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posted by  NDFLBO on 11/7/2009 9:54:36 AM  |  status: Closed  |  Earned Karma: 50

Rod

Course Textbook Chapter Problem Needs by
Calculus Based Physics Principles of Physics (3rd) by Serway, Jewett N/A N/A 11/11/2009 at 8:00:00 PM
Question Details:
A long thin rod of mass m=100g and length L=1m is pivoted about one end and oscillates in a vertical plane. Find the period (in s) of small oscillations.

Hint:

For a physical pendulum w^2 = mgd/I, where I is the moment of inertia.  The moment of inertia of a thin rod of length L about one of its ends is I = (1/3) m L^2, so w^2 = 3gd/L^2. (See the table in the book in the rotation chapter). Here m = 0.1 kg and d = L/2 = 0.5m. (d is the distance from the pivot point to the center of mass.) The period is T = 1/f = 2*pi/w.
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Oracle
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posted by watt (MNK) on 11/7/2009 10:06:51 AM  |  status: Live
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Response Details:
mass m = 100 g = 0.1 kg
length L = 1m
Moment of inertia  I = ( 1/ 3) m L ^ 2
                               = ( 1/ 3) * 0.1 * 1 ^ 2
                               = 0.03333 kg m ^ 2
distance from the pivot point to the center of mass d = L / 2 = 0.5 m
period of time  T = 2π √[ L ^ 2 / 3 gd ]
                          = 1.638 s
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