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\begin{large}\[ L_{i} = mv_{i}b \]\end{large}

\begin{large}\[ E_{\rm total} = KE + PE \]\end{large}

\begin{large}\[ E_{\rm initial\;total} = \frac{1}{2}m{v_{i}}^{2} - \frac{mMG}{r|

|!636px-Jupiter_showing_SL9_impact_sites.jpg!|
|Jupiter, showing the impact marks from fragments of Comet Shoemaker-Levy 9 in July 1994
Photo from Wikimedia Commons. Original by NASA/JPL|


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{excerpt:hidden=true}Calculation of Effective Cross-Section of a planet with Gravity{excerpt}

Because [gravity] will act to pull meteors, comets, and other space debris toward a planet, the effective cross-section for a planet to capture an object is larger than its geometrical cross-section. What is the size of this effective cross-section in terms of the physical qualities of the planet and the situation? What features of the impacting body is it independent of?

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}\[ IL_{f} = \fracmv_{1}{2}m r^2f}R \] \end{large}

{latex}

So the Moment of Inertia of the complete flywheel is:

{latex}\begin{large}\[ IE_{\rm final\;total} = \frac{1}{2}(m r^2 + M R^2 ) m{v_{f}}^{2} - \frac{mMG}{R} \]\end{large}{latex}

The expression for the angular velocity and the angle as a 

\begin{large}\[ mv_{i}b = mv_{f}R \]\end{large}

function of time (for constant angular acceleration) is given in the *Laws of Change* section on the [Rotational Motion] page:



{latex}\begin{large}\[ \omegav_{\rm f} = \frac{b}{R}v_{i} \]\end{large}

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

and

{latex}\begin{large}\[ \theta_{\rm f} b = R\theta_sqrt{\rm1 i} + \omega_{\rm i} ( t_{\rm f} - t_{\rm i} ) + \frac{1}{2} \alpha ( t_{\rm f} - t_{\rm i} )^2  frac{2MG}{{v_{i}}^{2}}\frac{r_{i}-R}{r_{i}R}} \]\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}\[ \omega_{\rm f}b = R\alpha t_{\rm f}\]\end{large}{latex}
and
{latex}\begin{large}\[ \theta_{\rm f} = \frac{1}{2} \alpha {t_{\rm f}}^2sqrt{1 + \frac{2MG}{R{v_{i}}^{2}}} \]\end{large}{latex}
where
{latex}

\begin{large} \[ Cross \alpha; Section = \fracpi b^{rF}{I_{\rm total}} =2} = \pi R^{2}\left(1 + \frac{2rF2MG}{mr^2 + MR^2 }{R{v_{i}}^{2}}\right) \]\end{large}{latex}

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