At what position should the force act?

Question

A slender rod AB of length L lies on the horizontal frictionless xy plane with origin O. Initially AB lies on the y axis with A at (0,a) and B at (0,a+L). A force F is then applied which acts normally to the rod at all times. At what position along the rod should the force act such that O, A and B remain colinear at all times?

Answer 1

linear acceleration of the rod due to the push:

a = F/m

The push is exerted at a distance x from the origin.  The rod will rotate about its center of mass if you push it at any distance other than its cm at x = L/2.

angular acceleration = torque / moment of inertia:

alpha = N / I

= (F * (x - L/2) ) / (mL^2 / 12)

= 6F(2x-L) / mL^2

The acceleration at the end of the end due to the rotation has to be equal to the linear acceleration due to the push:

a = alpha L/2 = 3F(2x-L) / mL = F/m

x = 2/3 L

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okay, thanks for the pic.  I've done the special case where a=0.  I don't especially like that you've chosen the variable a for the offset of the bar, which I like to use for linear acceleration.  I will call it D.

Since the bar is offset by a distance D, I want point A to accelerate tangentially by alpha * D to stay in line (rather than just having it stay still as I did above).

So the tangential acceleration at point A is:

a = F/m - alpha L/2 = alpha D

Substitute the angular acceleration from above (modifying it ever so slightly to incorporate the offset in the torque term):

F/m = (D + L/2) * 6F(2x-L-2D) / mL^2

And solve for the position you want to apply the force:

x = L^2 / 6(2D + L) + L/2 + D

It reduces to x = 2/3 L for D=0, as desired.

Vasek--here's what I don't get.  You say you are staying in the inertial frame (as I did), but you are taking derivatives of your unit vectors as if your reference frame is rotating.  So you pick up a coriolis term (a = 2 v omega) even though you say you don't want to use it and a centrifugal term.

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Anyway, your equations of motion look fine for the dynamic case.  I'm tracking you up through where you say that at time zero:

d - L/2 = L^2 / 12 / (a+L/2)

You lost a factor of L in your algebra:

d = L * (3a+2L) / (6a+3L)

But that's just at the starting point.  There's no reason that that relation needs to hold generally.  But then that's going to require your force to change over time.  Which requires going back to square one. :(

Dr. D., could you please state a little more clearly what is going on?  Did you really mean for this to be a dynamics problem?  Do you want the point of push to be constant?  Is the force constant or can it vary with time?  I'm not sure what to assume.  You can't assume too much without overconstraining the problem.

--yeah, after I wrote that note and looked at your work, I figured out what you were doing--basically just deriving the results most of us memorize.

--D, the problem is that once you get going, you have a centrifugal force pushing the rod out, and once it starts moving out, you get a coriolis force pushing back.  I'm not sure you can push in one spot what with all that going on.

Answer 2

I don't think it has a solution :-( But I have been a few years out of mechanics so my solution may be too rough and, of course, wrong. You may find some constructions below very weird :-D Well, let's dive in...

We may look at the problem in the rotating frame, but I have too little (no?) experience with Coriolis and especially Euler forces, so I chose the inertial frame of the origin.

Let's call the angle phi, k the unit vector in the direction A-O and n the unit normal vector in the direction of F. I'll denote the time derivative by a prime. The vectors k and n have time derivatives k' = phi' n and n' = -phi' k, for example.

The positions of A and B are a(t)*k(t) and (a(t)+L)*k(t), respectively. The mass of the rod is m.

