How does the angular velocity of a ferris wheel in radians/s compare to m/s?

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The angular velocity of a ferris wheel with a diameter of 18m is π/24 radians/s. To convert this to linear speed in m/s, the formula v = (2πr)/T is used, where r is the radius and T is the time for one revolution. The circumference is calculated as 2π(9m), resulting in a speed of approximately 0.157 m/s when considering a revolution time of 48 seconds. After confirming the calculations, the final speed is determined to be around 1.178 m/s. Understanding the relationship between angular and linear velocity is crucial for solving such problems.
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Homework Statement


The angular velocity of the ferris wheel with a diameter of 18m is pi/24 radians/s. How would that compare to m/s?

Homework Equations


v=(2pi(r))/360

The Attempt at a Solution


2pi(9)/360
=0.157m/s
 
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You know the circumference and you should know how find how long it will take to make one revolution.
Circumference (m)/ time for revolution (sec)= speed m/sec.
 
it takes 0.8mins to make one revolution
 
Right, so what do you get for speed in m/sec?
 
RUber said:
Right, so what do you get for speed in m/sec?
would that be 1.178m/s?

because 2pi(9m)/48s
 
Looks good.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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