A space station, in the form of a wheel 110 m in diameter, rotates to provide an "artificial gravity" of 3.00 m/s2 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.

Thank you

3 = v^2/r

3 * 55 = v^2

v = 12.8 m/s

Time for revolution = 2 pi r/v
= 2 pi * 55/12.8 = 26.9 seconds = .448 minutes

1/.448 = 2.23 revs/minute

Well, the first rule of artificial gravity is: what goes around, comes around! So let's get spinning!

To find the rate of rotation, we need to know the circumference of the wheel. The formula for circumference is C = π * d, where d is the diameter. So, C = π * 110 m.

Now, we also know that the centripetal acceleration (which is what creates the artificial gravity) is 3.00 m/s^2. The formula for centripetal acceleration is a = (v^2) / r, where v is the velocity and r is the radius.

Since the velocity is related to the rotational speed, we can say that v = (2π * r) / T, where T is the period (the time it takes for one revolution). Rearranging this equation, we get T = (2π * r) / v.

Now, if we combine these equations, we can find the period in terms of the circumference: T = (2π * r) / ((2π * r) / T). Canceling out the 2π * r, we get T^2 = T. Solving for T, we find T = 1.

So, the period is 1 minute. This means that the wheel completes one revolution every minute. In other words, the rate of rotation is 1 revolution per minute.

So there you have it! If you want an artificial gravity of 3.00 m/s^2, just spin the wheel at a rate of 1 revolution per minute. And remember, it's all about finding the right balance in space!

To find the rate of rotation of the wheel in revolutions per minute, we can use the formula for centripetal acceleration:

a = rω^2

where a is the acceleration, r is the radius, and ω is the angular velocity.

Given that the artificial gravity, or acceleration, is 3.00 m/s^2 and the radius is half the diameter, which is 110 m / 2 = 55 m, we can rearrange the formula to solve for ω:

ω = √(a / r)

Substituting the values, we have:

ω = √(3.00 m/s^2 / 55 m)

ω ≈ 0.18928 rad/s

To convert this to revolutions per minute, we need to convert the angular velocity to revolutions per second and then multiply by 60 to get revolutions per minute.

To convert from radians per second to revolutions per second, we divide by 2π:

ω_rev = ω / (2π)

ω_rev ≈ 0.18928 rad/s / 2π

ω_rev ≈ 0.030075 rev/s

Finally, to convert from revolutions per second to revolutions per minute, we multiply by 60:

ω_rev_min = ω_rev * 60

ω_rev_min ≈ 0.030075 rev/s * 60

ω_rev_min ≈ 1.8045 rev/min

Therefore, the rate of rotation of the wheel that will produce an artificial gravity of 3.00 m/s^2 is approximately 1.8045 revolutions per minute.

To find the rate of rotation of the wheel (in revolutions per minute) that will produce the desired artificial gravity, you can use the formula for centripetal acceleration:

a = ω²r

Where:
- a is the centripetal acceleration (which equals the artificial gravity, 3.00 m/s^2 in this case)
- ω is the angular velocity (rate of rotation in radians per second)
- r is the radius of the wheel (half the diameter, which is 110/2 = 55 m)

First, convert the given artificial gravity from m/s^2 to m/s^2:

3.00 m/s^2 = 3.00 m/s^2

Next, rearrange the formula to solve for angular velocity (ω):

ω² = a / r

ω² = (3.00 m/s^2) / (55 m)

ω² = 0.0545 s⁻² (rounded to four decimal places)

Now, take the square root of both sides to solve for angular velocity:

ω = √(0.0545 s⁻²)

ω = 0.2339 s⁻¹ (rounded to four decimal places)

To convert the angular velocity from seconds to minutes, multiply by 60 seconds:

ω = 0.2339 s⁻¹ * 60 s/min

ω ≈ 14.03 min⁻¹ (rounded to two decimal places)

Therefore, the rate of rotation of the wheel required to produce the desired artificial gravity is approximately 14.03 revolutions per minute.