Variation of gravity in a Rotating SpaceStation

[Algren]

This is very simple question, and i just need a 2nd opinion.

We have a Space Station (preferably a torus) with angular velocity ω and radius r. We have a car inside which OPPOSES the angular velocity and moves with the speed ωr . So, will the gravity felt in this car be Zero? Or will it be something else?

If we replace the car with Usain Bolt, and ask him to run at the speed of ωr , opposing Angular velocity, then, what will be the change in this case?

 

[A.T.]

We have a Space Station (preferably a torus) with angular velocity ω and radius r. We have a car inside which OPPOSES the angular velocity and moves with the speed ωr . So, will the gravity felt in this car be Zero?

Yes.

If we replace the car with Usain Bolt, and ask him to run at the speed of ωr , opposing Angular velocity, then, what will be the change in this case?

Nothing. Both will have problems propelling themselves at such a speed, because they will loose traction at the outer wall.

 

[tiny-tim]

in case you're wondering …

in the rotating frame of the spaceship, there is still a centrifugal force on the car of mΩ²r outward, but there's also the velocity-dependent Coriolis force of 2mΩ²r inward, net force of mΩ²r inward, supplying the centripetal acceleration which explains why the car appears to be rotating with speed Ωr even though there's no reaction force from the "floor" of the spaceship!

 

[truesearch]

Assume that the station is rotating to re-create gravitational effects on Earth.
ie the rotation is such that a 1kg mass placed on bathroom scales on the 'floor' of the station registers 10N
In the context of this post can anyone explain what the reading on bathroom scales would be when
a) a 1kg mass is placed on bathroom scales placed in the car when it is 'staionary'
b) a 1kg mass is placed on bathroom scales placed in the car when it is being driven around the space station...what happens to the reading... greater? smaller, ... the same?

 

[A.T.]

Assume that the station is rotating to re-create gravitational effects on Earth.
ie the rotation is such that a 1kg mass placed on bathroom scales on the 'floor' of the station registers 10N
In the context of this post can anyone explain what the reading on bathroom scales would be when
a) a 1kg mass is placed on bathroom scales placed in the car when it is 'staionary'

That is answered in the question, isn't it? 10N

b) a 1kg mass is placed on bathroom scales placed in the car when it is being driven around the space station...what happens to the reading... greater? smaller, ... the same?

Depends in which direction and how fast it drives, the centrifugal force will be modified by the Coriolis force.

 

[truesearch]

Absolutely correct for the first part.
so... suppose the car is traveling at the same speed as the ring when it is simulating gravity... what will the bathroom scales read?
(in each direction)

 

[A.T.]

so... suppose the car is traveling at the same speed as the ring when it is simulating gravity... what will the bathroom scales read?
(in each direction)

0N for opposite direction and 20N for outrunning the rotation

 

[truesearch]

if you are standing in such a space station how can you tell which way it is rotating?

 

[A.T.]

if you are standing in such a space station how can you tell which way it is rotating?

You just proposed the experiment that tells you the direction. If you mean just by standing still, you can't. But if you can move your arms, the Coriolis force will tell you how it rotates.

 

[truesearch]

So if I am standing 'still' in this space station I cannot tell which way it is rotating unless I use the coriolis force on my arms ?? How will the coriolis force reveal the direction of rotation.?

 

[A.T.]

How will the coriolis force reveal the direction of rotation.?

Just like in your scales example. You can do a similar test by swinging your arms and see which way they appear heavier. Maybe while holding a heavy object in your hand.

 

[truesearch]

That is not an acceptable answer ! wave your arms around... come on
HOW can you determine, 'in a physics way', which way the station is rotating?

 

[Whovian]

That is not an acceptable answer ! wave your arms around... come on
HOW can you determine, 'in a physics way', which way the station is rotating?

That is 'in a physics way.' If you mean in a controlled way, put to accelerometers (to measure gravity) into two cars, have them move in opposite directions, see which one measured the least acceleration.

 

[truesearch]

'Accelerometers to measure gravity'... that is what I mean by bathroom scales !
How would they tell you the direction of rotation??...if they are on a moving car?

 

[A.T.]

'Accelerometers to measure gravity'... that is what I mean by bathroom scales !
How would they tell you the direction of rotation??...if they are on a moving car?

See post #7

 

[truesearch]

I could not fully understand post 3, does your post agree with post 3?
I have to explain these things to 16 and 17 year old students.
Any help in this are would be valuable.

 

[A.T.]

I could not fully understand post 3, does your post agree with post 3?

Yes.

 

[truesearch]

If you hold a ball in your hand in this space station and let it go what will happen?

 

[A.T.]

If you hold a ball in your hand in this space station and let it go what will happen?

It will accelerate according to the sum of centrifugal and Coriolis forces.

 

[sophiecentaur]

0N for opposite direction and 20N for outrunning the rotation

I'm not sure of the precise scenario that your answer comes from but, if the car is doing the same 'speed' over the surface as the station peripheral speed then its resultant peripheral speed will be twice that of the station (?). The acceleration is given by v²/r so you would expect to measure 40N, not 20N.

