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The Cosmic Game ( PDFDrive.com )

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<strong>Cosmic</strong> <strong>Game</strong> © Douglass A. White, 2012 v151207 62<br />

18° .30902 .3141593<br />

21° .35837 .3665192<br />

45° .70711 .78539805<br />

90° 1.000 1.5707961<br />

We can describe the motion of a pendulum with a stiff, but massless arm in two<br />

dimensions with no losses to friction or air resistance using a differential equation:<br />

(d²θ/ dt²) = - (g /R) sin θ .<br />

θ<br />

R<br />

θ<br />

motion of bob<br />

F = - mg<br />

In this equation g is the acceleration caused by gravity and tends downward. R is the<br />

length of the arm, which we will set at 1 meter. <strong>The</strong> angle of displacement from vertical<br />

is θ. <strong>The</strong> motion of the bob at any moment is always along the tangent to the circular arc<br />

at the point the bob has reached along the arc, but the arm of the pendulum forces the bob<br />

to follow the circular path rather than continue along the tangent. We can represent the<br />

tangential <strong>com</strong>ponent of the gravitational force as:<br />

F = -mg sin θ = ma<br />

a = -g sin θ.<br />

<strong>The</strong>ta (θ) and the acceleration are in opposite directions, for when the bob swings to the<br />

right, it accelerates back to the left. If s is the length of the arc, we represent it in<br />

radians as R θ.<br />

<strong>The</strong> speed (NOT velocity) along the arc is then v = ds/dt = Rdθ/dt.<br />

<strong>The</strong> acceleration along the arc is then a = d²θ/ dt² = R d²θ/ dt².<br />

<strong>The</strong>refore, d²θ/ dt² = - (g/R) sin θ.<br />

We can also see the relation from the energy conservation principle. As the bob falls, it<br />

loses potential energy and gains kinetic energy. <strong>The</strong> change in potential energy is mgh,<br />

where h is the change in height of the bob. <strong>The</strong> change in kinetic energy from the bob at<br />

rest at its highest point when it starts to fall is ½mv². In our frictionless pendulum no<br />

energy is lost, so ½mv² = mgh, and v = √(2gh).

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