Angular Velocity of the Hubble Space Telescope

Homework Statement

The Hubble Space Telescope is poered by two solar panels as shown. The body of the telescope has a mass of 11 Mg and a radii of gyration kx = 1.64m and ky = 3.85m, whereas the solar panels can be considered as thin plates, eac having a mass of 54 kg. due to an internal drive, the panels are given an angular vlocity of .6j rad/s measured relative to the telescope. Determine the angular velocity of the telescope due to the rotation of the panels. Prior to roatin the panels, the telescope was originally traveling at vg = -400i + 250j + 175k m/s

The Attempt at a Solution

G--> m = 11,000 kg, kx = 1.64m, ky = 3.85m, vinitail = -400i + 250j + 175k m/s

Panels--> 6m X 1.5m, m = 54 kg each, w = .6j rad/s

mtotal = 11,000 + 2(54) = 11108 kg

Ix = mkx2 = 11108(1.64)2 = 29876.0768

Iy = mky2 = 11108(3.85)2 = 163794.1248

Should i then use principle of impulse and momentum by doing for each axis.

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Doc Al
Mentor
What's conserved?

by that do u mean using V = Wy

do i use that for each access,

like for Vx = Wy = 11108(-400)

because it is moving -400i

Doc Al
Mentor
by that do u mean using V = Wy

do i use that for each access,
Not sure what you are talking about.

What quantity is conserved when the panels rotate?

I'm confused, what do u mean by conserved

w = .6j when the panels rotate, does this somehow apply, should i use w = v/r

Doc Al
Mentor
I'm confused, what do u mean by conserved
Conserved meaning "doesn't change", as in "conservation of energy". What's conserved in this case?

w = .6j when the panels rotate, does this somehow apply, should i use w = v/r
You will certainly need the angular velocity of the plates, which is given as 0.6 rad/s with respect to the telescope. Which way will the telescope start to rotate?

the weights will not change

also the velocity in the k direction will remain

the velociy in the j direction changes becuase of the w = .6j, but will it also change because of the radii of gyration

will the velocity in the i direction change also because of the radii of gyration

do u just add the .6j to the 250j to get velocity = 250.6 j m/s in terms of just j

Doc Al
Mentor
the weights will not change

also the velocity in the k direction will remain

the velociy in the j direction changes becuase of the w = .6j, but will it also change because of the radii of gyration

will the velocity in the i direction change also because of the radii of gyration
The linear velocity is not going to change.

What's conserved? Big hint: It has something to do with angular velocity.
do u just add the .6j to the 250j to get velocity = 250.6 j m/s in terms of just j
No. You can't add an angular velocity (in rad/s) to a linear velocity (in m/s).

Ok so the tele scope will travel continiously on the linear velocity vector

the panels hoever will be rotating @ .6 rad/s so that the y axis is the axis of revolution

"You will certainly need the angular velocity of the plates, which is given as 0.6 rad/s with respect to the telescope. Which way will the telescope start to rotate? "

why would this cause the telescope to rotate. isnt the rotation independant

Doc Al
Mentor
why would this cause the telescope to rotate. isnt the rotation independant
The problem asks "Determine the angular velocity of the telescope due to the rotation of the panels." The rotation of the panels is caused by the telescope exerting a torque on them--thus the panels exert a torque back on the telescope, making it rotate.

During all this, what's conserved?

Ah i think i gotcha I need to use Conservation of Angular Momentum equation
H1 = H2

rmv = rmv

my radius for the panel is 3m
the mass is 54 kg but theres two of them so i think i should us 108kg
v = .6 rad/s

radius of the telescope is unknown
mass is 11,000 kg
v = -400i + 250j +175k m/s

do i have to convert the velocities somehow

Doc Al
Mentor
Ah i think i gotcha I need to use Conservation of Angular Momentum equation
H1 = H2
Yes, you need to use conservation of angular momentum. (Usually an L is used to represent angular momentum.)

rmv = rmv
"rmv" is the angular momentum of a point mass, not what you need here. The form you need here is L = Iω. You're given the rotational inertia of the telescope; you'll need to figure out the rotational inertia of the panels.

ok i see now

rotation inertia is k right

L = Iw = .6I

I = mk2 = 54k2 should i double the 54 since there are two of them

L = .6(54k2 = 32.4k2

is this the right track

Doc Al
Mentor
ok i see now

rotation inertia is k right
k is the radius of gyration; I = mk2.

