} = \frac{m_{t}v_{t,{\rm f}}}{m_{a}} \]\end{large}{latex}

Since

...

this

...

velocity

...

is

...

assumed

...

to

...

be

...

constant,

...

we

...

can

...

use

...

the

...

(sole)

...

Law

...

of

...

Change

...

from

...

One-Dimensional

...

Motion

...

with

...

Constant

...

Velocity

...

to

...

find

...

that

...

the

...

time

...

required

...

to

...

return

...

to

...

the

...

ship

...

is:

{
Latex
}\begin{large}\[ t = \frac{m_{a}x_{s}}{m_{t}v_{t,{\rm f}}} \] \end{large}{latex}

Similarly,

...

angular

...

momentum

...

is

...

conserved

...

since

...

there

...

are

...

no

...

external

...

torques.

...

We

...

can

...

choose

...

any

...

non

...

accelerating

...

axis

...

.

...

For

...

simplicity,

...

we

...

compute

...

the

...

angular

...

momentum

...

about

...

the

...

initial

...

location

...

of

...

the

...

astronaut's

...

center

...

of

...

mass.

Note

Since we have chosen an axis along the

{note}Since we have chosen an axis along the

astronaut's

center

of

mass'

line

of

motion,

the

translation

of

the

astronaut's

center

of

mass

will

not

contribute

to

the

angular

momentum.

Further,

since

we

are

treating

the

tool

as

a

point

particle,

it

has

no

moment

of

inertia

about

its

center

of

mass

and

so

its

rotations

will

not

contribute

to

the

angular

momentum.

Latex
{note}

{latex}\begin{large}\[ \frac{1}{12}m_{a}h^{2}\omega_{\rm f} - m_{t}v_{t,{\rm f}}h/2 = 0 \] \end{large}{latex}

giving:

{
Latex
}\begin{large}\[ \omega_{\rm f} = \frac{6 m_{t}v_{t,{\rm f}}}{m_{a} h} \]\end{large}{latex}

Then,

...

using

...

the

...

time

...

found

...

above

...

and

...

the

...

Law

...

of

...

Change

...

for

...

angular

...

kinematics

...

with

...

constant

...

angular

...

velocity,

...

we

...

can

...

find

...

the

...

total

...

angle

...

the

...

astronaut

...

rotates

...

through

...

before

...

reaching

...

the

...

ship.

{
Latex
}\begin{large}\[ \theta = \omega_{\rm f} t = \frac{6 x_{s}}{h} = \mbox{17 radians} = \mbox{2.7 revolutions}\]\end{large}{latex}




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Challenge

Suppose the clever astronaut decided to spin the tool in an attempt to reduce the astronaut's own resulting rotation rate. What direction should the astronaut spin the tool? Assuming the tool's center of mass still moves as described above, estimate the rotation rate the astronaut would have to impart to the tool to avoid any spin on their own part.

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