Bead on Circular Wire: Related Rates Reloaded

The differential equation of the particle is $m\binom{\sin\theta}{-\cos\theta} \ddot{\theta} + m\binom{\cos\theta}{\sin\theta}\dot{\theta}^2 \; = \; N\binom{\cos\theta}{\sin\theta} + \mu N\binom{-\sin\theta}{\cos\theta} + mg\binom{0}{-1}$ where $N$ is the magnitude of the normal reaction. Thus $m\ddot{\theta} \; = \; -\mu N + mg\cos\theta \hspace{2cm} m\dot{\theta}^2 \; = \; N - mg\sin\theta$ and hence, eliminating $N$ , $\begin{aligned} \ddot{\theta} + \mu\dot{\theta}^2 & = \; g(\cos\theta - \mu\sin\theta)\\ \frac{d}{d\theta}\left[\tfrac12\dot{\theta}^2e^{2\mu\theta}\right] & = \; g(\cos\theta - \mu\sin\theta)e^{2\mu\theta} \\ \tfrac12\dot{\theta}^2e^{2\mu\theta} & = \; \frac{g}{4\mu^2+1}\left[3\mu e^{2\mu\theta}\cos\theta + (1 - 2\mu^2)e^{2\mu\theta}\sin\theta -3\mu\right] + \tfrac18\\ \dot{\theta}^2 & = \; \frac{2g}{4\mu^2+1}\left[3\mu\cos\theta + (1 - 2\mu^2)\sin\theta - 3\mu e^{-2\mu\theta}\right] + \tfrac14e^{-2\mu\theta} \hspace{2cm} (\star) \end{aligned}$ With $g=10$ and $\mu=\tfrac12$ we want to find the value of $\theta$ attained at time $t=\tfrac12$ . Thus we need to solve the equation $\tfrac12 \; = \; \int_0^\theta \frac{2 du}{\sqrt{60\cos u + 20\sin u - 59e^{-u}}}$ and this can be done numerically, obtaining $\theta = \theta_0 = 1.0537654$ . We can now calculate $\dot{\theta}(\theta_0)$ and $\ddot{\theta}(\theta_0)$ , using $(\star)$ and the fact that $\ddot{\theta} = 10\cos\theta - 5\sin\theta - \tfrac12\dot{\theta}^2$ , obtaining $A \; = \; \frac{\ddot{\theta}(\theta_0)}{\dot{\theta}(\theta_0)} \; = \; -1.05447$ making the required answer equal to $\boxed{1054}$ .

This was a fun problem. Here are my two cents. Let $(x,v,a)$ be the tangential position, velocity, and acceleration, respectively.

$\large{A = \frac{d \dot{\theta}}{d \theta} = \frac{d \dot{\theta} / dt}{d \theta / dt} = \frac{\ddot{\theta}}{\dot{\theta}} = \frac{a}{v}}$

The rest is just number crunching. Code below:

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import math

R = 1.0   # Radius
m = 1.0   # Mass
u = 0.5   # Friction coefficient
g = 10.0  # Gravity

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

t = 0.0            # Time
dt = 10.0**(-6.0)  # Time step

x = 0.0   # Tangential position
v = 0.5   # Tangential velocity
a = 0.0   # Tangential acceleration

while t <= 0.5:         

    x = x + v * dt    # Euler integration
    v = v + a * dt

    theta = x/R       # Angle

    N = (m/R)*(v**2.0) + m*g*math.sin(theta)  # Normal force

    F = m*g*math.cos(theta) - u*N  # Force in tangential direction

    a = F/m   # Tangential acceleration

    t = t + dt

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

A = a/v
q1 = math.fabs(1000.0*A)
q2 = math.floor(q1)

print q2  # Answer is 1054

I will lay out a few brief steps for the approach to this question. I encourage the solvers to post more detailed solutions.

Using the laws of motion, the equation of motion is: $\ddot{\theta} = 10\cos(\theta) - 5\sin(\theta) - \frac{\dot{\theta}^2}{2}$

Now comes the important step: $\frac{d^2\theta}{dt^2} = \frac{d\dot{\theta}}{dt} = \frac{d\dot{\theta}}{d\theta}\frac{d\theta}{dt} = \frac{d\dot{\theta}}{d\theta} \dot{\theta}$

Replacing this in the EOM, we get: $\frac{d\dot{\theta}}{d\theta} \dot{\theta} = 10\cos(\theta) - 5\sin(\theta) - \frac{\dot{\theta}^2}{2}$

This differential equation can be solved to obtain a closed form solution where $\dot{\theta}$ is a function of $\theta$ . Solving for $\theta$ as a function of time would require numerical approaches. Recall that the initial conditions are:

$\dot{\theta}(0) = 0.5$ and $\theta(0) = 0$

Looking forward to seeing more elaborate solutions which demonstrate how to obtain the equation of motion and its solution.