What Factors Influence the Minimum Acceleration in Physics?

AI Thread Summary
The discussion focuses on the factors influencing minimum acceleration in physics, particularly in relation to pendulum motion. The equations presented show how angular frequency (ω) varies with different pivot distances, revealing a peak at 0.3m. It highlights the competing effects of restoring torque and moment of inertia when adjusting the pivot point. The conclusion identifies that the minimum acceleration occurs at this specific distance due to these dynamics. Understanding these interactions is crucial for analyzing pendulum behavior in physics.
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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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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Likes Lnewqban, hidemi and Delta2
Orodruin said:
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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