What is the frequency of oscillation for a human leg as a pivoting rod?

AI Thread Summary
The discussion focuses on calculating the frequency of oscillation for a human leg modeled as a uniform rod pivoting at one end. The moment of inertia for the leg is given as 1/3 m L², and the correct approach involves treating the leg as a physical pendulum rather than a simple pendulum. Initial calculations suggested a frequency of 0.55 Hz, but after clarification and using the appropriate formula, the final frequency determined is 0.675 Hz. The key takeaway is the importance of correctly identifying the pivot point and using the right equations for a physical pendulum. This emphasizes the need for precise definitions in physics problems.
goonking
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Homework Statement


Consider a human leg to be a rod of uniform density pivoting about one end. What will the frequency of oscillation be for a leg with a length of 0.82 meters?

Homework Equations



I believe we need the imagine the leg as a rod, so the moment of inertia would be = 1/3 m L2

The Attempt at a Solution


so is the the pivot point at the hips and the whole leg is moving?

or just the shin is moving and the knee is the pivot point? if this it the case, then do be divide .82m by 2?
 
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goonking said:
so is the the pivot point at the hips and the whole leg is moving?
Yes
or just the shin is moving and the knee is the pivot point? if this it the case, then do be divide .82m by 2?
No. It says 'pivoting about one end' so it's not pivoting about the knee.
 
goonking said:
so is the the pivot point at the hips and the whole leg is moving?
or just the shin is moving and the knee is the pivot point? if this it the case, then do be divide .82m by 2?
It doesn't matter whether it represents the shin or the whole leg, just do as it says: treat it as a rod of length .82m pivoted at one end.
 
haruspex said:
It doesn't matter whether it represents the shin or the whole leg, just do as it says: treat it as a rod of length .82m pivoted at one end.
so I would just use T = 2 pi sqrt(L/g)?

which comes out to be 1.81 seconds

frequency = 1/T = 1/1.81 = .55 Hz

is that correct?
 
goonking said:
so I would just use T = 2 pi sqrt(L/g)
No, I said a rod of that length, not a simple pendulum of that length. Use the moment of inertia you mentioned in the OP.
 
haruspex said:
No, I said a rod of that length, not a simple pendulum of that length. Use the moment of inertia you mentioned in the OP.
ok, I got the the new answer to be .675 Hz after using the physical pendulum formula . Thanks.
 
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