no Uniform Flywheel
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Uniform Flywheel

Consider a uniform flywheel of mass $M$ and radius $a$. Suppose that the flywheel rotates about a horizontal axes that passes through its center, and is perpendicular to the plane of the flywheel. Let the flywheel have a light inextensible cord wrapped around its circumference to one end of which is attached a mass $m$ that dangles below the flywheel. Let $T$ be the tension in the cord. See Figure 1.9. Let us determine the angular acceleration of the flywheel.

Figure 1.9: A flywheel.
\includegraphics[height=3.5in]{Chapter02/fly.eps}

The flywheel is rotationally symmetric about its axis of rotation, which implies that this axis is a principal axis of rotation. Suppose that the rotation axis corresponds to the $z$-axis (this axis is directed out of the paper in Figure 1.9). It follows that

$\displaystyle \mbox{\boldmath$\omega$}$ $\displaystyle = \omega\,{\bf e}_z,$ (1.203)
$\displaystyle {\bf L}$ $\displaystyle = I\,\omega\,{\bf e}_z,$ (1.204)

where $I$ is the associated principal moment of inertia, and use has been made of Equation (1.191). Now, it is clear that the tension in the cord exerts a torque

$\displaystyle \mbox{\boldmath$\tau$}$$\displaystyle = T\,a\,{\bf e}_z$ (1.205)

on the flywheel. Hence, the rotational equation of motion of the flywheel, (1.180), yields

$\displaystyle I\,\frac{d\omega}{dt} = \tau = T\,a.$ (1.206)

Consider the equation of motion of the dangling mass. We can write

$\displaystyle m\,\frac{dv}{dt} = m\,g-T,$ (1.207)

where $v$ is the mass's downward velocity, and $g$ is the acceleration due to gravity. Because the cord is inextensible, it follows that $v$ is also the downward velocity of the cord. Assuming that the cord unwraps from the flywheel without slipping, its downward velocity must match the tangential velocity of the flywheel's outer rim. In other words,

$\displaystyle v = a\,\omega,$ (1.208)

which implies that

$\displaystyle \frac{dv}{dt} = a\,\frac{d\omega}{dt}.$ (1.209)

Equations (1.206), (1.207), and (1.209) yield

$\displaystyle T$ $\displaystyle = \left(\frac{I}{I + m\,a^2}\right)m\,g,$ (1.210)
$\displaystyle \frac{dv}{dt}$ $\displaystyle = \left(\frac{m\,a^2}{I+ m\,a^2}\right)g,$ (1.211)
$\displaystyle \frac{d\omega}{dt}$ $\displaystyle = \left(\frac{m\,a^2}{I + m\,a^2}\right)\frac{g}{a}.$ (1.212)

It remains to calculate the principal moment of inertia, $I$, of the flywheel about its rotation axis. The moment of inertia is written

$\displaystyle I = \sum_i m_i\,r_i^{\,2}$ (1.213)

for all of the mass elements that make up then flywheel. Here, $r$ represents radial distance from the axis of rotation. See Figure 1.9. We can assume that the flywheel has a constant mass per unit area (in the plane perpendicular to the rotation axis), $\rho $. Consider the contribution of an annulus of inner radius $r$ and outer radius $r+dr$ to the moment of inertia. It is clear that the area of the annulus is $2\pi\,r\,dr$, Thus, its mass is $dm=2\pi\,r\,dr\,\rho$. Hence,

$\displaystyle dI= dm\,r^2 = 2\pi\,\rho\,r^3\,dr,$ (1.214)

which implies that

$\displaystyle I= 2\pi\,\rho\int_0^a r^3\,dr = \frac{\pi\,\rho\,a^4}{2}.$ (1.215)

However, the total mass of the flywheel is

$\displaystyle M = \int dm = 2\pi\,\rho\int_0^a r\,dr = \pi\,\rho\,a^2.$ (1.216)

It follows that

$\displaystyle I = \frac{1}{2}\,M\,a^2.$ (1.217)

Finally, according to Equations (1.212) and (1.217), the angular acceleration of the flywheel is

$\displaystyle \frac{d\omega}{dt} =\left(\frac{2\,m}{M+2\,m}\right)\frac{g}{a}.$ (1.218)

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Next: Gravitational Collapse of Star Up: Rigid Body Rotation Previous: Power