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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 (single-axis)].{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 (single-axis)], and what are the [angular velocity] and the angle[angular position] 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 (single-axis)].{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod} Rotational Motion and Constant [Torque (single-axis)].{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[angular position] 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 angular 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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