One problem for humans living in outer space is that they are apparently weightl

Question

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a cylindrical space station that spins about an axis through its center at a constant rate. (See the figure below. ) This spin creates "artificial gravity" at the outside rim of the station.

If the diameter of the space station is d= 680 m, how fast must the rim be moving in order for the "artificial gravity" acceleration to be "g" at the outer rim?


If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface. How fast must the rim move in this case?

Explanation / Answer

Artificial gravity would be caused by centripetal acceleration, which is: a(c) = v² / r We want the centripetal acceleration to equal g, so: g = v² / r Solved for v: v = vgr = v[(9.8m/s²)(340m) = 57.72m/s To convert to rad/s: ? = v / r = (57.72m/s) / 340m = 0.169rad/s For Mars, just replace g = 9.8m/s² with 3.7m/s². You should get 0.0962rad/s.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.