A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?
A. (6/5)½
B. (3/2)½
C. 6/5
D. 3/2
E. 5/3
(GR9277 #61)
Solution:
Angular frequency for physical pendulum:
ω = (MgL/I)1/2
Pendulum I
M = 2m
I = mr2 + mr2 = 2mr2
L = r + r = 2r
ωI = (2mg2r/2mr2)1/2 = (2g/r)1/2
Pendulum I
M = 2m
I = mr2 + mr2 = 2mr2
L = r + r = 2r
ωI = (2mg2r/2mr2)1/2 = (2g/r)1/2
Pendulum II
M = 2m
I = mr2 + m(r/2)2 = (5/4)mr2
M = 2m
I = mr2 + m(r/2)2 = (5/4)mr2
L = ½ r + r = (3/2)r
ωII =[ 2mg (3/2)r / (5/4)mr2]1/2 = (12g/5r)1/2
The ratio:
ωII / ωI = (12g/5r)1/2 / (2g/r)1/2 = 6/5
Answer: A
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