A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity. M = 4.6 kg L = 1.5 m x = 0.24 m a. In terms of M, L, and x, what is the rod’s moment of inertia I about the pivot point. b. Calculate the rod’s period T in seconds for small oscillations about its pivot point. c. In terms of L, find an expression for the distance xm for which the period is a minimum.
A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity. M = 4.6 kg L = 1.5 m x = 0.24 m a. In terms of M, L, and x, what is the rod’s moment of inertia I about the pivot point. b. Calculate the rod’s period T in seconds for small oscillations about its pivot point. c. In terms of L, find an expression for the distance xm for which the period is a minimum.
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter13: Rotation Ii: A Conservation Approach
Section: Chapter Questions
Problem 15PQ: A uniform disk of mass M = 3.00 kg and radius r = 22.0 cm is mounted on a motor through its center....
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A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity.
M = 4.6 kg
L = 1.5 m
x = 0.24 m
a. In terms of M, L, and x, what is the rod’s moment of inertia I about the pivot point.
b. Calculate the rod’s period T in seconds for small oscillations about its pivot point.
c. In terms of L, find an expression for the distance xm for which the period is a minimum.
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