# Tubular space ship rotates and falls

1. May 25, 2015

### Karol

1. The problem statement, all variables and given/known data
A space ship has a shape of a hollow tube with outer radius R1, internal radius R2 and mass "M". it rotates at angular velocity ω0. it has 4 short, mass less legs.
In it there is a bar of radius "r" and mass "m". it can be moved instantly to any distance "a" from the center.
The ship starts to fall from height H=10R2 with the legs up and the inner bar at the center. the computer immediately moves the bar to a certain distance "a" upwards, so when the ship touches the ground the legs are at the bottom. it made half a rotation.
Find a formula for "a" so that the ship will land with legs down.

2. Relevant equations
Moment of inertia of a cylinder round the center: $I_c=\frac{1}{2}MR^2$
Steiner's theorem: $I_a=I_c+Ma^2$
Free fall: $h=\frac{1}{2}gt^2$
Conservation of angular momentum: $I_c\omega_0=I_1\omega_1$

3. The attempt at a solution
Moment of inertia with the bar at the center:
$$I=\frac{1}{2}\left(R_1^2-R_2^2\right)$$
Moment of inertia with the bar at distance a:
$$I=\frac{1}{2}\left(R_1^2-R_2^2\right)+\frac{1}{2}mr^2+m\left(a+\frac{r}{2}\right)^2$$
$$I=\frac{1}{2}\left[ M\left( R_1^2-R_2^2 \right)+m\left( a^2+ra+\frac{5}{4}r^2 \right)\right]$$
The center of mass when the bar is off the center is at distance C:
$$m\left(a+\frac{r}{2}\right)=C(M+m)\rightarrow C=\frac{\left(a+\frac{r}{2}\right)}{M+m}$$
Free fall between the centers of mass at the top and bottom positions:
$$h=\frac{1}{2}gt^2\rightarrow t^2=\frac{2}{g}\left[10R_2-R_1+\frac{\left(a+\frac{r}{2}\right)}{M+m}\right]$$
The ship makes half a turn, i.e. π[rad]. The new angular velocity:
$$\omega_1=\frac{\pi}{t}=\frac{\pi}{\sqrt{\frac{2}{g}\left[10R_2-R_1+\frac{\left(a+\frac{r}{2}\right)}{M+m}\right]}}$$
Conservation of angular momentum:
$$I_c\omega_0=I_1\omega_1\rightarrow \frac{1}{2}\left(R_1^2-R_2^2\right)\cdot \omega_0=\frac{\frac{1}{2}\left[M\left(R_1^2-R_2^2\right)+m\left(a^2+ra+\frac{5}{4}r^2\right)\right]\pi}{\sqrt{\frac{2}{g}\left[10R_2-R_1+\frac{\left(a+\frac{r}{2}\right)}{M+m}\right]}}$$
It's hard to isolate a, i think i am wrong

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2. May 25, 2015

### rude man

Since no one has responded in over 12 hrs. I have a comment: in order to rotate the rocket about an axis perpendicular to the spin axis, a torque has to be developed about that perpendicular axis, resulting in a precession motion. But I fail to see how moving the bar provides such a torque.

3. May 25, 2015

### Karol

The rocket needn't rotate about an axis perpendicular to the spin axis, it should rotate about an axis parallel to the spin axis, or even the spin axis itself.
I think that moving the bar changes the angular velocity and distance of falling, that's what i made in my calculation.

4. May 26, 2015

### rude man

That doesn't make sense. Rotating about the spin axis will not "right" the rocket.
How do you visualize the bar? Is it coaxial with the spin axis or at rt. angles to that axis?

5. May 26, 2015

### haruspex

There is a diagram at the end of the OP. The rod is parallel to the spin axis, both being horizontal. The idea is to control the rate of spin so that when the tube hits the ground it has done one half rotation. Control is exercised by moving the rod nearer to or further from the spin axis.

6. May 26, 2015

### haruspex

That's not the right formula for an annulus (even if you plug the mass in!). Two solid cylinders of the two radii would have different masses.
Also, the bar has a small MoI even when at the axis of rotation.
I think 'a' is supposed to be the distance the rod's centre is displaced.
Not sure you are expected to allow for such niceties. If so, you also need to consider that the moment of inertia you just computed is not the one about its mass centre.

7. May 26, 2015

### rude man

Yes, I should have looked at the diagram; didn't.

