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Photo courtesy NASA images. |
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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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Astronaut as |
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There are no external influences on the system. |
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It is important to sketch the situation and to define linear and rotational coordinate axes.
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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:
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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 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 complete 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: 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
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gives:
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}\begin{large}\[ v_{a,{\rm f}} = \frac{m_{t}v_{t,{\rm f}}}{m_{a}} \]\end{large}{latex} |
Since
...
this
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velocity
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is
...
assumed
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to
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be
...
constant,
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we
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can
...
use
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the
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(sole)
...
Law
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of
...
Change
...
from
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One-Dimensional
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Motion
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with
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Constant
...
Velocity
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to
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find
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that
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the
...
time
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required
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to
...
return
...
to
...
the
...
ship
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is:
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}\begin{large}\[ t = \frac{m_{a}x_{s}}{m_{t}v_{t,{\rm f}}} \] \end{large}{latex} |
Similarly,
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angular
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momentum
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is
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conserved
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since
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there
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are
...
no
...
external
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torques.
...
We
...
can
...
choose
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any
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non
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accelerating
...
axis
...
.
...
For
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simplicity,
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we
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compute
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the
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angular
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momentum
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about
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the
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initial location of the astronaut's center of mass.
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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. |
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location_ of the astronaut's center of mass. {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:
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}\begin{large}\[ \omega_{\rm f} = \frac{6 m_{t}v_{t,{\rm f}}}{m_{a} h} \]\end{large}{latex} |
Then,
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using
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the
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time
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found
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above
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and
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the
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Law
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of
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Change
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for
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angular
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kinematics
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with
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constant
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angular
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velocity,
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we
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can
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find
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the
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total
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angle
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the
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astronaut
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rotates
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through
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before
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reaching
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the
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ship.
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}\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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