A spinning space station exhibits a change in moment of inertia.

Click For Summary
SUMMARY

The discussion revolves around calculating the new centripetal acceleration experienced by the remaining crew of a spinning space station after a portion of them moves towards the center. The space station has a radius of 95.0 m and an initial moment of inertia of 5.03 x 108 kg·m2. The initial total moment of inertia is calculated as 5.910 x 108 kg·m2 for 150 crew members, while the final moment of inertia for the remaining 50 crew members is 5.350 x 108 kg·m2. The calculated centripetal acceleration was found to be 1.859 m/s2, which was identified as incorrect due to a misapplication of the angular velocity formula.

PREREQUISITES
  • Understanding of angular momentum conservation principles.
  • Familiarity with moment of inertia calculations.
  • Knowledge of centripetal acceleration formulas.
  • Basic grasp of rotational dynamics.
NEXT STEPS
  • Review the derivation of angular velocity from centripetal acceleration.
  • Study the relationship between moment of inertia and angular velocity in rotating systems.
  • Explore the implications of mass distribution changes on rotational dynamics.
  • Practice solving problems involving conservation of angular momentum in rotating systems.
USEFUL FOR

Students studying physics, particularly those focusing on rotational dynamics and angular momentum, as well as educators seeking to clarify concepts related to centripetal acceleration in rotating systems.

MacRowan
Messages
1
Reaction score
0

Homework Statement



A space station shaped like a giant wheel has a radius 95.0 m and a moment of inertia of 5.03✕ 108 kg · m2. A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of 1g. When 100 people move to the center of the station for a union meeting, the angular speed changes. What acceleration is experienced by the managers remaining at the rim? Assume that the average mass of each inhabitant is 65.0 kg.

So,

r = 95.0

I = 5.03*108

and

m
ave = 65.0 kg

Δm = 150*m
ave - 100*mave, or mf - mi

Homework Equations



Li = Lf ---> (Iω)i = (Iω)f

ac = rω2, and therefor, w = √(acr)

I = mr2

The Attempt at a Solution



A reminder; I'm looking for the new centripetal acceleration felt by the remaining 50 New Earth colonizers.

I also want to clarify that I chose to determine the total initial moment of inertia by adding the I of the people to that of the provided I for the space station, because I reasoned I couldn't really find the difference without being provided the mass of the space-station, so I'm already unsure of my path here.

There is a change in moment of inertia of the system, so;

Li = Lf

Ii = (I150people) + (5.03*108) = (5.910*108

If = (I50people) + (5.03*108) = (5.350*108

but to find centripetal acceleration I need to relate ac and ω, and then put them into the conservation equation like so;

(5.910*10^8)*(√9.81*95.0) = (5.350*10^8)*(√ac*95.0)

When I isolated acceleration I got a value of 1.859, which my shitty electronic feedback java program says is more than 10% off. At least it's a smaller value, as it should be.

So, obviously I don't have the right answer, but can someone either point out my small error or point out how I have approached the problem entirely wrong?
 
Physics news on Phys.org
Welcome to PF!

MacRowan said:

The Attempt at a Solution



A reminder; I'm looking for the new centripetal acceleration felt by the remaining 50 New Earth colonizers.

I also want to clarify that I chose to determine the total initial moment of inertia by adding the I of the people to that of the provided I for the space station, because I reasoned I couldn't really find the difference without being provided the mass of the space-station, so I'm already unsure of my path here.

There is a change in moment of inertia of the system, so;

Li = Lf

Ii = (I150people) + (5.03*108) = (5.910*108

If = (I50people) + (5.03*108) = (5.350*108

but to find centripetal acceleration I need to relate ac and ω, and then put them into the conservation equation like so;

(5.910*10^8)*(√9.81*95.0) = (5.350*10^8)*(√ac*95.0)

First, ac=w2R, so your last equation is not correct. But you can simplify by R, so it does not really count.
I just wonder, how did you get 1.859 m/s2 for ac, or was it 1.859 g?. You miscalculated something.
 
Last edited:
You're on the right track but w= sqrt(acr) doesn't seem to be correct... which is affecting your last equation.
Your If seems a little off but should be fine for the purpose of this question.
I just got an answer using your values, only correcting the formula for w, and web assign accepted it as a correct answer.
 

Similar threads

Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K