Would the rotation of a space ship in vacuum slow it down

Firstly , a few disclaimers :

1. English is not my native language so I may use the wrong technical terms. Please notify me and I will correct the question.
2. As a CS major , I've only taken one class of college level Physics.
3. The topic of this question seems to be somewhere in between here and Quora , I apologise if you feel this is the wrong place to ask this question.

Okay , so I had a small debate with a friend , who claims that rotating a space ship (in order to simulate gravity) would cause it to slow down. His argument was that if you had to counteract gravitational pull (from a random angle) , the fact that you have to apply a force both in the plane in which the ship rotates (to prevent the rotation from slowing down) and in the plane of the forward momentum , the force you apply may cause the ship to lose forward momentum.

My argument was that since the two planes are perpendicular at all times (due to the ship's design) , the fact that the ship is rotating has no effect on the forward momentum of the ship. The force you apply in this plane is the same as the one you would use if the ship were not rotating. I failed to explain this argument in a convincing manner.

Can you help either to prove me wrong or provide a better explanation to my argument.

(Additional info): - We are assuming a cylinder-like ship with perfect weight distribution , an even number of equidistant thrusters for rotation

• this debate was started by the movie 'Interstellar' (specifically if it's better to rotate an entire ship or just the sleeping pods to simulate gravity)
• There is no bet involved (I wouldn't be asking for help otherwise)

If by "slow down" you mean "decelerate the rocket as it moves through space" the answer is no. Once the ship as begun traveling at a constant linear speed $v$ and begun rotation at a constant angular speed $\omega$ it will continue to move with a constant velocity (both linear and angular) in the absence of any net forces or torques. The object will have a constant linear momentum given by $p=mv$ and a constant angular momentum given by $L=I\omega$ where $I$ is the moment of inertia of the object. Conservation of both angular and linear momentum dictates that each of these values will remain constant, given the absence of any interactions with other objects.

However, it is possible to interpret your question another way. If by "slow down" you mean "travel at a slower linear velocity than a non-rotating ship would travel for the same amount of total kinetic energy" the answer is yes. A ship that is both rotating and traveling linearly will have two separate kinetic energy terms: the translational kinetic energy given by $KE_{trans}=\frac{1}{2}mv^2$ and the rotational kinetic energy given by $KE_{rot}=\frac{1}{2}I\omega^2$. If similar rocket ship is not rotating, a given amount of work done on the ship will all result in an increase of the ship's linear kinetic energy. However, a rotating ship with the same energy budget will have its work split between increasing the translational (linear) kinetic energy of the ship and its rotational kinetic energy of the ship.

• I hope you don't mind if I dumb-down your answer to bring it closer to my level of understating. From what I can tell , you are saying that if you want to rotate , you have to devote some of your energy budget (fuel) to the rotation . That much is clear. By "slow-down" I mean - "does having to rotate (in a gravitational environment) around the axis of linear momentum cause you to lose some of that linear momentum" – user1689207 Nov 20 '14 at 15:01

I think there is no influence of rotation, regardless of rotation plane and direction vector. The center of mass will move with the same speed.

Newton's First Law of motion should be enough to prove it:

When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force.

If you consider some kind of propulsion-based rotation initiation, then yes, that initiation may change the velocity of the center of mass - but only because some gas/fuel has been used and therefore an external force has been applied. Should you apply an equal force to accelerate both ships the acceleration would be the same.

No, I have no math to prove it.

• Yes, we are considering a propulsion based rotation initiation , but the external forces should cancel each other like in @Hypnosifl 's answer – user1689207 Nov 20 '14 at 13:59

It can be proven that if you just want to know the motion the center of mass of a system of particles (or a continuous extended body), you can just calculate the vector sum of the force vectors on each particle (or the integral of the infinitesimal force on each infinitesimal bit of mass dm in a continuous extended body), and then treat this total force vector as if it were acting on a point particle located at the center of mass (with this imaginary point particle having the same mass as the total mass of all the particle in the system, or the total mass of the extended body). See this page for a derivation.

So with that in mind, I think it's sufficient to point out that it's possible to rotate an object without any change in the total force acting on the center of mass--just apply equal and opposite force vectors to points on opposite sides of the object at the same radius from its center of mass, with the force vectors at right angles to the radial direction, and with both exerting a torque in the same angular direction (both clockwise or both counter-clockwise). I looked for an image online, here is an example with forces on opposite sides of a wheel: The point is that although these forces both exert a torque (clockwise, in this case) and should therefore cause the object to rotate, the two force vectors have the same magnitude but opposite directions in space, so their vector sum is zero and they don't have any effect on the net force acting on the center of mass. Therefore if the center of mass is accelerating under the influence of gravity (in orbit, for example), it should make no difference to the path of its center of mass whether you choose to apply equal and opposite forces to rotate it, or whether you choose not to rotate it at all.