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\begin{large}\[ \vec{\tau} = \vec{r} X \vec{F} = rF = I_{\rm total} \alpha \]\end{large}

\begin{large}\[ I_{\rm total} = I_{\rm small} + I_{\rm large} \]\end{large}

\begin{large}\[ I = \frac{1}{2}m r^2 \] \end{large}

\begin{large}\[ I_{\rm total} = \frac{1}{2}(m r^2 + M R^2 ) \]\end{large}

\begin{large}\[ \omega_{\rm f} = \omega_{\rm i} + \alpha (t_{\rm f} - t_{\rm i}) \] \end{large}

\begin{large}\[ \theta_{\rm f} = \theta_{\rm i} + \omega_{\rm i} ( t_{\rm f} - t_{\rm i} ) + \frac{1}{2} \alpha ( t_{\rm f} - t_{\rm i} )^2  \]\end{large}

\begin{large}\[ 

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|!400px-Schwungrad02.jpg!|
|Flywheel from Witten, Germany
Photo from Wikimedia Commons by Markus Schweiss|



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{excerpt:hidden=true}The dangers of angular momentum in outer space.{excerpt}
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.)

h4. Solution

{toggle-cloak:id=sys} *System:*  {cloak:id=sys}Astronaut as [rigid body] plus tool as [point particle].{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}There are no external influences on the system.{cloak}

{toggle-cloak:id=mod} *Model:*  {cloak:id=mod}[Constant Linear Momentum|Momentum and External Force] and [1-D Angular Momentum and Torque] plus [One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)].{cloak}

{toggle-cloak:id=app} *Approach:*  

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{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}

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

!spacewalkcoord.jpg!

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{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

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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}\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 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.{note}

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

giving:

{latex}

\begin{large}\[ \omegatheta_{\rm f} = \frac{6 m_1}{2} \alpha {t}v_{t,{\rm f}}}{m_{a} h} ^2 \]\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}\[ \thetaalpha = \omegafrac{rF}{I_{\rm ftotal}} t = \frac{6 x_{s}}{h} = \mbox{17 radians} = \mbox{2.7 revolutions}\2rF}{mr^2 + MR^2 }\]\end{large}{latex}

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h4. 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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