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2.5.1 The Simple Pendulum Lets consider a simple pendulum, which consists of a particle of mass m (called the bob of the
pendulum) suspended from one end of an unstretchable, massless sting of length L that is fixed at the
other end (See Figure 6). The bob is free to swing back and forth in the plane of the page, to the left and
right of a vertical line through the pendulum’s pivot point.
r
r
The forces acting on the bob are the force T from the string and the gravitational force Fg as r shown in Figure 7 where the string makes an angle θ with the vertical. We resolve Fg into a radial r r component Fg cos θ and a component Fg sin θ that is tangent to the path taken by the bob. Figure 6 Diagram of a Simple Pendulum.
http://xeon.concord.org:8080/modeler1.3/mirror/mechanics/pendulum.gif Figure 7 Diagram of the Forces Acting on a
Simple Pendulum.
http://img.sparknotes.com/figures/1/103aba3edae
345165f4823058d6a1e6f/pendulum3.gif This tangential component produces a restoring torque about the pendulum’s pivot point because the
component always acts opposite the displacement of the bob so as to bring the bob back towards its 6 central location. That location is called the equilibrium position (θ=0) because the pendulum would be at
rest there were it not swinging.
Using sin (35)
(36) From equation (35) and Figure 7, we find that that the restoring torque is sin (37) where the minus sign indicates that the torque acts to reduce θ and L is the moment arm of the force
r
component Fg sin θ about a pivot point. Substituting equation (37) into Newton’s Second Law of Motion (38)
where I is the pendulum’s rotational inertia about the pivot point and α is its angular acceleration about
that point, we then obtain sin (39) We can simplify equation (39) with an approximation if we assume that the angle θ is small
and that θ IS IN RADIANS AND NOT DEGREES, sin (40) With this approximation and some rearranging, we then have (41)
This equation is the angular equivalent of equation (24), the hallmark o...

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