A meter stick is pivoted at a point. Which results in the shortest period of oscillation?

  • Thread starter hidemi
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  • #1
hidemi
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Homework Statement:
A meter stick is pivoted at a point a distance a from its center and swings as a physical pendulum. Of the following values for a, which results in the shortest period of oscillation?

A. a = 0.1m
B. a = 0.2m
C. a = 0.3m
D. a = 0.4m
E. a = 0.5m

The answer is C.
Relevant Equations:
T = 2 π √(I/k)
-mg*sinθ*d =(1/12 * m^2* 1^2 + ma^2) θ"
(since θ is very very small, sin θ = θ)
θ" + [ga/ (1/12 + a^2)] θ =0

T (period) = 2π / ω

ω (0.1m) = 9.8 * 0.1 / (1/12+0.1^1) = 10.5
ω (0.2m) = 9.8 * 0.2 / (1/12+0.2^1) = 15.9
ω (0.3m) = 9.8 * 0.3 / (1/12+0.3^1) = 17.0
ω (0.4m) = 9.8 * 0.4 / (1/12+0.4^1) = 16.1
ω (0.5m) = 9.8 * 0.5 / (1/12+0.5^1) = 14.7

Therefore, C is the correct answer. However, is there a way to explain this phenomenon of why a=0.3m hits the minimum?
 

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  • #2
Orodruin
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You have two competing effects. Putting the pivot further out results in a greater restoring torque, but putting it further in reduces the moment of inertia.
 
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  • #3
hidemi
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You have two competing effects. Putting the pivot further out results in a greater restoring torque, but putting it further in reduces the moment of inertia.
Thank you so much.
 

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