The rotational period for a cylinder with a 10 mile diameter?

Question

In order to experience earth like gravity (one G) on its inter- surface, what would the rotational period of a cylinder with a diameter of 10 miles (17 kilometers) be? The cylinder is 25 miles long (40 kilometers) and is rotating in space outside the influence of other gravatational influences. The cylindar has a mass of 18,247,680,000 metric tons. Would you notice coriolis forces while walking in such a cylindar?

Answer 1

> Would you notice coriolis forces while walking in such a cylindar?
Yes
The force would be equal to the acceleration sideways of the velocity vector. 1 g is 32 ft/sec/sec. The surface velocity at 60 rpm (1 rps) is (radius*PI) 26400*3.14159 = 82937.976 ft/sec and the rate of change would be 2D per sec or 5280*2 or 10560 feet
32 ft/sec/sec results in a drop of 16 feet in the first second, so we have a right triangle (roughly) 16 feet on the far side and the actual distance per second on hypotenuse, so for 1 G we want 16/10560 rev. per second or 0.00151515152, or 0.0909090912 rpm or 10.999999964 - 11 minutes per revolution or 5.4546 rev./hour
I think. Probably more precise with arc radians, etc. and subtracting the gravity due to the mass of the cylinder and the difference in OD and ID for an assumed material density not given in the question.

Answer 2

Centripetal acceleration ac = R*om^2 (where om is angular velocity omega). If we want ac = g = 9.81 m/s^2 then:

om = 0.0339723070611759 rad/s

or the period of rotation:

P = 2*pi/om = 185 s = 3 min 5 s.

Coriolis acceleration ak = 2*om*v. Normally walking v = 5 km/h man will experience ak = 0.0944 m/s^2 which is about 100 times less than the gravity.