Circular Motion and artificial gravity

[mkwok]

Homework Statement

A space station, in the form of a wheel 135 m in diameter, rotates to provide an "artificial gravity" of 3.80 m/s$$^{2}$$ for persons who walk around on the inner wall of the outer rim. Find the rate of rotation of the wheel (in revolutions per minute) that will produce this effect.
answer must be in rev/min

Homework Equations

a$$_{c}$$=$$\frac{v^{2}}{R}$$
VT=2piR

The Attempt at a Solution

3.8=$$\frac{v^{2}}{135/2}$$
v=$$\sqrt{3.8/67.5}$$=16.0156m/s

then since VT=2piR
T=2piR/v = 2pi(67.5)/16.0156 = 26.4813 rev/sec
converting that to rev/min is (26.4813 rev/sec)*(60sec/min) = 1588.88rev/min

however, I think I am doing something wrong, my online homework website tells me this is incorrect

 

[hage567]

T=2piR/v = 2pi(67.5)/16.0156 = 26.4813 rev/sec

This gives you seconds/revolution, not revolutions/second. Check your units. You're calculating the period here, not the frequency.

 

[ogb p]

.

I would like to provide a more detailed explanation of the solution to this problem.

First, let's start with the concept of circular motion and artificial gravity. In this scenario, the space station is rotating, creating a centrifugal force that acts as a form of artificial gravity. This is similar to the feeling of being pushed to the side when a car takes a sharp turn. The faster the rotation, the stronger the centrifugal force and the stronger the artificial gravity.

To calculate the rate of rotation of the wheel, we can use the equation for centripetal acceleration, a_c = v^2/R, where a_c is the centripetal acceleration, v is the tangential velocity, and R is the radius of the circle (in this case, half the diameter of the wheel). We are given that the artificial gravity is 3.80 m/s^2, so we can plug this into the equation and solve for v.

3.80 = v^2/(135/2)
v = √(3.80 * 135/2) = 14.39 m/s

Now, we can use the equation for tangential velocity, v = 2πR/T, where v is the tangential velocity, R is the radius of the circle, and T is the period (time for one full rotation). We know the tangential velocity and the radius, so we can solve for T.

14.39 = 2π(67.5)/T
T = 2π(67.5)/14.39 = 9.88 seconds

To convert this to revolutions per minute, we can use the fact that there are 60 seconds in a minute, so we can divide the period by 60.

9.88 seconds / 60 = 0.1653 minutes

Finally, we can calculate the rate of rotation in revolutions per minute by dividing 1 by the period in minutes.

1/0.1653 = 6.05 rev/min

Therefore, the rate of rotation of the wheel should be 6.05 rev/min to produce an artificial gravity of 3.80 m/s^2 on the outer rim. It is important to note that this is an ideal calculation and does not take into account any external factors or variations in the rotation.