In a popular amusement-park ride, a cylinder of radius 2.00 m is set in rota

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In a discussion about calculating the minimum coefficient of friction needed to prevent riders from slipping in a rotating amusement park ride, the initial calculation was incorrect due to improper parentheses in the formula. The correct formula is mu = g/(omega^2*R), leading to a revised answer of approximately 0.136. After some back-and-forth, the final value of 0.13611111 was confirmed to be correct for the problem. The conversation highlighted the importance of careful attention to mathematical notation and the possibility of errors in answer keys. Ultimately, the correct coefficient of friction was established, ensuring rider safety.
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[SOLVED] In a popular amusement-park ride, a cylinder of radius 2.00 m is set in rota

Homework Statement



In a popular amusement-park ride, a cylinder of radius 2.00 m is set in rotation at an angular speed of 6.00 rad/s, as shown in Figure 7-20 (http://www.webassign.net/sfhs99/7-20.gif). The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall of the cylinder is needed to keep the rider from slipping?

Homework Equations



mu = g/omega^2*R

The Attempt at a Solution



mu= 9.8/36*2 = .54

but it says I am incorrect..
 
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You have it parenthesized wrong. mu=g/(omega^2*R), mu=9.8/(36*2). That's not 0.54.
 
Dick said:
You have it parenthesized wrong. mu=g/(omega^2*R), mu=9.8/(36*2). That's not 0.54.

so it's .14

i got that answer before, but its still incorrect.
 
It looks pretty right to me. Answer keys are not always correct. But then I'm not always correct either. Did you try 0.13611111...
 
Dick said:
It looks pretty right to me. Answer keys are not always correct. But then I'm not always correct either. Did you try 0.13611111...

i typed in 0.13611111, and it worked! thank you!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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