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

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In summary, the rotor ride at the amusement park presses you against the walls and 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?
  • #1
delta_mu
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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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  • #2
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:
  • #3
thanks a lot...it helped out a lot
 
  • #4
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 Centripital Force(Fc=mass times the velocity squared divided by the radius. Thanks again to whoever answers.
 
  • #5
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:
 

What is a "problem rotor ride" and why is it important to have help with it?

A "problem rotor ride" refers to an issue with the rotation of a mechanical rotor, which can occur in various scientific and engineering applications. It is important to have help with this problem because it can lead to malfunctioning equipment, inaccurate data, and potential safety hazards.

What are some common causes of a problem rotor ride?

There are several potential causes of a problem rotor ride, including misalignment of the rotor, worn out bearings or shafts, imbalance of the rotor, and issues with the motor or power supply.

How can I troubleshoot and fix a problem rotor ride?

The first step in troubleshooting a problem rotor ride is to identify the cause by inspecting the rotor and related components. This may involve checking for misalignment, wear and tear, or imbalance. Once the cause is identified, it can be addressed through methods such as realigning the rotor, replacing worn out parts, or balancing the rotor.

Can I prevent a problem rotor ride from occurring?

While some factors that contribute to a problem rotor ride may be unavoidable, there are certain measures that can be taken to prevent it. These include regular maintenance and inspection of the rotor and related components, balancing the rotor during installation, and ensuring proper alignment and lubrication.

How can I find professional help with a problem rotor ride?

If you are experiencing a problem rotor ride and are unable to troubleshoot or fix it on your own, it is best to seek professional help. This can be done by contacting a qualified engineer or technician who specializes in rotor systems and can provide expert assistance in diagnosing and resolving the issue.

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