 

[K^2]

I could not fully understand post 3, does your post agree with post 3?
I have to explain these things to 16 and 17 year old students.

In order to explain something, you need to understand it yourself at least one level higher, so that's a serious problem right there.

So first, let's try to explain it to you on a level above of what you'll need to explain to students.

Suppose, I have an object in an inertial frame of reference. For inertial frame of reference, we have Newton's Second Law.

$$F = ma$$

Now, acceleration is the second derivative on position. Furthermore, it is a vector. For a rotation problem 2D is sufficient. So let me rewrite it.

$$F_x = m \frac{d^2x}{dt^2}$$
$$F_y = m \frac{d^2y}{dt^2}$$

In a rotating reference frame, it's much easier to use polar coordinates.

$$x = r cos(\theta)$$
$$y = r sin(\theta)$$

Lets rewrite the equations for forces again.

$$F_x = m \frac{d^2}{dt^2}\left(r cos(\theta)\right)$$
$$F_y = m \frac{d^2}{dt^2}\left(r sin(\theta)\right)$$

Taking derivative once.

$$F_x = m \frac{d}{dt}\left(\frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt}\right)$$
$$F_y = m \frac{d}{dt}\left(\frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt}\right)$$

Taking derivative twice.

$$F_x = m \left(\frac{d^2r}{dt^2} cos(\theta) - 2 \frac{dr}{dt} sin(\theta)\frac{d\theta}{dt} - r cos(\theta) \left(\frac{d\theta}{dt}\right)^2 - r sin(\theta)\frac{d^2\theta}{dt^2}\right)$$
$$F_y = m \left(\frac{d^2r}{dt^2} sin(\theta) + 2 \frac{dr}{dt} cos(\theta)\frac{d\theta}{dt} - r sin(\theta) \left(\frac{d\theta}{dt}\right)^2 + r cos(\theta)\frac{d^2\theta}{dt^2}\right)$$

Suppose, now, that you are standing on the station rotating around r=0 with angular velocity ω. Suppose that you measure all angles and x/y relative to yourself. I'll call these θ', x', and y' coordinates. Clearly θ=θ' + ωt. (To within a choice of origin for θ) The equations for x' and y' are same polar equations as before.

$$x' = r' cos(\theta ')$$
$$y' = r' sin(\theta ')$$

The r'=r, of course, since we chose the coordinate system with the same center, and the center of rotation does not move.

Let us substitute θ=θ' + ωt into force equations.

$$F_x = m \left(\frac{d^2r}{dt^2} cos(\theta' + \omega t) - 2 \frac{dr}{dt} sin(\theta' + \omega t)\frac{d\theta}{dt} - r cos(\theta' + \omega t) \left(\frac{d\theta}{dt}\right)^2 - r sin(\theta' + \omega t)\frac{d^2\theta}{dt^2}\right)$$
$$F_y = m \left(\frac{d^2r}{dt^2} sin(\theta' + \omega t) + 2 \frac{dr}{dt} cos(\theta' + \omega t)\frac{d\theta}{dt} - r sin(\theta' + \omega t) \left(\frac{d\theta}{dt}\right)^2 + r cos(\theta' + \omega t)\frac{d^2\theta}{dt^2}\right)$$

Of course, we can also expand dθ/dt the same way.

$$F_x = m \left(\frac{d^2r}{dt^2} cos(\theta' + \omega t) - 2 \frac{dr}{dt} sin(\theta' + \omega t)\left(\frac{d\theta'}{dt}+\omega\right) - r cos(\theta' + \omega t) \left(\frac{d\theta'}{dt} + \omega\right)^2 - r sin(\theta' + \omega t) \frac{d^2\theta'}{dt^2}\right)$$
$$F_y = m \left(\frac{d^2r}{dt^2} sin(\theta' + \omega t) + 2 \frac{dr}{dt} cos(\theta' + \omega t)\left(\frac{d\theta'}{dt}+\omega\right) - r sin(\theta' + \omega t) \left(\frac{d\theta'}{dt} + \omega\right)^2 + r cos(\theta' + \omega t) \frac{d^2\theta'}{dt^2}\right)$$

Now, I'm going to simplify a bit, using the fact that x/r = cos(θ).

$$F_x = m \left(\frac{d^2r}{dt^2}\frac{x}{r} - 2 \frac{dr}{dt}\frac{y}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{x}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 - r^2 \frac{y}{r} \frac{d^2\theta'}{dt^2}\right)$$
$$F_y = m \left(\frac{d^2r}{dt^2}\frac{y}{r} + 2 \frac{dr}{dt}\frac{x}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{y}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 + r^2 \frac{x}{r} \frac{d^2\theta'}{dt^2}\right)$$

Since x/r and y/r simply give us directions, we go over completely to primed coordinates.

