Wire Reaction Force

Apologies for the messy solution. The solution also has a few step jumps (steps that I find trivial). I find myself feeling lazy to type it out in a Latex format right now.

It is a very nice problem! As always, I welcome feedback and comments on the solution.

First I tried analytically, here's about how far I got:

Because of the shape and using dot notation for time derivatives, we have $y=x^2$ $\dot{y}=2x\dot{x}$

Because of conservation of energy $E=E_{kin}+E_{pot}=\frac12m(\dot{x}^2+\dot{y}^2)+mgy$ and the initial condition ( $x=0, y=0, \dot x=5, \dot y=0$ ) we have (putting $m=1$ and $g=10$ ): $\dot{x}^2+\dot{y}^2+20y=25$ At $t_f$ , when $\dot{x}= \dot{y}=0$ , we have $y_f=\frac54$ and $x_f=\frac12\sqrt5$ .

Combining the above equations, we arrive at the differential equation $(1+4x^2) \dot{x}^2+20x^2-25=0$ or, maybe simpler: $(\frac{1}{4}+y) \dot{y}^2+20y^2-25y=0$

Differentiating the expression for $\dot y$ once again: $\ddot{y} = 2(\dot{x}^2+x\ddot{x})$ and using Newton's law, we can write $\ddot{y}=F_y/m-g$ from which follows (putting $m=1$ and $g=10$ ): $F_y = 10+ 2(\dot{x}^2+x\ddot{x})$ Because the wire is smooth, the force is just the normal force, which is perpendicular to the curve, so that $F_x/F_y= -dy/dx=-2x$ and $F_x=-2xF_y$ $F_x^2+F_y^2=(4x^2+1)F_y^2$ $|F_w|=\sqrt{F_x^2+F_y^2} =\sqrt{4x^2+1}( 10+ 2(\dot{x}^2+\ddot{x}))$

We want $\int_0^{t_f} |F_w|dt =\int_0^{\frac12\sqrt5}\frac{|F_w|}{\dot x}dx$

At this point I saw no way to do the differential equation and integral, so I resorted to programming, (which in effect is a numerical integration). I used some of the results above (also for checking the sanity of the output). The last lines of the output are shown below the code.

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import math

def WireReactionForce():
    step=0;
    x=0;
    y=0;
    vx=5;
    vy=0;
    v2=25;     # square of velocity
    t=0;
    dt=0.00001;
    Integral=0;
    while v2>0:
        step+=1;     # just used to print once in a while
        oldvx=vx;
        oldvy=vy;
        # use dy/dx =2x to find vx and vy from v^2
        vx=math.sqrt(v2/(1+4*x**2));
        vy=math.sqrt(v2*(4*x**2)/(1+4*x**2));
        ax=(vx-oldvx)/dt;      # acceleration in x and y direction
        ay=(vy-oldvy)/dt;
        F=math.sqrt(ax**2+(ay+10)**2);
        Integral+=F*dt;
        t+=dt;
        x+=dt*vx;
        y=x**2;
        v2=25-20*y;
        if (step==100 or v2<0):
            print("t={:5.3f},x={:5.3f},y={:5.3f},vx={:5.3f},vy={:5.3f},ax={:5.3f},ay={:5.3f}".format(t,x,y,vx,vy,ax,ay));
            step=0;
    print ("∫|F|dt={:5.3f}".format(Integral));
WireReactionForce();

The last lines of the output were:

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t=0.633,x=1.118,y=1.250,vx=0.000,vy=0.000,ax=-4.884,ay=-10.921
∫|F|dt=8.603