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Jimmy25
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Homework Statement
A uniform disk of radius R = 1.40 m and a 6.0 kg mass has a small hole a distance d from the disk's center that serves as a pivot point.
What should be the distance d so that this physical pendulum will have the shortest possible period?
Homework Equations
T = 2π[tex]\sqrt{\frac{I}{MgL}}[/tex]
The Attempt at a Solution
Using the parallel axis theorem and moment of inertia of a disk I found the period as:
T = 2π[tex]\sqrt{\frac{\frac{1.4^{2}}{2}+d^{2}}{9.8d}}[/tex]
When I find the minimum of this function I get d = 0.990 which is the right answer.
What I don't understand is why the d isn't zero. Won't T approach zero as d approaches zero?
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