$$F_x' = m \left(\frac{d^2r}{dt^2}\frac{x'}{r} - 2 \frac{dr}{dt}\frac{y'}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{x'}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 - r^2 \frac{y'}{r} \frac{d^2\theta'}{dt^2}\right)$$
$$F_y' = m \left(\frac{d^2r}{dt^2}\frac{y'}{r} + 2 \frac{dr}{dt}\frac{x'}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{y'}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 + r^2 \frac{x'}{r} \frac{d^2\theta'}{dt^2}\right)$$

And we extract from it the parts that come directly from differentiating x' and y'.

$$F_x' = m \left(\frac{d^2 x'}{dt^2} - 2 \frac{dr}{dt} \frac{y'}{r} \omega - 2 r \frac{x'}{r} \frac{d \theta '}{dt}\omega - r \frac{x'}{r}\omega^2\right)$$
$$F_y' = m \left(\frac{d^2 x'}{dt^2} + 2 \frac{dr}{dt} \frac{y'}{r} \omega - 2 r \frac{x'}{r} \frac{d \theta '}{dt}\omega - r \frac{x'}{r}\omega^2\right)$$

The above already contains the centrifugal and Coriolis terms. In vector form, it can be equivalently written like this.

$$F = m(a' + 2 \omega \times v - \omega^2 r \hat{r})$$

So long as you define vector ω as vector perpendicular to x/y plane.

Finally, suppose you want to continue using Newton's 2nd in this coordinate system. Then you would have to define.

$$F' = F - 2m\omega \times v + m\omega^2 r \hat{r}$$

Where F is the real forces acting on the object, and the two additional terms are Coriolis and centrifugal fictitious forces respectively.

Obviously, this is not how you are going to explain the subject to high school kids. But this is level of understanding you should have before teaching centrifugal forces.

 

[A.T.]

I'm not sure of the precise scenario that your answer comes from but, if the car is doing the same 'speed' over the surface as the station peripheral speed then its resultant peripheral speed will be twice that of the station (?). The acceleration is given by v²/r so you would expect to measure 40N, not 20N.

Yes you are right. In the station frame you have:
centrifugal : 10N outwards
Coriolis : 20N outwards
But the net force must be : 10N inwards to provide the centripetal acceleration, so the wall must provide 40N inwards.

 

[A.T.]

If you hold a ball in your hand in this space station and let it go what will happen?

If you stand on the outer wall and try to drop the ball on your foot, it will deviate from the radial direction opposite to the station rotation.

 

[Algren]

0N for opposite direction and 20N for outrunning the rotation

It depends on How fast. If we conside ωr as the speed. Then it will be 0N and 40N, since the V is doubled, or you can say ω is doubled, the acc. will become fourfold.

So if I am standing 'still' in this space station I cannot tell which way it is rotating unless I use the coriolis force on my arms ?? How will the coriolis force reveal the direction of rotation.?

Or you could throw a ball, and see how far it goes.

That is not an acceptable answer ! wave your arms around... come on
HOW can you determine, 'in a physics way', which way the station is rotating?

Try and see, :). Or you could run really fast both ways, and see in which way you feel lighter.

BUT, there is a problem, if we have ω<<<1 and r>>1km, we need accelerometers, or perhaps a window.

CONCLUSION:

Taking into consideration this post, and a previous post of mine, and some formulae deducted taking into these errors into account:

We will be unable to replicate Earth-like conditions in space until and unless we sell Mars and build a humongous Station.

 

[HallsofIvy]

This thread was started by Algren. Why has truesearch now taken it over?

 

[Algren]

This thread was started by Algren. Why has truesearch now taken it over?

People have their doubts, and he was clearing his, =)

 

[sophiecentaur]

That is not an acceptable answer ! wave your arms around... come on
HOW can you determine, 'in a physics way', which way the station is rotating?

To be fair, that is the first and only 'arm waving' answer I have read on this forum that has been both practical and correct. It would be up to the student to fill in the details, of course, but it gets my vote.

 

[A.T.]

It depends on How fast. If we conside ωr as the speed. Then it will be 0N and 40N, since the V is doubled, or you can say ω is doubled, the acc. will become fourfold.

Yes, see post #22 for how it is explained in the station frame.

 

[sophiecentaur]

Yes you are right. In the station frame you have:
centrifugal : 10N outwards
Coriolis : 20N outwards
But the net force must be : 10N inwards to provide the centripetal acceleration, so the wall must provide 40N inwards.

It's a much easier problem when you choose the appropriate reference frame! (The one I would have used at School )

 

[truesearch]

Tiny Tims post3 is the clearest explanation here. As it relates to the motionof the car against the rotation of the space station. When the car has the same speed (ωr) as the station the car appears to be in circular motion in the rotating frame yet there is no reaction force with the side of the station. To explain this in the rotating frame you need to 'invent' a 'force' (=mω^2r). This is the resultant of 2 other imaginary forces..coriolis force and centrifugal force.
Viewing this from the inertial frame gives a clear picture of what is going on. The car is stationary in the inertial frame and the station is rotating past the car. There is no reaction force with the floor and youcould say that the car was weightless...beautifully simple when viewed from the inertial frame.