Design Space Station Homework: Rotation Speed & Gravitational Acceleration

AI Thread Summary
To design a space station that simulates gravity without artificial means, it must rotate at a speed calculated using the formula a = v^2/r, resulting in a required rotation speed of approximately 56.17 m/s and an angular velocity of 0.1745 rad/s. When an astronaut jogs at an additional speed of 5 m/s in the direction of rotation, their total speed becomes 61.17 m/s. This increased speed leads to a simulated gravitational acceleration of about 11.62 m/s². Rounding this value for precision, it may be presented as 11.6 m/s², depending on instructor preferences. The calculations demonstrate how rotation can effectively simulate gravitational forces in a space environment.
PrideofPhilly
Messages
37
Reaction score
0

Homework Statement



Part 1:

You want to design a large, permanent space station so that no artificial gravity is necessary. You decide to shape it like a large coffee can of radius 322 m and rotate it about its central axis. The acceleration of gravity is 9.8 m/s2. What rotation speed would be required to simulate gravity?

Part 2:

If an astronaut jogged in the direction of the rotation at 5 m/s, what is the simulated gravitational acceleration the astronaut would feel?

Homework Equations



a = v^2/r

ω = v/r

a = vω

The Attempt at a Solution



For Part 1:
9.8 = v^2/322 m
v^2 = 3155.6
v = 56.17472741 m/s

vω = v^2/r
ω = v/r
ω = (56.17472741 m/s)/(322 m)
ω = 0.174456 rad/s


For Part 2:

v = rω

I don't really know where to begin for this part.
 
Physics news on Phys.org
a = v2/r = ω2r = g

ω = (g/r)1/2

... which yields your answer.

In part 2 they want you to determine what happens when Kier Dullea (astronaut from 2001:Space Odyssey) is jogging, thereby increasing his speed (and the g force) by 5 m/s in the direction of rotation.
 
Oh...okay, so:

astronaut's speed: 56.17472741 m/s + 5 m/s

a = v^2/r
a = (61.17472741 m/s)^2/322 m = 11.62223624 m/s^2
 
PrideofPhilly said:
Oh...okay, so:

astronaut's speed: 56.17472741 m/s + 5 m/s

a = v^2/r
a = (61.17472741 m/s)^2/322 m = 11.62223624 m/s^2

Pretty much.

Since 9.8 is the given precision of gravity, maybe the correct form of your answer would be 11.6, depending on your instructor.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top