# Uniform Circular Motion amusement park ride

1. May 7, 2014

### collegelife101

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!

2. May 7, 2014

Where is the diagram?

3. May 7, 2014

### collegelife101

The diagram is in my book so I can't add it.

4. May 7, 2014

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!

5. May 7, 2014

### collegelife101

I have attached a picture.

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6. May 7, 2014

### dauto

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.

7. May 7, 2014

### Andrew Mason

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

8. May 7, 2014