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    NB: This is intentionally a similar thread to that which I started a little while back, but I wanted to discuss a few extra ideas and hence thought it warranted starting a new thread.


    Here is a thought experiment:
    a rigid body - a 3 dimensional rod - with a rocket booster aimed perpendicular to its displacment form the rod's centre of mass (CoM), is sat in deep space and not experiencing any gravity etc.
    What motion does the rod undergo when the rocket is fired for a short length of time?


    After some thinking I wanted to ask whether my conclusions are correct:

    (1)A force F applied to a rigid body at a distance d from the center of mass has the same effect as: the same force applied directly to the centre of mass *and* a couple centred on the CoM such that the couple forces, C, satisfy: CL = Fd?

    (2) Following from (1): would the aforementioned rod undergo (whilst rocket is firing) a. accelerating rotation about its CoM with b. its CoM undergoing either circular or elliptical translational motion?

    (3) Once the rocket stops firing, will the rod translate linearly whilst undergoing constant rotation ad infinitum?

    (4) Is a couple the only physical situation by which a body can undergo rotation *without* translation?

    (5) Does the couple definition mean that the rotation axis will always be at the midpoint of the straight line joining their points of application?

    (6) Extending the above to a physical pendulum (i.e. not a particle on a light string): Whilst the derivation of a physical pendulum undergoing SHM about an arbitrary pivot uses rotational motion, isn't the CoM undergoing translational motion too? Or if measured wrt pivot, actually isn't the CoM undergoing translational motion? I can't decide on this one!

    If someone can address the above points one at a time, that would be much appreciated.
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    (Original post by Sm0key)
    NB: This is intentionally a similar thread to that which I started a little while back, but I wanted to discuss a few extra ideas and hence thought it warranted starting a new thread.


    Here is a thought experiment:
    a rigid body - a 3 dimensional rod - with a rocket booster aimed perpendicular to its displacment form the rod's centre of mass (CoM), is sat in deep space and not experiencing any gravity etc.
    What motion does the rod undergo when the rocket is fired for a short length of time?
    Let me try to analyse a simpler scenario. It's quite tricky to do this from first principles, so I'm not 100% sure of my argument, but it may help you.

    Let's consider a dumbbell floating in space, with masses m,M connected by a massless rigid rod. We hit mass m with an impulsive force. What happens?

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    1. Since we have given the body some momentum to the right, it moves, as a whole, to the right with some speed v_\text{com}

    2. However, mass m felt the impulsive force, and since m < m+M, its initial instantaneous velocity v > v_\text{com} to the right.

    3. m is connected to M by a rigid rod, and M has inertia, so an instantaneous force F arises in the rod, since it is rigid, and can't stretch (it's trying to stretch, since m is moving away from the rest of the body).

    4. By Newton III, both m and M feel the same force F directed as shown in the diagram above.

    5. So instantaneously, m has a velocity transverse to the rod, and feels a force pulling it down the rod. Thus, instantaneously, it must move on the arc of a circle of some radius r_1, centre C.

    6. But m and M are connected via the rigid rod. So if m turns about C with some instantaneous angular velocity \omega then so does M, since rigid rods can't bend. Suppose that M has a radius of r_2 about C.

    7. So far, nothing stops C from being at any particular distance from m, though we know that it must lie above M and below m, from the direction of the forces, which both point toward C (so C must lie somewhere on the rod).

    We also know that both masses feel the same instantaneous force F. So from circle motion, we have that:

    F=mr_1\omega^2 = Mr_2\omega^2 \Rightarrow \frac{r_1}{r_2} = \frac{M}{m}

    So if M=3m then r_1 = 3 r_2 i.e. C lies closer to M than to m

    We can see that each mass starts off rotating about a point between them, with the distances to the point in the inverse ratio of the masses. This happens to coincide with the centre of mass of the body. So the c-o-m doesn't rotate, and must move purely translationally with the velocity v_\text{com}.

    8. And then by conservation of energy, we conclude that this instantaneous rotational and translational motion continues until another external force acts on the body (well, assuming that the rotational and translational modes can't exchange energy between them)
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    That's excellent - just the sort of first principles justification I was hoping for. I really appreciate it.
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    (Original post by Sm0key)
    That's excellent - just the sort of first principles justification I was hoping for. I really appreciate it.
    I've edited point 7 above, re: the location of the point C - we can, in fact, tell that it must lie on the rod somewhere.
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    (Original post by atsruser)
    I've edited point 7 above, re: the location of the point C - we can, in fact, tell that it must lie on the rod somewhere.
    Thanks - I did consider that when I read through it initially. Thanks again.
 
 
 
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