Lovely Dynamics of Friction in Semicircle

Let the $x$ and $y$ coordinates of the particle be:

$x = \sin{\theta} \implies \dot{x} = \cos{\theta} \dot{\theta} \implies \ddot{x} = \cos{\theta} \ddot{\theta} - \dot{\theta}^2 \sin{\theta}$ $y = -\cos{\theta} \implies \dot{y} = \sin{\theta} \dot{\theta} \implies \ddot{y} = \sin{\theta} \ddot{\theta} + \dot{\theta}^2\cos{\theta}$

Applying Newton's second law along the X and Y directions gives:

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

Solving for $\ddot{\theta}$ by eliminating $N$ gives after simplification:

$\ddot{\theta} = \mu \dot{\theta}^2 -g \sin{\theta} + \mu g \cos{\theta}$ $\implies \ddot{\theta} = \frac{\dot{\theta}^2}{2} -10 \sin{\theta} + 5\cos{\theta}$ $\implies \ddot{\theta} - \frac{\dot{\theta}^2}{2} = 5\left(\cos{\theta} - 2\sin{\theta}\right)$

$\implies \dot{\theta} \frac{d\dot{\theta}}{d\theta} - \frac{\dot{\theta}^2}{2} = 5\left(\cos{\theta} - 2\sin{\theta}\right)$

Multiplying both sides by $\mathrm{e}^{-\theta}$ and simplifying:

$\implies \frac{d}{d\theta}\left(\mathrm{e}^{-\theta}\frac{\dot{\theta}^2}{2} \right) = 5\mathrm{e}^{-\theta}\left(\cos{\theta} - 2\sin{\theta}\right)$

When $\t = 0$ then $\theta = \pi/2$ and $\dot{\theta}=0$ . Using these initial conditions, separating the variables and integrating gives:

$\implies \mathrm{e}^{-\theta}\frac{\dot{\theta}^2}{2}= \int_{\pi/2}^{\theta}5\mathrm{e}^{-\theta}\left(\cos{\theta} - 2\sin{\theta}\right) \ d\theta$ $\implies \mathrm{e}^{-\theta}\frac{\dot{\theta}^2}{2} = 5\left(\dfrac{\mathrm{e}^{-{\theta}}\left(3\sin\left({\theta}\right)+\cos\left({\theta}\right)\right)}{2}-\dfrac{3\mathrm{e}^{-\frac{{\pi}}{2}}}{2}\right)$ $\dot{\theta} = -\sqrt{5\left(3\sin\left({\theta}\right)+\cos\left({\theta}\right)-3\mathrm{e}^{-\frac{{\pi}}{2}+\theta}\right)}$

The minus sign is introduced since as $t$ increases, $\theta$ decreases. By plotting the RHS of the above equation, one can easily see that it becomes zero at $\theta=\theta_o \approx -0.151428 \ \mathrm{rad}$

Now, separating variables and integrating gives:

$\boxed{T=\int_{\theta_o}^{\pi/2}\frac{d\theta}{ \sqrt{5\left(3\sin\left({\theta}\right)+\cos\left({\theta}\right)-3\mathrm{e}^{-\frac{{\pi}}{2}+\theta}\right)}} \approx 1.02 \ s}$

Hey mate, I just solved your problem. Nice one! Loved it😁👍🏽 There are already two pretty comprehensive solutions out there. I used the computational approach that Steven Chase used. I want to know what method you used @Neeraj Anand Badgujar . Tell me in the comment section.

Firstly, there are 3 forces acting on the object;

The Tangential Gravitational Force

Where $\theta$ is the angle of the slope/tangent line to the circle, the force acting tangentially to the circle (due to gravity) is:

$\displaystyle \bold{F}_{gt} = mg \sin(\theta)$

The value of $\theta$ can be given by finding $\tan ^{-1} ()$ of the slope, given by $\displaystyle \tan^{-1} \left(\frac{x}{\sqrt{1-x^2}} \right)$

The Normal Force

The normal force acting down a ramp dependent on the cosine of the ramp angle. $\displaystyle \bold{N} = mg \cos(\theta) +\frac{m v^2}{R}$

The Frictional Force

The frictional force is $\bold{F}_{friction} = \mu N$ , and is positive or negative depending on the velocity of the object.

Once you carry out these computations and numerically integrate the acceleration, you can get the result. Here's my code:

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import math


time = 0
deltaT = 10**-6 #simulation accuracy


#constants and initial conditions
alpha = 3.1415926/2

x = 1
y = 0

xDot = 0
yDot = 0

m = 1
g = 10

mu = 0.5

velocity = 0



while velocity <= 0:  #simulate till the object comes to stop
    if x == 1:  #handling of exception (division by zero)
        theta = alpha
    else:
        theta = math.atan(x/math.sqrt(1-x**2))


    Fgt = -m*g*math.sin(theta)  #gravity down ramp
    normalForce = m*g*math.cos(theta) + m*(velocity**2) #normal force on ramp


    #friction force is dependent on velocity direction
    if xDot < 0:
        frictionForce = mu*normalForce
    else:
        frictionForce = -mu*normalForce


    totalTangentialForce = Fgt + frictionForce #total force down ramp


    #numerical integration
    acceleration = totalTangentialForce/m 
    velocity += acceleration*deltaT

    #resolving of velocity components
    xDot = velocity*math.cos(theta)
    yDot = velocity*math.sin(theta)

    #distance/position computation
    x += xDot*deltaT
    y += yDot*deltaT

    #progressing simulation
    time += deltaT

print(time)

Nice problem. Simulation code is below, with comments. The block takes $1.02$ seconds to come to a stop, and it travels through an angle of $98.7$ degrees

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import math

# Constants

R = 1.0
u = 0.5
M = 1.0
g = 10.0

deg = math.pi/180.0

dt = 10.0**(-6.0)

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

t = 0.0
theta = 0.0

S = 0.0       # S is linear distance traveled
Sd = 0.0
Sdd = 0.0

while Sd >= 0.0:  # simulate until block stops

    S = S + Sd*dt       # numerical integration
    Sd = Sd + Sdd*dt

    thetad = Sd/R    # angular speed

    theta = theta + thetad*dt  # cumulative angle traversed

    Fc = M*(Sd**2.0)/R  # centripetal force

    # Fc = N - M*g*math.sin(theta)  

    N = Fc + M*g*math.sin(theta)   # normal force

    F = M*g*math.cos(theta) - u*N  # linear accelerating force

    Sdd = F/M  # linear acceleration along circumference

    t = t + dt

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

print dt
print t
print (theta/deg)

#>>> ================================ RESTART ================================
#>>> 
#1e-05
#1.02051
#98.6761041383
#>>> ================================ RESTART ================================
#>>> 
#1e-06
#1.02049700001
#98.6762165773
#>>>