Explain translation and rotation in empty space

Question

There is a previous question of mine which asks about rotation in empty space.

Please look at previous question for greater understanding of this question.

Explain rotation in empty space

My question now is:

If I have a rod in empty space initially at rest and now if I apply a constant force acting at one end (fixed point)of the rod how would the motion of the rod be?

My guess:

• I was told in the previous question that when a force is applied at one end,the rod rotates about com along with translation. In this situation we have a constant force acting at one end.

So,initially the rod would move in north (let's say) and also has rotation and as the rod rotates, it moves the translation towards north west and when it rotated by 90 degrees it would,at that instant,have translational motion caused by the force in West direction.

As the rod makes 360 degree it would come back to the same initial position.

Is this understanding of mine wrong?

After doing some research and developing greater understanding of concepts involved I arrived at the following answer to the problem:

Note:To have a greater understanding of my answer please have a look at Luke Pritchett's answer!

My answer is as follows: In my answer, the force on rod acts at a distance towards the right of COM and in the picture given initially I assumed that force to be acting at COM and the body with some rotation (with constant torque)

Don't think from the perspective that rod comes back to the same position, what I mean is that in the picture I don't mean to convey that rod makes a rotation and comes back to same position i.e the picture is just showing different orientations at different times. Since we have time differences for each 90degree rotation we can have the following cases:

Before discussing the cases have a look at the general concept which is seen in each case: considering the time differences we can say that, the rod is accelerated along negative x direction for more time than it is decelerated along the same direction with same magnitude.Hence we can conclude that the final position of the rod will definitely lie towards the left of initial position.

We can get different cases when we consider motion along y direction.If we look at motion from A to C we would observe that the rod has acceleration along (+y)from A to B and along (-y) from B to C and since (t1-0) > (t2-t1) we can say that the rod would still have velocity along (+y) just after orientation C also.

The different cases that are possible are: These cases arise depending on magnitudes in so and so intervals.

Case(I):

We consider that the deceleration between C to D is good enough to Change velocity of rod from (+y) to zero exactly at orientation D.And later on(after D) it has acceleration along (+y) and hence the rod would move along (+y) with increasing velocity.

In this case trajectory of body would be similar to the one(diagram)given in Luke Pritchett's answer.

Case(II):

In this case we consider that the deceleration is good enough to reduce velocity from (+y) direction to zero before it reaches orientation D and after reaching zero velocity we would have some more cases(I'm not discussing here).

In this case trajectory of body would definitely not match the one given in Luke Pritchett's answer.

Case(III):

In this case we consider that the deceleration is NOT good enough to reduce velocity of rod from (+y) direction to zero before (or) at orientation D.In this case rod would have velocity along (+y) direction just after D also and later on its velocity along (+y) would increase as it is accelerated after D.

In this case the trajectory of body would be similar to the one given in Luke Pritchett's answer.

Is this new understanding of mine correct and are the cases mentioned right?

Answer 1

No, the rocket does not return to where it started, especially if it starts from rest.

You are correct that the rod will spin around in a circle. In fact it will spin faster and faster because it experiences a constant torque.

But you are not applying the relationship between acceleration, velocity, and position correctly. Try sitting down and drawing the object, its velocity vector, and its acceleration (or force) vector at a few consecutive points in time. At each step the acceleration vector will tell you how to draw the next velocity vector, and the velocity vector will tell you how to draw the next position. If you do it right, you will find that the rod does not move in a circle, but zooms off in close to the direction the it started facing. Remember to make the rod rotate a little at each step as well.

[Edit] The mistake you seem to consistently be making is to make the object travel in the direction of the force. That is not how force works. The force tells you how the velocity changes in time, not how the position changes in time. In the picture you have drawn you have not drawn the velocity vector of the rod, which will make it very hard to get it right. Draw the velocity vector as it changes in time, and then make sure that the way it changes is consistent with the force/acceleration you give it.

I attached my own sketch done below, though I integrated the equations of motion so this is the exact solution. Notice that by the time the rocket is turned around the forward velocity is too large to be reversed.

[More edit] Ultimately, diagrams aside you have to do the math of solving Newton's second law for the position of the rod as a function of time. If we say that the orientation of the rod is given by the angle $\phi(t)$, with $\phi(0)=0$, the rod horizontal, the force pointing in the positive $y$ direction, and the position is given by $\vec{x}(t)$, then Newton's laws tell us $$\ddot{\phi} = \frac{3F\ell}{m\ell^2}$$ $$\ddot{\vec{x}} = \frac{F}{m} (-\sin\phi(t)\hat{x} + \cos(\phi(t)) \hat{y})$$

Integrating the first equation twice gives $\phi(t) = \frac{1}{2}\frac{3F\ell}{m\ell^2}t^2 \equiv \frac{1}{2}\alpha t^2$

Integrating the second equation twice gives

$$\vec{x}(t) = \vec{x}_0+\frac{F}{m}\int_0^t\left( \vec{v}_0 +\int_0^{t'} (-\sin(\alpha t''^2/2)\hat{x} +\cos(\alpha t''^2/2) \hat{y})dt'' \right) dt'$$

We can write this out in terms of named functions $$ \vec{x}(t) = (\vec{x}_0+\ell/3 \hat{x}) + t\left(\vec{v}_0 + \frac{\ell\sqrt{\alpha\pi}}{3}(-\operatorname{FresnelS}(t\sqrt{\alpha/\pi})\hat{x}+\operatorname{FresnelC}(t\sqrt{\alpha/\pi})\hat{y}) \right) + \ell/3(-\cos(\alpha t^2/2)\hat{x} + \sin(\alpha t^2/2)\hat{y})$$

The Fresnel functions both oscillate but approach the limit $+1/2$ at infinity, so after a long time the rod will be moving mostly in a straight line with average velocity $$\vec{v}_\infty = \vec{v}_0 + \sqrt{\pi\frac{F\ell}{3 m}}(-1/2,1/2)$$

There's no other solution as long as the force is constant and always perpendicular to the rod. The rod will always zoom off in a straight line, plus small oscillations. Properly tracing out the changes in velocity and position will tell you this.

Answer 2

Yes your understanding is correct as the rod is rotating force is always perpendicular to one of end its so the rod is rotating but when the rod has rotated 180 degree the force is acting in opposite direction with respect to its initial direction like this we can show that the if after rotating thita force vector is F then after rotating 180+thita degree force vector will be -F, so the rod will return to its original position after completing the circle as net force on rod after rotating 360 degree is 0