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|!SL9ImpactGalileo.jpg!|
|Shoemaker-Levy Comet Fragment 9 impacting Jupiter July 22 1994
Photo from Wikimedia Commons by NASA/JPL|

|!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 Hubbel Space Telescope Comet Team and NASA|


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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 [rigidPoint bodyparticle] rotatingsubject aboutto a[gravity] fixedbut pointmoving underwith constant Torque[angular momentum].{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}The fixed axis keeps the Flywheel from Accelerating. The Externally applied Torque[Gravity].{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod} Rotational Motion and Constant TorqueXXXXX.{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!The Force Diagram of the Meteor approaching the Planet

|Force Diagram of Meteor and Planet|

The [single-axis torque] about the center of the planet is zero, because the force of [gravity] acts along the same direction as the radius *r*. About this point, therefore, [angular momentum] is conserved.

|Sketch showing Torque|

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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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