Physics Rolling Momentum?

Question

A space station shaped like a giant wheel has a radius 105 m and a moment of inertia of 5.05 x108 kgm2. A crew of 150 are living on the rim, and the station's rotation causes the crew to experience an acceleration of 1 g (Fig. P11.40). 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.

Comet Halley moves about the Sun in an elliptical orbit, with its cloest approach to the Sun being about 0.580 AU and its greatest distance from the sun being about 35.0 AU (1 AU = the average Earth-Sun distance). If the comet's speed at its closest approach is 54.0 km/s, what is its speed when it is farthest from the Sun?
The angular momentum of the comet about the Sun is conserved b/c no torque acts on the comet. The gravitational force exerted by the sun on the comet has a moment arm of ____.

Answer 1

Angular acceleration A has two components ac(normal) and at (tangential to the motion). During steady state at=0 and A=ac
ac=V^2/R

Torque is
T=IA where
I - total moment of inertia

When 100 people move to the center we have a reduction in moment of inertia and increase in A.

I= Iw+ nIp
Iw- I wheel
Ip- 150 point masses (equally distributed I hope)
n- number of people on the 'rim'.
Iw=4.96 E+8 kg•m2.
Ip= mR

Torque before
T1= (Iw + 150 mR) g

T2=T1= (Iw + 50 mR^2) ac2
V2^2/R=ac2= (Iw + 150 mR*2) V^2/[R(Iw + 50 mR^2)]
Let's simplify
K1=(Iw + 150 mR*2)
K2= (Iw + 50 mR^2)
we have
ac2= K1 g/(K2)
K1=5.86E+8 kg•m2
K2=5.26E+8 kg•m2
now
ac2= K1 g/(K2)
ac2=5.86E+8 x 9.81/ 5.26E+8=10.93 or

10.93/9.81=1.11g


One question at the time please

Answer 2

This is an interesting question, and one that has made me curious for quite a long time.