Conical Pendulum: Solving the Height Issue

AI Thread Summary
The discussion centers on the height issue in conical pendulum problems, specifically when the height exceeds the string length, which is deemed impossible. Participants suggest that if height (H) is greater than length (L), it should be treated as equal for practical purposes, indicating potential errors in the problem's specifications. The period of a conical pendulum is confirmed to be less than or equal to that of a planar pendulum, with the formula T=2π√(h/g) supporting this. Additionally, a critical condition for the conical pendulum's existence is established as Lω² ≥ g, preventing the radius from becoming zero or imaginary. The conversation emphasizes the relationship between circular motion and harmonic motions, reinforcing the constraints on angular frequency.
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Homework Statement
A 3m long string is attached to a 5kg mass and rotates at 5 rpm. Calculate the height of the pendulum.
Relevant Equations
H=g/w2 (Height = gravity divided by angular velocity squared)
= 9.81/ .27415 = 35.78m
So in this instance height is longer than the 3m string which is impossible. Do we just say that when ever H>L, H=L?
 
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I suspect that the given values for this problem are incorrectly specified for a real solution to exist. Consider that at 5 rpm the bob would have a period per revolution of 12 seconds. The same setup but swinging like a normal (planar) pendulum would have a period of
$$2 \pi \sqrt{\frac{3 m}{g}} \approx 3.5 \text{ seconds}$$
Pretty sure the period of the conical pendulum should be less than or equal that of the planar version (haven't proved this yet, so it's just intuition at the moment).
 
I know the formula is correct as is the math...just must be missing some sort of rule?
 

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gneill said:
I suspect that the given values for this problem are incorrectly specified for a real solution to exist. Consider that at 5 rpm the bob would have a period per revolution of 12 seconds. The same setup but swinging like a normal (planar) pendulum would have a period of
$$2 \pi \sqrt{\frac{3 m}{g}} \approx 3.5 \text{ seconds}$$
Pretty sure the period of the conical pendulum should be less than or equal that of the planar version (haven't proved this yet, so it's just intuition at the moment).
Yes, the period of the conical pendulum is ##T=2\pi \sqrt{\frac{h}{g}}## where ##h## is the "height". Since ##h < L##, it is less than that of the planar pendulum unless, of course, the bob isn't moving.
 
Makes sense, thank you so much!
 
I find that the criteria is

$$L \omega^2 \ge g$$
else r becomes either zero or imaginary.
 
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gneill said:
I find that the criteria is

$$L \omega^2 \ge g$$
else r becomes either zero or imaginary.
Yes. Circular motion can be decomposed as a superposition of linear harmonic motions at right angles with a 90-degree phase shift. In the limit of small angles for the conical pendulum, the harmonic motions are just pendulum swings of small amplitude which we know have angular frequency ##\sqrt{g/L}##. You cannot make the conical pendulum have a smaller angular frequency than this.
 
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That is great...thanks again.
 
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