Finding the Radius of a Spinning Rotor: Can You Help Me Solve This Problem?

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Homework Help Overview

The problem involves a scenario related to Uniform Circular Motion and the Law of Universal Gravitation, specifically focusing on calculating the radius of a spinning rotor ride based on the forces acting on riders during the ride.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between frictional force and gravitational force, questioning how to determine the normal force and radial acceleration. There is an exploration of the equations related to centripetal force and friction.

Discussion Status

Some participants have provided equations and reasoning related to the problem, while others express uncertainty about the equations being used. There is acknowledgment of different approaches to finding the radius, but no consensus has been reached on the correct method or final value.

Contextual Notes

Participants note the specific parameters given in the problem, such as the coefficient of static friction and the number of rotations over a set time, while also indicating a lack of clarity on certain equations and their applications.

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Help with problem..."rotor ride"

OK I was assigned this homework sheet on Uniform Circular Motion and Law of Universal Gravitation...and there is only one problem I can not get. If you could, please help me. Thanks in advance.

Here it is:
The "Rotor" ride is the one which presses you against the walls of the spinning rotor as the floor drops away. The coefficient of static friction between the wall and the riders is 0.0580. If the floor dropped away for 2.00 minutes, and during this time the rotor has spun 100 times at a constant rate, what must be the radius of the rotor?

I received .006 meters for the radius
 
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You know that the friction needs to cancel the gravity force so the person doesn't fall:
[tex]f = mg = N\mu[/tex]
But what is N? You can find it by finding the radial acceleration of the device:
[tex]\Sigma F_r = ma_r = m\omega ^2r[/tex]
But what is [tex]\omega[/tex]? You know that the device rotates 100 times in 120 seconds, which makes for a frequency of 0.833Hz. And you know that:
[tex]\omega = 2\pi f[/tex]
And therefore:
[tex]\Sigma F_r = ma_r = 4\pi ^2f^2mr[/tex]
This is equal to the normal force, so:
[tex]mg = N\mu = 4\pi ^2f^2mr\mu[/tex]
And finally:
[tex]r = \frac{g}{4\pi ^2f^2\mu}[/tex]
I get a radius of about 6.17 meters.
 
Last edited:
thanks a lot...it helped out a lot
 
Wait...I am so sorry, I realized that I have no idea as to what equations you are using. The equations I was given are the Force of Friction(Ff=mu times the normal force) and the centripetal Force(Fc=mass times the velocity squared divided by the radius. Thanks again to whoever answers.
 
Fc=mass times the velocity squared divided by the radius is:
[tex]F_c = ma_r = m\frac{v^2}{r}[/tex]
But since [tex]v = \omega r[/tex] this becomes:
[tex]F_c = ma_r = m\omega ^2r[/tex]
Which is what I used. :smile:
 

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