L = Iw = .6I
Careful: The rotational speed of the panels is not 0.6--that's its speed with respect to the telescope.

Hint: Call the unknown rotational speed of the telescope ω. What would be the rotational speed of the panels in terms of ω? Use that to write the conservation of angular momentum.

You aren't given the k for the panels, but you know their mass and dimensions so you can figure out their rotational inertia.

I = mr2
I = 108(32) = 972 I double the weight since there are two panels

k=$$\sqrt{I/m}$$ = $$\sqrt{972/108}$$ = 3

u used equation I = mr2 & k = $$\sqrt{I/m}$$, but doens that mean that r = k, so im assuming this is wrong

Doc Al
Mentor
I = mr2
I = 108(32) = 972 I double the weight since there are two panels
The panels are rectangular plates. What formula would you use to calculate their rotational inertia? (Hint: Compare to a rod.)

oooooooo

I = 1/12ML2 = 1/12(108)(6)2 = 324

I double the mass since theres two plates

therefore k for the plates is

k = $$\sqrt{324/108}$$ = 1.732

so this is in the y directino so do i add this to the k for the telescope in the y direction

Doc Al
Mentor
oooooooo

I = 1/12ML2 = 1/12(108)(6)2 = 324

I double the mass since theres two plates
Good.

therefore k for the plates is

k = $$\sqrt{324/108}$$ = 1.732
The only purpose for needing k would be to find I. But you just found I!

so this is in the y directino so do i add this to the k for the telescope in the y direction
No.

This problem is the rotational analog to a problem like this: Say two kids were on ice. The boy shoves the girl so that the girl moves at speed X with respect to him. How fast is the boy moving after the shove?

This is a similar problem but with rotation. The telescope turns the panels. But angular momentum of the system (telescope + panels) cannot change. It started at zero and will remain at zero. If the panels move clockwise, the telescope has to move counterclockwise.

Ok im confused as to what to do now,

do i plug it into the conservation of angular momentum

Doc Al
Mentor
Here's how conservation of angular momentum would apply to this problem:
Itelescopeωt,i + Ipanelsωp,i = Itelescopeωt,f + Ipanelsωp,f

LTi = Iw = 0
I = mk2 = 11000(3.85)2 i use ky because w for the panels is in terms of j aka y
w = 0

LPi = Iw = 324(.6) = 194.4

LTf = 163047.5w this w is the solution to the problem right?

LPf = 324w should w still equal .6

Doc Al
Mentor
LTi = Iw = 0
I = mk2 = 11000(3.85)2 i use ky because w for the panels is in terms of j aka y
w = 0
Good.

LPi = Iw = 324(.6) = 194.4
No. Initially, nothing is rotating. Then the panels and the telescope end up rotating.

LTf = 163047.5w this w is the solution to the problem right?
Right.

LPf = 324w should w still equal .6
This is a bit tricky. They didn't say that the panels rotate at 0.6 with respect to an inertial frame (imagine someone floating in space right next to the telescope); instead they said that the panels rotate at 0.6 with respect to the telescope. So if we call the rotational speed of the telescope ω (from that inertial frame), what would we call the rotational speed of the panels?

would it be...... wP = wT + .6

Doc Al
Mentor
would it be...... wP = wT + .6
Yes! You're almost done.

0 = 324(w+.6) + 163047.5w = 163371.5w = -194

w = -.001874776