Conical Pendulum, Mass and Period

AI Thread Summary
The discussion centers on the relationship between mass and the period of a conical pendulum, with the understanding that mass does not affect the period. The centripetal force equation indicates that while a heavier mass results in a larger force, it also requires a proportionally larger force to achieve the same acceleration, leading to a cancellation of effects. This means that all swinging objects, regardless of mass, exhibit the same period. The confusion arises from the need to clarify why mass does not appear in the period equation. Ultimately, the conclusion is that the period remains constant for different masses in a conical pendulum.
deanchhsw
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Hello.

I am having trouble understand the varying mass's effect on the period of a conical pendulum. Well, I understand that there is no effect. However, I am having trouble verifying that in a centripetal force equation for circular motion. Most generally, conical pendulum's centripetal force is the sine component of its tension: therefore Fc = Sin theta * Mass * Tension. Therefore, in theory, the greater the mass, the greater Fc, thereby altering velocity and consequentially affecting the period.

However, the I found an explanation here in some web source:

We can state that a heavier object with the larger (heavy) mass will experience a larger driving force in the direction of the equilibrium position. However, this heavier object will have a larger inert mass and will need a larger force to experience the same acceleration. Because heavy mass and inert mass are strictly proportional to each other both effect cancel. All swinging objects, which differ only by their mass, show the same period.

I still have trouble understanding this, however,
can somebody explain this to me more clearly, so to speak?
Thank you very much
 
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To simplify the question,

"All swinging objects, which differ only by their mass, show the same period."

Why is this so?
 
Because the mass doesn't plays a role in the period equation, why don't do get an expression for \omega and remember the period T = \frac{2 \pi}{\omega}.
 
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