What force is required to rotate a space ship?

  • Thread starter Darrin
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Hi,

I was researching the force required to spin a space ship. There is an old thread here but I'm not quite understanding the post.

https://www.physicsforums.com/archive/index.php/t-151000.html

To move a spaceship forward we have something like

F = ma.
1000 kilogram space ship
100 meters in length
Assume a cylinder shape
30 m/s of acceleration

F = 1000 x 30 = 30,000 newtons

The old post talks about the Moment of Inertia

For rotation, the equation is \tau = I \alpha which says that the torque produces an acceleration alpha, which is inversely porportional to the moment of inertia I. Just as in F=ma we are saying that the force accelerates the mass at an acceleration that is in inverse proportion to its mass. Same force, twice the mass, half the acceleration. Same torque, twice the I, half the rotational acceleration alpha.
He also provides a link to the moment of http://en.wikipedia.org/wiki/List_of_moments_of_inertia where if we assume our spaceship is a cyclinder like a wire.

I = (m * L^2)/12

So what is the force formula to rotate the ship at 30 meters per second give the same spaceship listed above, assuming that we can put perpendicular thrusters at the very tips of the ship?

I don't quite understand how torque and moment of intertia help me calculate how powerful of engines spin it at a desired rate?

thanks!
 

D H

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This looks a bit too much like homework, so I'm going to treat it as such (i.e., I won't give you the answer -- yet).

Saying that you want to "rotate the ship at 30 meters per second" doesn't quite make sense. Think about it this way: The center of the ship will not be moving at all. I gather that what you mean is that the ends of the spacecraft are moving at 30 meters/second solely due to rotation. It is better to talk about angular velocity rather than translational velocity when talking about rotation. One obvious choice is degrees/second. A better choice is radians/second, better because the velocity of some point on this long, cylindrical spacecraft that is a distance [itex]r[/itex] from the center of mass is just [itex]v=r\omega[/itex]. If you used rotations/second (e.g., frequency) you would need to multiply by [itex]2\pi[/itex]. With degrees/second, you also need to divide by 360.

So, the first thing you need to do is to convert the velocity to angular velocity.

There is a similar relation between torque and force. The torque is [itex]\tau=rF[/itex]. (Note: This is true only if the offset is perpendicular to the force.) You will want two thrusters firing in opposite directions so that the total force is zero but the total torque is doubled the single engine torque.

You can use the rotational analog to F=ma, [itex]\tau = I\alpha[/tex] to compute the angular acceleration in terms of the torque. So, how to fire the jets? Go back to the translational state. Suppose you have a thruster that exerts a force along the long axis of the cylinder and suppose you want to make the entire vehicle speed up by 30 meters/second. How are you going to do that? What are the relevant equations?


One last item: The equation [itex]\tau=I\alpha[/itex] is only correct under some very special circumstances. This problem is one of those special circumstances.
 
hmm, well it is not home work. Been a decade or two since I've had that. I'm just outline a physics model for a space game. Try to maintain some realism while being fun. Perhaps should have said that at the beginning but did in another post.

I want to keep accelerations in meters per second because of the human factor. Humans can only take about 3g (~30 meters/second) and might be moving anywhere on the ship at the time of rotation.

The ships in the model will only have to rotate up to 180 degrees to decelerate or 90 to change direction. Then use the primary engine carry on in the new direction.

anyway, I'm sorry I don't get your hints but thanks for posting, I just don't remember the details of classes I took quite a few years ago.
 

D H

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You can't say 30 meters/second is the equivalent of 3g without specifying the size of the vehicle! Suppose the vehicle is cylindrical and is rotating about an axis passing through the center of the cylinder and normal to the cylinder axis and the ends are moving at 30 meters/second (i.e., what you have specified in the original post). For a 20 meter long cylindrical vehicle that 30 meters/second velocity from rotation represents an acceleration of over 9g. For a 200 meter long vehicle, the acceleration is only 0.9g, and for a 2 kilometer long vehicle, only 0.09g.
 
You can't say 30 meters/second is the equivalent of 3g without specifying the size of the vehicle! Suppose the vehicle is cylindrical and is rotating about an axis passing through the center of the cylinder and normal to the cylinder axis and the ends are moving at 30 meters/second (i.e., what you have specified in the original post). For a 20 meter long cylindrical vehicle that 30 meters/second velocity from rotation represents an acceleration of over 9g. For a 200 meter long vehicle, the acceleration is only 0.9g, and for a 2 kilometer long vehicle, only 0.09g.
Sorry I don't follow you. I'm probably not being clear. I specified the size of the vehicle above.

Maybe a better example is a merry-go-round. Lets say we want the merry-go-round to spin 180 degree with a kid on the outside rim taking a maximum of 2gs of force. Up to 90 degrees he is accelerating and from 90 to 180 he is decelerating.

The merry-go-round is 10 meters in diameter and weighs 100kg including the kid.

Ignoring friction, how much force does it to push the kid from the outside lip 180 degrees to have him feel 2gs of acceleration/deceleration the whole time?

I can figure out how long it takes him to get from one side to the other (based on the average velocity and 1/2 circumference of the merry-go-round) but I'm unsure as to the force required to get him there. Any ideas?
 
bump for help.
 

D H

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Darrin, it looks like you may have a misunderstanding even of linear motion, let alone rotational motion. 30 m/s is a velocity, not an acceleration. 30 m/s2 is an acceleration.

I gather that what you want is a 100 meter long spaceship rotating such that at the ends a crewmember will experience a 30 m/s2 centripetal acceleration: they will feel a bit more than 3g. That means the spacecraft is rotating at 292 revolutions per minute.

So, what kind of force is needed to keep the spacecraft rotating at that rate?

None. A torque is required to change that motion. No torque is required to maintain it. Once you have the spacecraft rotating at 292 RPM will keep rotating at 292 RPM forever -- unless some external torque acts on the spacecraft.
 
Darrin, it looks like you may have a misunderstanding even of linear motion, let alone rotational motion. 30 m/s is a velocity, not an acceleration. 30 m/s2 is an acceleration.

I gather that what you want is a 100 meter long spaceship rotating such that at the ends a crewmember will experience a 30 m/s2 centripetal acceleration: they will feel a bit more than 3g. That means the spacecraft is rotating at 292 revolutions per minute.

So, what kind of force is needed to keep the spacecraft rotating at that rate?

None. A torque is required to change that motion. No torque is required to maintain it. Once you have the spacecraft rotating at 292 RPM will keep rotating at 292 RPM forever -- unless some external torque acts on the spacecraft.
I meant 30 m/s^2 ... can't seem to find the edit button above. Typo...

So now that I think I'm using the correct terms.

Could someone please tell me the formula to determine the torque required to get the above space craft up to 292 revolutions per minute?
 

rcgldr

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Could someone please tell me the formula to determine the torque required to get the above space craft up to 292 revolutions per minute?
Angular velocity = time x torque / (angular inertia) = time x (angular acceleration).
 

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