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

AI Thread Summary
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!
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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