Circular Motion -- Swinging keys on a string in a vertical circle

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Homework Help Overview

The discussion revolves around a problem involving circular motion, specifically the dynamics of swinging keys attached to a string in a vertical circle. The original poster presents calculations related to the minimum speed required to maintain circular motion and the tension in the string at the bottom of the swing.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of uniform circular motion versus varying speeds as the keys move through different points in the vertical circle. There are questions about the assumptions made in the calculations, particularly regarding the use of velocity at the bottom of the swing.

Discussion Status

Some participants express confusion regarding the assumptions of uniform circular motion and the application of results from one part of the problem to another. There is acknowledgment of potential mistakes in the calculations, and the discussion is ongoing with participants seeking clarification on the dynamics involved.

Contextual Notes

Participants note that the problem may involve changing kinetic energy as the keys move through the vertical circle, which could affect the assumptions about uniform circular motion. There is also mention of the original poster's struggle with the problem, indicating a need for further exploration of the concepts involved.

dylanjames
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Homework Statement


Assuming the keys are moving in uniform circular motion![/B]
Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the vertical plane.

a)What is the slowest speed that the keys can swing and still maintain a circular path?

b)What is the tension in the string at the bottom of the circle?

Homework Equations


Given: V =10m/s, R=25m, Fc = mv^2/r, Ff = UFn[/B]
a) Centripetal Force is Fg, so Fg = MV^2/r
So, mg = mV^2/r
0.1(9.8) = (0.1)(V^2/0.25)
Mass cancels out..
9.8 = V^2/0.25
So, (0.25)(9.8) = V^2
Therefore, V^2 = 2.45, so V = 1.56m/s
The slowest speed the keys can travel and still maintain circular motion is 1.56m/s.

b) b) Fnet = Ft - Fg or Ft = Fnet + Fg (assuming this is uniform circular motion!)
Ft = mV^2/r + mg
Ft = [ (0.1)(1.56)^2 / 0.25 ] + (0.1)(9.8)
Ft = 0.97N
Therefore, assuming that the keys are moving in uniform circular motion, the Ft at the bottom of the circle is 0.97N.

Ok.. Really having a tough time with this one.
Been looking over it for too long and eyes are gone blurry.. any help is appreciated!
 
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dylanjames said:
Ft = [ (0.1)(1.56)^2 / 0.25 ] + (0.1)(9.8)
Ft = 0.97N
Some mistake there, surely.
 
At the top it has a certain velocity, but you need to think about what happens as it goes downward (following the circular path).
 
you can't assume uniform circular motion in this case . it gains kinetic energy at the bottom
 
throneoo said:
you can't assume uniform circular motion in this case . it gains kinetic energy at the bottom
Judging from the calculations, dylanjames was not really assuming uniform circular motion, merely moving in a circle.
 
haruspex said:
Judging from the calculations, dylanjames was not really assuming uniform circular motion, merely moving in a circle.
if he wasn't why would he use the result from a) in b) for the keys' velocity at the bottom? I'm a bit confused here .
 
throneoo said:
if he wasn't why would he use the result from a) in b) for the keys' velocity at the bottom? I'm a bit confused here .
Sorry, you're right.
 

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