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Old Flywheel in Witten, Germany |
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Acceleration of a symmetric object about a fixed axis under constant torque |
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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 angular position as a function of time?
Solution
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Flywheel as |
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The fixed axis keeps the Flywheel from Accelerating. The Externally applied |
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Rotational Motion and Constant |
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It is important to sketch the situation and to define linear and rotational coordinate axes.
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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
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{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 [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:* {cloak:id=app} {toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color} {cloak:id=diag} It is important to sketch the situation and to define linear and rotational coordinate axes. !Accelerating Flywheel 01.PNG! {cloak:diag} {toggle-cloak:id=math} {color:red} *Mathematical Representation* {color} {cloak:id=math} 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} |
The Moment of Inertia of combined bodies about the same axis is simply the sum of the individual Moments of Inertia:
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\begin{large}\[ I_{\rm total} = I_{\rm small} + I_{\rm large} \]\end{large}{latex}
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The
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Moment
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of
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Inertia
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of
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a
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solid
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disc
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of
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radius
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r
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and
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mass
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m
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about
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an
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axis
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through
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the
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center
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and
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perpendicular
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to
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the
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plane
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of
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the
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disc
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is
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given
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by:
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}\begin{large}\[ I = \frac{1}{2}m r^2 \] \end{large}{latex} |
So
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the
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Moment
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of
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Inertia
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of
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the
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complete
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flywheel
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is:
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}\begin{large}\[ I_{\rm total} = \frac{1}{2}(m r^2 + M R^2 ) \]\end{large}{latex} |
The
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expression
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for
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the
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...
...
and
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the
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...
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as
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a
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function
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of
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time
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(for
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constant
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...
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)
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is
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given
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in
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the
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Laws
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of
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Change
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section
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on
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the
...
...
...
page:
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{latex}\begin{large}\[ \omega_{\rm f} = \omega_{\rm i} + \alpha (t_{\rm f} - t_{\rm i}) \] \end{large} |
and
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{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
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assume
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that
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at
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the
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start,
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t
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i
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=
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0
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,
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we
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have
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both
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angular
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position
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and
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angular
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velocity
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equal
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to
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zero.
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The
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above
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expressions
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then
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simplify
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to:
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}\begin{large}\[ \omega_{\rm f} = \alpha t_{\rm f}\]\end{large} |
and
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{latex}
and
{latex}\begin{large}\[ \theta_{\rm f} = \frac{1}{2} \alpha {t_{\rm f}}^2 \]\end{large} |
where
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{latex} where {latex}\begin{large}\[ \alpha = \frac{rF}{I_{\rm total}} = \frac{2rF}{mr^2 + MR^2 }\]\end{large}{latex} {cloak:math} {cloak:app} {td} {tr} {table} {live-template:RELATE license} |
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