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Homework 11 Solution

# If the pivot is a distance x from the rods center for

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Unformatted text preview: .wikipedia.org/wiki/Fluidized_bed_reactor.) 3. A pendulum is formed by pivoting a long thin rod of mass M and length L about a point on the rod. If the pivot is a distance x from the rod’s center, for what x is the period of the pendulum 1 minimum? The moment of inertia for a thin rod about its center of mass is I = 12 ML2 . Solution: In the end, we only have a physical pendulum, and we already know that the period is given by T = 2π I mgh where I is the moment of inertia of the rod (of mass m) about the pivot point, and h is the distance between the rod’s center of mass and the pivot point. Let the pivot be a distance x from the end of the rod, making it a distance l/2 − x from the center of mass. The moment of inertia is then I = Icom + m l −x 2 2 = 1 ml2 + m 12 l −x 2 2 The distance between the center of mass and the pivot is h = l/2 − x, so I= 1 ml2 + mh2 12 The period is thus Problems ions ured , reperlope ring. tion, With ured lues ding and each and eastalue e for alue 483 12 l + h2 l2 h T = 2π 12 = 2π + gh 70. Consider a damped o...
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