A space station is
designed so that people can walk on the
inside of the
outer wall and feel their
normal weight. The station
is to be 750
m in diameter. What is the period of the station's
rotation?
9.8m/s^2=w^2 *r
but w=2PI/period so
w= sqrt(9.8/375)
period=2PI/sqrt( )=38.9 sec
check allthat.
To determine the period of rotation of the space station, we need to use the concept of centripetal acceleration.
The centripetal acceleration of an object moving in a circular path can be given by the formula:
ac = (v^2) / r
where ac is the centripetal acceleration, v is the linear velocity, and r is the radius of the circular path.
In this case, the linear velocity is the distance traveled per unit time. Since the space station completes one full rotation in a period of time, the linear velocity can be calculated as the circumference of the circular path divided by the period of rotation.
v = (2πr) / T
where π is a mathematical constant approximately equal to 3.14159.
Rearranging the formula, we get:
T = (2πr) / v
Substituting the given values:
r = 750 m (radius of the space station)
v = normal walking speed
We can assume a normal walking speed to be approximately 1.4 m/s.
T = (2π * 750 m) / 1.4 m/s
Simplifying this equation, we get:
T ≈ 1075.36 seconds
Hence, the period of rotation for the space station is approximately 1075.36 seconds.