# Final Angular Momentum of a Space Station

• marjine
I suggest that you read post #5 by @haruspex and redo the calculation to see if you still get 3.99 rad/s.f

#### marjine

Homework Statement
A space station has the form of a hoop of
radius R = 15 m, with mass M = 1000 kg.
Initially its center of mass is not moving, but
it is spinning with angular speed ωi = 4 rad/s.
A small package of mass m = 19 kg is thrown
at high velocity by a spring-loaded gun at an
angle θ = 19 ◦
toward a nearby spacecraft
as shown. The package has a speed v =
310 m/s after launch. What is the space
station’s rotational speed ωf after the launch?
You may ignore the mass of the package in
calculating the moment of inertia of the space
station.
Relevant Equations
Conservation of angular momentum: Lf=Li
Rotational angular momentum: Iω
Translational angular momentum: mvrsinθ
Li = Lrf +Ltf
Iωo = Iωf + mvRsinθ
I = MR^2
(MR^2)ωo = (MR^2)ωf + mvRsinθ
ωf = (MR^2ωo -mvRsinθ)/MR^2 = 3.99

Is there a diagram that goes with this problem? What is the angle of the the package measured with respect to?

Is there a diagram that goes with this problem? What is the angle of the the package measured with respect to?

I can't figure out how to upload the diagram so I'll do my best to explain it: the hoop is drawn with a horizontal line through the center, theta is the angle of the velocity vector above that dotted line. Hope that helps, sorry.

#### Attachments

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I can't figure out how to upload the diagram so I'll do my best to explain it: the hoop is drawn with a horizontal line through the center, theta is the angle of the velocity vector above that dotted line. Hope that helps, sorry.
Click "Attach files", lower left at the same height as "Post reply" and follow instructions. It's easy and leaves nothing to our imagination.

ωf = (MR^2ωo -mvRsinθ)/MR^2 = 3.99

I think it's because OP entered "19" in the argument of a sine algorithm that expected radians. I got 3.94 rad/s under that assumed mistake.

To @marjine: Make sure your calculator is set to "degrees" and redo the calculation.

I think it's because OP entered "19" in the argument of a sine algorithm that expected radians. I got 3.94 rad/s under that assumed mistake.
Good thought, but the difference from initial angular velocity is 0.01rad/s according to @marjine and 0.06 according to that presumed error (which I confirm).

Good thought, but the difference from initial angular velocity is 0.01rad/s according to @marjine and 0.06 according to that presumed error (which I confirm).
It is interesting to note that if the (more correct) form for the initial angular momentum ##L_0= (M+m)R^2\omega_0## is considered despite the problem statement's suggestion not to, the final answer (with the correct sine) is 3.94 rad/s. The discrepancy introduced by ignoring the initial angular momentum of the package is significant in this case.

An additional assumption is that the speed of the ejected package is relative to the center of the hoop, not relative to the point of launch. The problem is unclear on that.

Finally, this problem crosses over to the twilight zone and enters the realm of incredulity. A solid iron hoop of radius 15 m and mass 1000 kg will have a radius of 2 cm. Some space station ##\dots##

An additional assumption is that the speed of the ejected package is relative to the center of the hoop, not relative to the point of launch.
I guess you mean a non-rotating frame at the hoop's centre.
Isn't that largely equivalent to the choice of including the package's mass in the initial angular momentum? If we include it, the thrower does not need to supply the tangential momentum, so taking the velocity as relative to the ejection point is equivalent to excluding it and taking the velocity in the inertial frame.
Then again, we could take either of the other two combinations.
A solid iron hoop of radius 15 m and mass 1000 kg will have a radius of 2 cm.
The space station was constructed by an advanced civilisation of ants.

I think it's because OP entered "19" in the argument of a sine algorithm that expected radians. I got 3.94 rad/s under that assumed mistake.

To @marjine: Make sure your calculator is set to "degrees" and redo the calculation.
The calculator mode is always the first thing I check when I get a physics problem wrong hahaha. 3.94 rad/s is not the correct answer, unfortunately.

The calculator mode is always the first thing I check when I get a physics problem wrong hahaha. 3.94 rad/s is not the correct answer, unfortunately.
I didn't say or imply that 3.94 rad/s is the correct answer. It is the answer one gets if one uses the wrong argument for the sine.

I suggest that you read post #5 by @haruspex and redo the calculation to see if you still get 3.99 rad/s.