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|!400px-Schwungrad02.jpg500px-Lissajous_curve_3to4.svg!|
|Old Flywheel in Witten, Germany
Photo Lissajous Figure
from Wikimedia Commons: Image by Peter MarkusD SchweissReid|



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{excerpt:hidden=true}AccelerationImage ofgenerated a symmetric object about by a fixedPendulum axiswith undertwo constantnatural torqueFrequencies.{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 solid 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}

The Moment of Inertia of combined bodies about the same axis is simply the sum of the individual Moments of Inertia:

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

The Moment of Inertia of a solid disc of radius *r* and mass *m* about an axis through the center and perpendicular to the plane of the disc is given by:

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

So the Moment of Inertia of the complete flywheel is:

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

The expression for the angular velocity and the angle as a function of time (for constant angular acceleration) is given in the *Laws of Change* section on the [Rotational Motion] page:



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

and

{latex}\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}{latex}

We assume that at the start, *t{~}i{~} = 0* , we have both position and angular velocity equal to zero. The above expressions then simplify to:

{latex}\begin{large}\[ \omega_{\rm f} = \alpha t_{\rm f}\]\end{large}{latex}
and
{latex}\begin{large}\[ \theta_{\rm f} = \frac{1}{2} \alpha {t_{\rm f}}^2 \]\end{large}{latex}
where
{latex}\begin{large}\[ \alpha = \frac{rF}{I_{\rm total}} = \frac{2rF}{mr^2 + MR^2 }\]\end{large}{latex}

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