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



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{excerpt:hidden=true}Acceleration of a symmetric object about a fixed axis under constant torque.{excerpt}

A *Flywheel* is a large symmetrical wheel that is used to store kinetic energy. It is also used to "even out" the rate of rotation, making it less susceptible to variations in the driving force. Potter's wheels and Drop Spindles are millenia-old examples of the latter case -- the large angular momentum of the spinning disc making it less likely that small interruptions or changes in the driving force will have a large effect on the angular velocity . In the 19th century large flywheels were used to store the large amounts of kinetic energy of water-driven machinery in factories, as in the photo above.

Assume that a flywheel consists of two joined discs of differing diameter, and that the force is applied tangentially to the smaller of these. What is the torque, and what are the angular velocity and the angle as a function of time?

h4. Solution

{toggle-cloak:id=sys} *System:*  {cloak:id=sys}Flywheel as [rigid body] rotating about a fixed point under constant Torque.{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}The fixed axis keeps the Flywheel from Accelerating. The Externally applied Torque.{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod} Rotational Motion and Constant Torque.{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.

!Accelerating Flywheel 01.PNG!

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

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The force is supplied by a belt around the smaller wheel of radius *r* (in a 19th century factory, it would probably be a circular leather belt attached to the water wheels). This means that the direction the force is applied along is always tangential to the circumference of the wheel, and hence *Torque = r X F = rF*

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

{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}\[ tI = \frac{m_{a}x_{s}}{m_{t}v_{t,{\rm f}}}1}{2}m r^2 \] \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}\[ I_{\rm total} = \frac{1}{2}(m r^2 + M R^2 ) \]\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}\[ \frac{1}{12}m_{a}h^{2}\omega_{\rm f} - m_{t}v_{t,{= \alpha t_{\rm f}}h/2 = 0 \] \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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