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Question

http://i37.tinypic.com/1051y4w.jpg

A slender rod AB with length L and mass m lies unconstrained on the horizontal frictionless xy plane. Initially AB lies on the x axis with A at the origin and B at (L, 0). The rod has a propeller which produces a force of constant magnitude F on a point P where AP = 2L/3. The propeller can rotate on the rod so that the angle φ at which the force acts changes with time. The objective is for the point A never to leave the origin. Initially φ = 0. How many revolutions does the rod rotate before it attains its maximum angular velocity?

Answer 1

I was thinking of a problem a lot like this with an inward component of force to kill the centrifugal problem.



I'm guessing the answer is infinite. Omega will approach its maximum value assymptotically. I'm also guessing that as usual, the problem will require MS excel and Euler to at least get a rough idea what's going on.



I'll just cutnpaste and slightly edit Vasek's equations of motion:



The rod has one degrees of freedom, theta. It's center of mass starts at L/2 and stays there. The initial values of theta and theta' are zero.



(1) r'' = (L/2) theta'^2 - sin(phi) F/m = 0

(2) (L/2) theta'' = cos(phi) F/m

(3) theta'' mL^2/12 = cos(phi) * F * L/6



(1) The rod doesn't accelerate outward, so the inward component of the propeller's push must offset the centrifugal force.

(2) The tangential push of the propeller is equal to the tangential pseudoforce (we've killed the coriolis force by keeping the rod from moving out).

(3) The torque on the rod (tangential push * lever arm) is equal to the moment of inertia times the angular acceleration.



Solve (1) for phi:

phi = arcsine (mL theta'^2 / 2F)



And what we really need to plug into (2) and (3) is cos(phi):

cos(phi) = sqrt (1 - sin(phi)^2)

= sqrt (1 - m^2 L^2 theta'^4 / 4F^2)



Solve (2) and (3) for theta'' and eliminate:



theta'' = cos(phi) 2F/ mL = cos(phi) 2F/mL



So (2) and (3) are redundant. The propeller is well-placed or else (2) and (3) would be contradictory. If we had not been given the fact that the propeller acts at the point P, 2/3 of the way down the length, we could have used this to solve for it (as we did in a previous problem). So our equation of motion for theta is:



theta'' = sqrt (1 - m^2 L^2 theta'^4 / 4F^2) 2F/mL

= sqrt (4F^2 / m^2L^2 - theta'^4)

= sqrt (theta'max^4 - theta'^4)



So theta' starts at zero and will approach its maximum value, theta'max=sqrt(2F/mL), assymptotically. Now Dr. D can make us some pretty pictures.



Or since Dr D is hinting that more analysis may be fruitful, let's get cute and just try to get an expression for theta:



omega' = sqrt (omegamax^4 - omega^4)

= d (omega) / d(theta) * d(theta)/dt

= d(omega) / d(theta) * omega



theta = integral from omega=0 to omegamax of:

d omega * omega / sqrt (omegamax^4 - omega^4)



**********Made a typo in mathematica--everything after here is wrong, but I'll leave it

http://integrals.wolfram.com/index.jsp?expr=x%2FSqrt%5Bc%5E4+%2B+x%5E4%5D&random=false



That factor of omega on top makes this integral much nicer than the one you get when you integrate for omega (see Remo's link).



theta = 1/2 log (omega^2 + sqrt (omegamax^4 + omega^4) )



evaluated at omega = 0 and omega = omegamax



The difference of logs is the log of the ratio, so the numbers all cancel,

and your answer is:

theta = 1/2 log (1 + sqrt(2) )



That is an answer I never would have guessed in a gajillion years.



Remo--you made a plug-in mistake when you substituted in sin(phi) from the centrifugal term. The radius of the CM is L/2, not L/3. Fixing that will make that funny factor of 4/9 go away in your final equation of motion.



DrD. Pretty please, could you run the equation of motion through the MS Excel / Euler solution. You can use omegamax = 1 and some time steps that seem reasonable. Since we have an analytic solution in our back pockets, I'm curious to see how well the simulation performs. Plot theta vs time, omega vs. time, and omega vs theta. Zero that last plot in on the terminal bit so we can see the solution.



Was that a book problem or did you make it up? That's one of the cleanest solutions to a seemingly intractable mess I've ever seen.



*********

DAMN--I liked the logs. Okay, so it's an arctangent.

http://integrals.wolfram.com/index.jsp?expr=x%2FSqrt%5Bc%5E4+-+x%5E4%5D&random=false



1/2 (arctangent of infinity - arctangent of zero)

= pi/4

Well I guess that's an even cleaner answer than the logs gave us. I still want to see simulation :). You didn't say where you got this or if you made it up.

Answer 2

Let w(t) be the angular velocity of the rod about the origin at time t. Then the radial force F_c(t) required to keep A at the origin is -1/2*m*L*w(t)^2. This leaves a force of magnitude sqrt(F^2-F_c(t)^2)

=sqrt(F^2-1/4*m^2*L^2*w(t)^4) applied perpendicularly at P to angularly accelerate the rod about the origin. Thus the maximum angular velocity is obtained when this quantity reaches zero, i.e., when w(t)=sqrt(2*F/m/L). w(t) will satisfy the differential equation

w'(t)

=c*sqrt(F^2

-1/4*m^2*L^2*w(t)^4) for some constant c, hence t=1/2/c*2^(1/2)/

(1/F*m*L)^(1/2)

*(4-2/F*m*L*w(t)^2)^(1/2)

*(4+2/F*m*L*w(t)^2)^(1/2)/

(4*F^2-m^2*L^2*w(t)^4)^(1/2)

*EllipticF(1/2*w(t)*sqrt(2)

*sqrt(1/F*m*L),I)

Thus the limiting value of sqrt(2*F/m/L) is not attained by w(t) for any finite t, thus, I believe the answer is the rod never quite attains its maximum angular velocity, so I think the answer is an infinite number of rotations.

Answer 3

Angular velocity = ω = dθ/dt

Angular accleration = α = d ω/dt

Centripetal force =m ω² r = F*sin φ

Torque = τ = 2/3L * F *cos φ

Moment of Inertia = I = m r²/3



α = τ/I

α = (2/3 L * F * cos φ ) / (m L²/3)

α = 2 F * cos φ / m L



F² = (F *cos φ)² + (F*sin φ)²

F² = (α mL/2)² + (m ω² L/3)²

F²= m² (α L/2)² + (ω² L/3)²

F² =m² (dω/dt L/2)² + (ω² L/3)²

F²= (m² L² /4)[ (d²θ/dt²)² + 4/9(dθ/dt)^4]

4F²/ m² L² = [ (d²θ/dt²)² + 4/9(dθ/dt)^4]



This is your equation of motion for theta. It will never attain max angular velocity because the jerk will never go to zero.

It is also possible to sove for t based on ω and using the third to the above equation, i.e.



F² =m² (dω/dt L/2)² + (ω² L/3)²

dt = 3/2m * √[1/(9F²/L² - ω^4)] dω



The solution is elliptical and messy. http://integrals.wolfram.com/index.jsp?expr=1%2FSqrt%5BC+%2B+x%5E4%5D&random=false