Effects of offcenter forces...

[carrotstien]

If you have a rod, or more simply 2 very small masses connected by a massless rod of length L (each mass being equal; M kgs.) You then connect a massless rocket thruster an x distance to the right of the center of the rod so that the thrust is always F newtons perpendicular to the rod. How would the system behave? O, and before people start saying how massless things don't work, then say that the mass of the rod is .0000000001 * M and the material behind ejected out of an equally light rocket is lighter a million times lighter than that, but is going at a very very large velocity close to the speed of light so that its dP/dt = F

 

This is how I tried to solve it:

 

α - angular accel

α1 - angular accel of mass 1

α2 - angular accel of mass 2

 

τ - torque

a - linear accel

a1 - lin accel of mass 1

a2 - lin accel of mass 2

M - mass

M1 - mass1

M2 - mass2

L - length of rod

R1 - distance that the center of mass 1's (left mass) motion is to the left of 1 along the line of the rod where mass 2 is always relatively to the right of mass 1 by distance of L meters

R2 - '' '' '' for mass 2 (right mass)

Rf = distance to the point left of the thurster that the thurster creates a torque around

F - force from rocket

 

I - moment of inertia of system

 

I = (M1)*(R1^2) + (M2)*(R2^2)

M1 = M2 = M

R2 = R1 + L

let R = R1

I = MR^2 + M(R+L)^2

I = M*(R^2 + R^2 + 2RL + L^2)

 

I = M*(2R^2 + 2RL + L^2)

 

τ = F*Rf

τ = F*(x+L/2 + R)

 

assumption: α2 = α1

 

let α2 and α1 = α (in the case that assumption above is correct)

 

α = τ/I

α = [F*(x+L/2 + R)] / [M*(2R^2 + 2RL + L^2)]

 

 

 

right from the starting point where nothing is moving or spinning or anything

 

F = (a1)*(M1) + (a2)*(M2) (conservation of momentum)

 

a1 = α1*R1

a1 = α*R

 

a2 = α2*R2

a2 = α*(R+L)

 

F = (α*R)*(M) + (α*(R+L))*(M)

F = Mα(2R + L)

 

α = [F*(x+L/2 + R)] / [M*(2R^2 + 2RL + L^2)]

F = M*[F*(x+L/2 + R)] / [M*(2R^2 + 2RL + L^2)]*(2R + L)

M's cancel, F's cancel

1 = [(x+L/2 + R)] / [(2R^2 + 2RL + L^2)]*(2R + L)

2R^2 + 2RL + L^2 = (x+L/2 + R) * (2R + L)

2R^2 + 2RL + L^2 = 2Rx + RL + 2R^2 + Lx + L^2/2 + LR

L^2 = 2Rx + Lx + L^2/2

L^2 - Lx - L^2/2 = 2Rx

L^2/2 - Lx= 2Rx

(L^2/2 - Lx)/2x = R

 

 

R = (L^2 - 2Lx)/4x

 

So far, this works for when X = 0, because you would expect the system to travel straight (or equavilently saying that it would travel a circular path of radius infiniti). Also, the further you get from the center (or as X approaches + or - infiniti, R becomes -.5L (which would mean that the center of rotation becomes the center of the system) -> also expected.

 

However, i have some difficulty expected that any force far enough from the center would add nothing to the linear momentum of the system (for the short time that it is going more or less straight (the first dθ).

This would be true, if, for a similiar case, if the center of the rod was attached to a pivot - if the force that the pivot applies onto the rod when a torque/force is applies reduces to 0 as the distance of the application of the force approaches infiniti.

 

Even if I made no incorrect calculations or assumption, I can't imagine how to even apply this to the case after the first dtime where there is already and angular and linear motion because the formula F = (α*R)*(M) + (α*(R+L))*(M) is only true if both masses are traveling around the same center point and this seems to only be true for the very first moment.

 

So...um...how would I do this

[thedarkshade]

Could you LaTeXise it please. I find it very difficult to follow!

[carrotstien]

i attached a Word document, maybe its easier to read with the sub and super scripts and *'s

math.doc

[thedarkshade]

Got it.. reading now!

 

Thnx

 

and search the forum for a LaTeX tutorial