A space station, in the form of a wheel 197 m in diameter, rotates to provide an “artificial gravity” of 2.5 m/s2 for persons who walk around on the inner wall of the outer rim. The rate of the wheel’s rotation in revolutions per minute that will produce this effect is
rev/min
My online course is telling me the answer is 1.52, but I can't get this answer no matter how I try.*
*actually its telling me v=1.52, but I believe this to still be rpm
Most recent work
a=v^2/r 2.5=v^2/197
v=Sqrt(2.5*197)
v=22.19234102
Circumfrence= 2*pi*r c=2*pi*197
c=1237.787506
c/v=55.77543641 seconds
s/(1 min=60 seconds)
55.77543641/60= 0.929590607 minuets
1/0.929590607= 1.075742367 rev/min
I don't see where my math has gone wrong.
Thank you
radius is half the diameter
To find the rate of the wheel's rotation in revolutions per minute (rpm) that will produce an artificial gravity of 2.5 m/s^2, you need to use the formula for centripetal acceleration:
a = v^2 / r
where a is the centripetal acceleration, v is the linear velocity, and r is the radius of the wheel.
In this case, you are given the desired centripetal acceleration as 2.5 m/s^2, and you are asked to find the rate of the wheel's rotation in rpm.
Let's go through the steps again:
First, rearrange the formula to solve for v:
v^2 = a * r
v = sqrt(a * r)
Substitute the given values into the equation:
v = sqrt(2.5 * 197) ≈ 22.192 m/s
Next, calculate the circumference of the wheel:
C = 2 * π * r
C = 2 * π * 197 ≈ 1237.787 m
Now, calculate the time it takes for one revolution:
t = C / v ≈ 1237.787 / 22.192 ≈ 55.776 s
Finally, convert the time to minutes:
1 minute = 60 seconds
t_min = t / 60 ≈ 55.776 / 60 ≈ 0.9296 min
Now, to find the rate of rotation in revolutions per minute:
rpm = 1 / t_min ≈ 1 / 0.9296 ≈ 1.0757 rev/min
Therefore, the correct answer is approximately 1.08 rev/min, not 1.52 rev/min as stated in your online course. It seems that your calculations are correct, and there might be an error in the course material.