TheRealTL said:
I used to write out equations for things like this in high school when I was bored in statistics. Imparting angular momentum requires your propulsion to be EXACTLY halfway away from the center of mass.
No. If the line of force does not pass through the center of mass you will always be applying a non-zero torque -- and you will still be applying a non-zero force that results in translation as well. It is a simple matter of conservation of linear and angular momentum.Speaking of which,
comment said:
D H said:
End result: The three forces F1=F acting at a point r, F2=F acting at the center of mass, and F3=-F acting at the center of mass result in an acceleration of F/m and a torque of r×F. These three forces are indistinguishable from the single force F acting at a point point r, so the response to this single force is exactly the same as the response to the three forces above.
This is not too intuitive. This tells that force F acting at rigid body not at the center of mass makes the body move and
at the same time rotate, both with the same force (F).
What about the conservation of energy?
This is admittedly non-intuitive. Conservation of linear momentum dictates that the linear acceleration that results from applying a force to an object is independent of the point of application. At the same time, conservation of angular momentum dictates that a non-zero angular acceleration will result if the line of force does not pass through the center of mass. So what gives? In the case of an external force, the answer is simple: Force and work (energy) are different concepts. More work is required to apply the same force off-center than through the center.
In the case of a rocket, the resolution to the problem lies in looking at the entire system, that is, the vehicle plus the cloud of exhaust left behind while the vehicle is thrusting. This is a constant mass system, so all the problems associated with variable mass systems vanish. Suppose the rocket is operating in deep space, well removed from any gravitational influences. I'll denote the rate at which the rocket is ejecting mass into space as \dot m_e, with a positive value indicating that the rocket is losing mass and the exhaust cloud is gaining mass. I will denote the velocity of this newly-ejected exhaust relative to the vehicle as \mathbf u_e, emboldened to indicate that this is a vector quantity.
Without derivation (you can see a somewhat simplistic derivation [thread=199087]here[/thread]), the rate at which the rocket+exhaust system is changing kinetic energy is
\dot T_{r+e} = \frac 1 2 \dot m_e u_e^2
Conservation of energy only has two things to say about this situation:
- That energy didn't just magically appear out of nowhere. It has to be balanced by a reduction in potential energy. That potential energy source is of course the unburnt fuel. (Aside: The reduction in potential energy is actually greater than the gain in kinetic energy. Some of that consumed potential energy is wasted in the form of hot exhaust gas.)
- The energy gain is partitioned between the exhaust gas and the vehicle. All conservation of energy has to say about this partitioning is that the change in the exhaust gas's kinetic energy plus the change in the vehicle's translational kinetic energy plus the change in the vehicle's rotational kinetic energy must sum to equal this total. Conservation of energy is moot regarding the partitioning of the change in kinetic energy.
That partitioning is highly sensitive to both circumstances and even to the observer. To demonstrate the latter, consider a rocket that is thrusting such that all the energy is going into translational energy. Let's look at what four different inertial observers see, call them observer #0 to #3, where
- Observer #0 is motionless with respect to an inertial frame that is instantaneously co-moving with the rocket. In other words, observer #0 sees the rocket as (at least instantaneously) having a velocity of zero.
- Observer #1 sees the rocket's instantaneous velocity as equal but opposite to [\itex]\mathbf u_e[/itex].
- Observer #2 sees the rocket's instantaneous velocity as -2\mathbf u_e.
- Observer #2 sees the rocket's instantaneous velocity as -3\mathbf u_e.
Observer #0 is equivalent to a person watching a launch. Imagine a Shuttle launch. A huge cloud of exhaust is created at ignition. The vehicle starts to rise very slowly at first. All of the energy appears to be going into that exhaust cloud. That is because, from your perspective, at the time of launch all of that energy
is going into the exhaust cloud.
Observer #1 sees a very different situation. From their perspective, the exhaust is being left behind with zero velocity relative to the observer. All of the energy is going into the vehicle.
Observer #2 sees the same thing as observer #0, but for a different reason. In this case, the gain in momentum due to an increasing velocity is exactly counterbalanced by the loss in momentum due to momentum transfer to the exhaust cloud.
Finally, observer #3 sees the rocket as losing momentum as it gains speed! So, four different observers of the same event see three very different outcomes as far as energy is concerned.