8. May 26, 2015

### Karol

Moment of inertia of an annular ring:
$$I=\frac{1}{2}M\left(R_1^2+R_2^2\right)$$
If i take 'a' to be the distance to the center of the rod:
$$I=\frac{1}{2}M\left(R_1^2+R_2^2\right)+\frac{1}{2}mr^2+ma^2$$
The center of mass when the bar is off the center is at distance C:
$$ma=C(M+m)\rightarrow C=\frac{m}{M+m}$$
Free fall between the centers of mass at the top and bottom positions:
$$h=\frac{1}{2}gt^2\rightarrow t^2=\frac{2}{g}\left(10R_2-R_1+\frac{a}{M+m}\right)$$
The ship makes half a turn, i.e. π[rad]. The new angular velocity:
$$\omega_1=\frac{\pi}{t}=\frac{\pi}{\sqrt{\frac{2}{g}\left(10R_2-R_1+\frac{a}{M+m}\right)}}$$
Moment of inertia round the new center of mass:
$$I_1=\frac{1}{2}\left[M\left(R_2^2+R_1^2\right)+mr^2\right]+M\left(\frac{a}{M+m}\right)^2+m\left(a-\frac{a}{M+m}\right)$$
Conservation of angular momentum:
$$I_c\omega_0=I_1\omega_1\rightarrow \frac{1}{2}\left[M\left(R_1^2+R_2^2\right)+mr^2\right]\cdot \omega_0=I_1\omega_1$$
And i have to isolate 'a' from here.

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9. May 27, 2015

### haruspex

Several places where you have $\frac m{M+m}$ or $\frac a{M+m}$ that should be $\frac {ma}{M+m}$. Check the dimensionalities throughout.
I don't know how you get those last two terms. They're both dimensionally wrong.
Remember, if you have a moment of inertia about a point other than the mass centre you need to apply the parallel axis theorem in reverse to get it about the mass centre. Anyway, I get:
$I_1=\frac{1}{2}\left[M\left(R_2^2+R_1^2\right)+mr^2\right]+\frac{Mma^2}{M+m}$

10. May 29, 2015

### Karol

$$I_1=\frac{1}{2}\left[M\left(R_2^2+R_1^2\right)+mr^2\right]+\frac{Mma^2}{M+m}$$
Is that all? i think i finished this problem, right?
Thank you Haruspex

11. May 29, 2015

### haruspex

Well, you still have the task of isolating a, no?

12. May 29, 2015

### Karol

Well i think i will give it up.

13. May 29, 2015

### rude man

Why? you've done the hard work. Now just invoke conservation of angular momentum, figure out what your ω has to be to right the ship in the time of fall to earth, subtract the initial ω0 and solve easily for a.
EDIT: sorry, ignore the bit about "..."subtract the initial ω0 ...".
Just go I(a) = I0ω01 after you determined ω1.
Solve for a. Use equation of your post 10 .

Last edited: May 30, 2015
14. May 30, 2015

### Karol

$$I_c\omega_0=I_1\omega_1\rightarrow \frac{1}{2}\left[M\left(R_1^2+R_2^2\right)+mr^2\right]\cdot \omega_0=I_1\omega_1$$
$$\frac{1}{2}\left[M\left(R_1^2+R_2^2\right)+mr^2\right]\cdot \omega_0=\left\{ \frac{1}{2}\left[M\left(R_2^2+R_1^2\right)+mr^2\right]+\frac{Mma^2}{M+m}\right\} \frac{\pi}{\sqrt{\frac{2}{g}\left(10R_2-R_1+\frac{ma}{M+m}\right)}}$$
$$\sqrt{\frac{2}{g}\left(10R_2-R_1+\frac{ma}{M+m}\right)}=\frac{\pi \left\{ \frac{1}{2}\left[M\left(R_2^2+R_1^2\right)+mr^2\right]+\frac{Mma^2}{M+m}\right\}}{\frac{1}{2}\left[M\left(R_1^2+R_2^2\right)+mr^2\right]\cdot \omega_0}$$
If i raise both sides to square it will be complicated

15. May 30, 2015

### haruspex

My suspicion is that you were expected to discard the occurrence of a on the left of that last equation on grounds of its being insignificant, but I don't see any basis for saying that it is. Indeed, it could be more significant than the a on the right.
Arguably "a formula for a" does not have to mean isolating it as "a = ....". The ship's computer should be capable of finding a from the equation you arrived at.

16. May 30, 2015

### Karol

Thanks

17. May 30, 2015

### rude man

OK this is my last butt-in; talking to myself I know. My apologies to the OP and haruspex. I rejoined only when the OP declared his or her intention to quit. Besides,there's not much else doing on PF right now ...

And so my summary:
$$I_1=\frac{1}{2}\left[M\left(R_2^2+R_1^2\right)+mr^2\right]+\frac{Mma^2}{M+m}$$
or I1 = I0 + I(a)
I1ω1 = I0ω0
(I0 + I(a))ω1 = I0ω0
π = ω1T, T = time of descent
10R2 = (1/2)gT2
Solve for I(a) and thus for a.

18. May 30, 2015

### haruspex

The snag is that, as Kaspis showed, the time of descent is also a function of a. This is why it appears on the left hand side of K's final equation. As I wrote, I was expecting to be able to argue that this could be ignored, but I see no basis for doing so. Indeed, it may be the dominant effect of a.
It would be eliminated if the question were to state that the rod returned to its original position at the last moment.

19. May 30, 2015

### rude man

I am pretty certain the intent, justifiable or not, was to ignore this complication.