What Is the Minimum Frequency for a Mass in Vertical Circular Motion?

AI Thread Summary
To find the minimum frequency for a mass in vertical circular motion, the centripetal force must equal the gravitational force at the top of the circle when tension is zero. The equation F = 4(pi^2)(r)(f^2)(m) relates centripetal force to frequency, mass, and radius. At the minimum frequency, the weight of the mass provides the necessary centripetal force, meaning the tension in the rope is zero. Understanding the forces acting on the mass at the top of the circle is crucial for solving the problem. This approach will help determine the minimum frequency required for the mass to maintain circular motion.
ninetyfour
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Homework Statement



A 0.2kg mass attached to a rope 1.6m long is being spun in a vertical plane. Find the minimum frequency.

Homework Equations



F = 4(pi^2)(r)(f^2)(m)
Ft min = Fc - mg

- minimum frequency occurs at minimum tension


The Attempt at a Solution



I had this question on a test and spend like 45 minutes on it. I have no idea what to do, and I am curious as to what I should have done. I tried inserting numbers and re arranging formulas and substituting here and there and nothing seemed to work.

HELP? D:
 
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ninetyfour said:


F = 4(pi^2)(r)(f^2)(m)
Ft min = Fc - mg

- minimum frequency occurs at minimum tension



What kind of force is F in your first equation?
How much is the minimum tension?

ehild
 
F = 4(pi^2)(r)(f^2)(m)

That F is centripital force. It can also be written as:

F = m(v^2) / r
 
Thinking physically, if the frequency is not high enough what happens? The mass will not make it to the top of the circle, right? It will execute free-fall motion (i.e., a parabola) until the string catches it again. So the key is to consider the top of the circle.

What are the forces acting? Weight and tension. These combine together to provide the centripetal force maintaining the circular motion. The smallest the tension can be is zero. So we want to find the frequency in which the weight alone is able to provide the sufficient centripetal force.

I think you can probably take it from here...
 
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