Uniform Circular Motion amusement park ride

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Homework Help Overview

The discussion revolves around a problem related to uniform circular motion, specifically involving an amusement park ride with a carriage attached to a mechanical arm. The question focuses on calculating the normal force experienced by the carriage at the top of its swing when it is in motion at a specified speed.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the setup of the problem, questioning the inclusion of tension in the mechanical arm as a contributing force to the centripetal force. There is also a suggestion to clarify the relationship between the normal force and the mass of the occupants in the carriage.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved. Some guidance has been offered regarding the focus on the normal force in relation to the mass of the carriage's occupants, but no consensus has been reached on the role of tension in the mechanical arm.

Contextual Notes

Participants note the absence of a diagram, which is considered crucial for understanding the problem setup. There are also references to formatting equations correctly, indicating a potential barrier to clear communication of the mathematical aspects involved.

collegelife101
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Hello everyone,

I'm really stuck on this question. The diagram shows an amusement park ride which contains a carriage attached to a mechanical arm. This arm spins around full circle. The carriage has a mass of 300 kg and a maximum occupancy of 300 kg.

The question asks: With its carriage full, the second ride goes through the top of its swing. What is the value of the normal force if speed is 20 m/s?

The solution sets it up as Fc= Fg + Fn= mv(squared)/r. I understand the equation, but why is the tension in the mechanical arm not included? I would think that it would also be a force contributing to the Fc.

Thanks!
 
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Where is the diagram?
 
The diagram is in my book so I can't add it.
 
collegelife101 said:
The diagram is in my book so I can't add it.

Can you draw it using paint and upload it? Or you can just take a photo of it.
We can't answer questions without seeing it. Ya know!
 
I have attached a picture.
 

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collegelife101 said:
Hello everyone,

I'm really stuck on this question. The diagram shows an amusement park ride which contains a carriage attached to a mechanical arm. This arm spins around full circle. The carriage has a mass of 300 kg and a maximum occupancy of 300 kg.

The question asks: With its carriage full, the second ride goes through the top of its swing. What is the value of the normal force if speed is 20 m/s?

The solution sets it up as Fc= Fg + Fn= mv(squared)/r. I understand the equation, but why is the tension in the mechanical arm not included? I would think that it would also be a force contributing to the Fc.

Thanks!
Isn't Fn the tension? A diagram would help.

Next time try mv2/r instead of mv(squared)/r. It's really not that hard. just press on advanced and than on the superscript button.
 
collegelife101 said:
Hello everyone,

I'm really stuck on this question. The diagram shows an amusement park ride which contains a carriage attached to a mechanical arm. This arm spins around full circle. The carriage has a mass of 300 kg and a maximum occupancy of 300 kg.

The question asks: With its carriage full, the second ride goes through the top of its swing. What is the value of the normal force if speed is 20 m/s?

The solution sets it up as Fc= Fg + Fn= mv(squared)/r. I understand the equation, but why is the tension in the mechanical arm not included? I would think that it would also be a force contributing to the Fc.

Thanks!
Welcome to PF collegelife101!

The reason the tension in the mechanical arm is not included is that the question is asking for the force between the carriage and the people. That force depends only on the mass of the contents of the carriage (i.e the mass of the people in it) and its speed. The tension in the arm provides the that normal force, but the mass of the people will determine the magnitude of that normal force (for a given rotational speed).

AM
 
dauto said:
Next time try mv2/r instead of mv(squared)/r. It's really not that hard. just press on advanced and than on the superscript button.
Or learn Latex. It's million times better than BBcodes(Sup ,sub scripts etc...)
 

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