1) p' = F:

Momentum p is m times the derivative of position of the centre of mass, p = mq', where q = (a+L/2)*k, so that

p = ma' k + m(a+L/2)phi' n

The derivative of p is

p' = ma'' k + ma'phi' n + ma'phi' n + m(a+L/2)phi'' n - m(a+L/2)phi'^2 k

The force is perpendicular to the rod, i.e.,

F(vector) = F*n

Since k and n are orthogonal, the k component of p' must be zero:

m(a+L/2) phi'^2 = ma''

The n component is F:

m(2a'phi' + (a+L/2)phi'') = F

Dividing both equations by m gives

a'' = (a+L/2) phi'^2

2a'phi' + (a+L/2) phi'' = F/m

2) L' = N:

The momentum of inertia I of the rod with respect to O is, according to Steiner theorem, I = 1/12*m*L^2 + m(a+L/2)^2. Its angular momentum L is I*phi':

L = 1/12*m*L^2*phi' + m(a+L/2)^2*phi'

However, L' is not simply I*phi'' since I also depends on time through a:

L' = 1/12*m*L^2*phi'' + 2m(a+L/2)a'phi' + m(a+L/2)^2*phi''

The torque N is found easily to be

N = F*(a+d),

where d is the point on the rod where F acts.

Putting the equations together, we easily get

1/12*m*L^2*phi'' = F*(d-L/2)

So far, we have three differential equations for a and phi:

(1) a'' = (a+L/2) phi'^2

(2) 2a'phi' + (a+L/2) phi'' = F/m

(3) 1/12*L^2*phi'' = F/m*(d-L/2)

If the initial condition is rest, i.e., phi(t=0) = phi'(t=0) = 0, we can promptly solve (3) explicitly:

phi(t) = 6/L^2 * F/m * (d-L/2) * t^2

then compute back

phi'(t) = 12/L^2 * F/m * (d-L/2) * t

phi''(t) = 12/L^2 * F/m * (d-L/2)

and plug into (1) and (2). From (2), we get

2a' * 12/L^2 * F/m * (d-L/2) * t + (a+L/2) * 12/L^2 * F/m * (d-L/2) = F/m

2a't + a+L/2 = L^2/12/(d-L/2)

a' = (L^2/12/(d-L/2) - a - L/2) / (2t)

The solution of the last equation diverges in t=0, and thus is in contradiction with the initial conditions, except for

L^2/12/(d-L/2) = a + L/2

d - L/2 = L^2 / 12 / (a+L/2)

d = (3a+2L) / (6a+3L) * L,

when a = const.

The solution so far

d = (3a+2L) / (6a+3L) * L

phi(t) = 1/(2a+L)*F/m*t^2

a = const.

is OK with respect to (2) and (3), but not with (1) except for the trivial case F = 0.

Please fix me if you find any errors.

Edit: Thanks, Mistress Bekki, for correcting my typo. After fixing it, we two have the same result: your x is my d+a, after unifying the letters.

Other d's lead to nonsenses. You can note I continued with this one and found it satisfies only two of my three equations. This reflects in the facts that the collinearity is conserved only at t=0. The equations hold to the first order only and break soon. As Dr D asked for conservation "at all times", we need more than looking at t=0, which I think is what you did.

In other words, a (or D) begins sooner or later changing too, and then the definition of d (or x-D) can't be valid anymore in the form given because it is not a constant.

Finally, let me please clarify what "method" I used. I do take an inertial frame and stay in it. In this frame, all of the coordinates would be full of goniometric functions. So, for simplicity, I define a basis in each instant,

k = (cos phi, sin phi)

n = (-sin phi, cos phi),

and express all vectors using it. This "method" helped me many times in the first year. It gives the same result as noninertial approach, but one does not need to remember ANY formulas for pseudoforces.

-- Yes, that's it, some formulas simply failed to find the way into my memory.

I should tell what I assumed: to find a solution such that both the force and the point on the rod are constant and we are interested in arbitrarily long time intervals. If some condition was loosened, I could possibly use less complicated way of solution.

Result: as Mistress Bekki said, overconstrained. The only solution is a trivial one when F is zero (then d can be of course whatever; I divided by F at one moment).

Answer 3

I think this problem have 4 solutions.

If you push at the ends parallel with the length axis of the body,

Then it will move with stable direction of the length axis.

You might also push with 90 deg angle on the rod from both sides, and also then get a movement with stable direction of the rod.