Sliding block on a circular track with friction II

Let the coordinates of the block be:

$x = -R\cos{\phi}$ $y = -R\sin{\phi}$

Computing time derivatives:

$\dot{x} = R\dot{\phi}\sin{\phi}$ $\dot{y} = -R\dot{\phi}\cos{\phi}$

Therefore, the speed of the block is:

$v = \sqrt{\dot{x}^2 + \dot{y}^2} = R\dot{\phi}$

$\ddot{x} = R\ddot{\phi}\sin{\phi} + R\dot{\phi}^2\cos{\phi}$ $\ddot{y} = -R\ddot{\phi}\cos{\phi} + R\dot{\phi}^2\sin{\phi}$

Let the normal force acting on the block due to contact with the circular surface be $N$ at point $P$ . The friction force acting on the block is $f = \mu N$ acting tangentially to the circular surface against the block's motion.

The equations of motion along X and Y can be computed as such:

$m\ddot{x} = N\cos{\phi} - \mu N \sin{\phi}$ $m\ddot{y} = N\sin{\phi} + \mu N \cos{\phi}-mg$

Replacing the double derivative terms and eliminating $N$ from the above two equations after some simplification gives the equation of motion:

$\ddot{\phi} + \mu \dot{\phi}^2 = \frac{g}{R}\left(\cos{\phi}-\mu \sin{\phi}\right) = f(\phi)$

Since the block starts from rest, $\phi(0) = \dot{\phi}(0) = 0$ .

The differential equation:

$\ddot{\phi} + \mu \dot{\phi}^2 = f(\phi)$ $\dot{\phi}\frac{d\dot{\phi}}{d\phi}+ \mu \dot{\phi}^2 = f(\phi)$ $2\dot{\phi}\frac{d\dot{\phi}}{d\phi} + 2\mu \dot{\phi}^2 = 2f(\phi)$

Multiplying both sides by $e^{2\mu\phi}$ gives:

$e^{2\mu\phi}2\dot{\phi}\frac{d\dot{\phi}}{d\phi} + 2\mu e^{2\mu\phi} \dot{\phi}^2 = 2f(\phi)e^{2\mu\phi}$

$\frac{d}{d\phi}\left(\dot{\phi}^2 e^{2\mu\phi}\right) = 2f(\phi)e^{2\mu\phi}$

Now, separating and integrating (and applying initial conditions) gives:

$\dot{\phi}^2 e^{2\mu\phi} = \int_{0}^{\phi} 2f(x)e^{2\mu x}dx$

Where $f(x) = \frac{g}{R}\left(\cos{x}-\mu \sin{x}\right)$

Solving this tedious integral and simplifying gives:

$\dot{\phi} = \left(\frac{2g}{R}\left(-\frac{\left(2{\mu}^2-1\right)\sin\left({\phi}\right)-3{\mu}\cos\left({\phi}\right)}{4{\mu}^2+1}-\frac{3{\mu}\mathrm{e}^{-2{\mu}{\phi}}}{4{\mu}^2+1}\right)\right)^{1/2}$

Now, given that $R =1$ the speed of the block $v = \dot{\phi}$ . For this to be zero at $\phi = \pi/2$ , the above expression on the right hand side must be evaluated at $\phi = \pi/2$ and equated to zero. On simplification, this given the following non-linear equation:

$\boxed{\frac{3\mu}{1 - 2 \mu^2} = e^{\pi \mu}}$

This equation can be solved numerically to obtain the solution of approximately:

$\boxed{\mu \approx 0.603}$

This one requires only a slight modification of the previous code. Vary the friction coefficient over multiple trials. Run the simulation until the speed of the block goes to zero. When the speed goes to zero, check to see the fraction of the quarter arc that the block has slid. The desired friction coefficient corresponds to the block sliding a full quarter arc before stopping. My simulation says that this occurs somewhere between $\mu = 0.603$ and $\mu = 0.604$ .

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import math

m = 1.0
R = 1.0
g = 9.8
u = 0.604

dt = 10.0**(-6.0)

xf = (math.pi/2.0)*R

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

t = 0.0

x = 0.0
v = 0.0
a = 0.0

while v >= 0.0:

    x = x + v*dt
    v = v + a*dt

    theta = x/R

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

    F = m*g*math.cos(theta) - u*N

    a = F/m

    t = t + dt

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

print dt
print u
print (x/xf)

#1e-06
#0.603
#1.00029688593

#1e-06
#0.604
#0.999433679088

Applying Newton 2nd law to this problems, yields an interesting equation!... When the velocity of the block increases as it goes down, the normal force exerted by the surface increases (since it depends on $v^{2}$ ), hence the friction force on the block increases as well, reducing thus its velocity!... I solved numerically the equation for its velocity as function of the angle $\phi$ for a given $\mu_k$ , assuming its tangential acceleration is constant within a very small interval $\Delta t$ . Then I varied by trail and error the coefficient of friction until its final velocity at B is zero.