Rotational Freq of Wheel for "Artificial Gravity" of 9.3 m/s^2

[rrodriguez119]

1. A space station in the form of a large wheel, 124 m in diamete, rotates to provide and "artificial gravity" of 9.3 m/s^2 for people located on the outer rim.

Find the rotational frequency of the wheel that will produce this effect.
(answer in units of rpm)

2. v^2=gr, w=v/r, w/2pi = frequency


3. To find velocity I used v^2=gr. I know that w=v/r so with my found velocity I divided it by the radius (62). to find the frequency=> w/2pi = frequency. I must be missing a step because the answer I got is wrong. Help would be fancied. thank you!

 

[PhanthomJay]

The number you calculated for frequency is probably expressed in revolutions (cycles) per second (I'm guessing, since you did not show how you arrived at whatever answer you got). The question asks for revolutions per minute (rpm). You must watch units!

 

[rrodriguez119]

The number you calculated for frequency is probably expressed in revolutions (cycles) per second (I'm guessing, since you did not show how you arrived at whatever answer you got). The question asks for revolutions per minute (rpm). You must watch units!

what would you do?

 

[rrodriguez119]

nvm, I understand what i have to do. thanks!

 

[rwisz]

I would like to commend your effort in trying to solve this problem. Your approach is correct, but there may be a small error in your calculation. Let me walk you through the steps again to help you find the correct answer.

First, we need to understand the concept of "artificial gravity" and how it is created in this scenario. The people on the outer rim of the wheel experience a centrifugal force that is equal to the force of gravity on Earth (9.8 m/s^2). This force is created due to the rotation of the wheel, and it is given by the equation F = mv^2/r, where m is the mass of the person, v is the linear velocity, and r is the radius of the wheel.

Now, we know that the desired "artificial gravity" is 9.3 m/s^2, so we can set up the equation as follows:

9.3 = v^2/r

Next, we need to find the linear velocity, v. This can be done by rearranging the equation as v = √(9.3r). Now, we can substitute this value into the equation w = v/r, where w is the angular velocity (in radians per second) and r is the radius of the wheel. The equation becomes w = √(9.3/r).

To convert this angular velocity into rotations per minute (rpm), we can use the formula w/2π = frequency. Substituting our value for w, we get:

√(9.3/r)/2π = frequency

Now, we can plug in the value for r, which is given as 62 m, to find the frequency in rpm:

√(9.3/62)/2π = frequency

Frequency = 0.094 rpm

Therefore, the rotational frequency of the wheel that will produce an "artificial gravity" of 9.3 m/s^2 is approximately 0.094 rpm.

I hope this explanation helps you understand the problem better and find the correct answer. Keep up the good work!