Pendulum with Oscillating Force

@Karan Chatrath has given a very interesting solution showing the Cartesian version of Newton's Second Law. The code below shows the polar version. Specifially:

$\vec{\tau} = \vec{r} \times \vec{F} \\ |\vec{\tau}| = I \ddot{\theta}$

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import math

deg = math.pi/180.0

dt = 10.0**(-6.0)

m = 1.0
L = 1.0

I = m*(L**2.0)

##########################

t = 0.0
count = 0

theta = 0.0
thetad = 0.0
thetadd = 0.0

while t <= 6.0:

    theta = theta + thetad*dt
    thetad = thetad + thetadd*dt

    x = L*math.cos(theta)
    y = L*math.sin(theta)

    Fx = 0.1*math.cos(0.2*math.pi*t)
    Fy = 0.1*math.sin(0.2*math.pi*t)

    T = x*Fy - y*Fx

    thetadd = T/I

    t = t + dt
    count = count + 1

    #if count % 1000 == 0:
        #print t,(theta/deg)

##########################

print ""
print ""
print dt
print t
print (theta/deg)

#1e-05
#6.00000999994
#71.5404169143
#>>> ================================ RESTART ================================
#>>> 

#1e-06
#6.00000000048
#71.5406387753
#>>> 

The position vector of the particle, and hence its acceleration, is $\mathbf{r} \; = \; \binom{\cos\theta}{\sin\theta} \hspace{2cm} \ddot{\mathbf{r}} \; = \; \binom{-\sin\theta}{\cos\theta}\ddot{\theta} - \binom{\cos\theta}{\sin\theta}\dot{\theta}^2$ so we have the equation of motion $\binom{-\sin\theta}{\cos\theta}\ddot{\theta} - \binom{\cos\theta}{\sin\theta}\dot{\theta}^2 \; = \; \tfrac{1}{10}\binom{\cos\frac15\pi t}{\sin\frac15\pi t} - T\binom{\cos\theta}{\sin\theta}$ where $T$ is the magnitude of the tension in the rod, and hence $\ddot{\theta} \; =\; -\tfrac{1}{10}\sin(\theta - \tfrac15\pi t)$ If we write $\varphi = \theta - \tfrac15\pi t$ , we see that $\ddot{\varphi} \; = \; -\tfrac{1}{10}\sin\varphi \hspace{2cm} \varphi(0) = 0\,,\,\dot{\varphi}(0) = -\tfrac15\pi$ and hence $\tfrac12\dot{\varphi}^2 \; = \; \tfrac{1}{50}\pi^2 + \tfrac{1}{10}(\cos\varphi - 1) \; = \; \tfrac{1}{50}\big[\pi^2 - 10\sin^2\tfrac12\varphi\big]$ Thus $\varphi$ oscillates between $\Phi = 2\sin^{-1}\big(\tfrac{\pi}{\sqrt{10}}\big)$ and $-\Phi$ with period $T \; =\; 2\int_{-\Phi}^\Phi \frac{5}{\sqrt{\pi^2 - 10\sin^2\frac12u}}\,du \; = \; 20\int_{-\frac12\pi}^{\frac12\pi} \frac{d\psi}{\sqrt{10 - \pi^2\sin^2\psi}} \; =\; 4\sqrt{10}\mathsf{K}\big(\tfrac{1}{10}\pi^2\big)$ where $\mathsf{K}$ is the complete elliptic integral of the first kind.

For $0 < t < \tfrac14T \approx 11.2722$ we see that $\varphi$ varies between $0$ and $-\Phi$ , and we have $t \; = \; t(\varphi) \; = \; \int_{\varphi}^0 \frac{5\,du}{\sqrt{\pi^2 - 10\sin^2\frac12u}}$ We can solve this equation numerically to deduce that $\varphi = -2.52129056$ when $t=6$ , and so $\theta = \tfrac65\pi - 2.52129056$ , which is $\boxed{71.5407}^\circ$ , when $t=6$ .

Since $\theta = \tfrac15\pi t + \varphi$ , this is not an unstable oscillation, but simply superposition of the oscillation of $\varphi$ with the constant angular velocity $\tfrac15\pi$ imposed by the driving force.

The coordinates of the particle are:

$(x,y) = (\cos{\theta},\sin{\theta})$

Where $\theta$ the rod makes with the +X axis. The velocity and acceleration components are:

$(\dot{x},\dot{y}) = (-\sin{\theta} \ \dot{\theta},\cos{\theta} \ \dot{\theta})$ $(\ddot{x},\ddot{y}) = (a_x,a_y)=(-\sin{\theta} \ \ddot{\theta} - \cos{\theta} \ \dot{\theta}^2,\cos{\theta} \ \ddot{\theta}- \sin{\theta} \ \dot{\theta}^2)$

Let the tension in the rod be $T$ directed along the rod and towards the origin. Applying Newton's second law along the X and Y directions gives:

$a_x = F_x - T\cos{\theta}$ $a_y = F_y - T\sin{\theta}$

Plugging in all values and re-arranging everything in a matrix form gives the set of linear equations in $\ddot{\theta}$ and $T$ as follows:

$\left[\begin{matrix} -\sin{\theta}&\cos{\theta}\\\cos{\theta}&\sin{\theta}\end{matrix}\right]\left[\begin{matrix} \ddot{\theta}\\T\end{matrix}\right]=\left[\begin{matrix} F_x+\dot{\theta}^2\cos{\theta}\\F_y+\dot{\theta}^2\sin{\theta}\end{matrix}\right]$

$\implies \left[\begin{matrix} \ddot{\theta}\\T\end{matrix}\right] = \left[\begin{matrix} -\sin{\theta}&\cos{\theta}\\\cos{\theta}&\sin{\theta}\end{matrix}\right]^{-1} \left[\begin{matrix} F_x+\dot{\theta}^2\cos{\theta}\\F_y+\dot{\theta}^2\sin{\theta}\end{matrix}\right]$

A little bit of inspection allows us to see that the matrix to be inverted is the inverse of itself. So it follows that:

$\implies \left[\begin{matrix} \ddot{\theta}\\T\end{matrix}\right] = \left[\begin{matrix} -\sin{\theta}&\cos{\theta}\\\cos{\theta}&\sin{\theta}\end{matrix}\right] \left[\begin{matrix} F_x+\dot{\theta}^2\cos{\theta}\\F_y+\dot{\theta}^2\sin{\theta}\end{matrix}\right]$

Using this, an expression for $\ddot{\theta}$ can be conveniently derived. The rest involves numerical integration: $\ddot{\theta} = -\sin{\theta}(F_x+\dot{\theta}^2\cos{\theta}) + \cos{\theta} (F_y+\dot{\theta}^2\sin{\theta})$ $\implies\ddot{\theta} = -F_x \sin{\theta} + F_y \cos{\theta}$ $\implies \boxed{\ddot{\theta} = 0.1 \sin(0.2 \pi t - \theta)}$

$\theta(6) \approx 71.54^{\circ}$

The system in consideration is clearly unstable.