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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.  

h3. Part A 

First, 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 Acceleration|1-D Motion (Constant Acceleration)].

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 gives:

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



Similarly, angular momentum is conserved since there are no external torques.  For simplicity, we compute the angular momentum about the astronaut's center of mass.

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