Magnetism Series #4

Here is my attempt. Consider a coordinate system placed at the point at which the particle just starts moving along the incline. The X-axis faces rightward and the Y-axis faces upwards.

Firstly, we consider the motion of the particle along the incline. As the particle slides down the incline, the magnetic force acting on it is perpendicular to the incline, so the acceleration of the particle along the downward slope of the incline is:

$a = g \sin{\theta}$ $\implies v = g \sin{\theta} t$

Balancing forces along the perpendicular direction of the incline gives:

$mg \cos{\theta} = N + qvB$

At the instant when the particle loses contact, the above equation becomes:

$mg \cos{\theta} = qvB$ $\implies v = \frac{mg \cos{\theta}}{qB}$

The velocity along the X direction at this instant is:

$v_{ox}= v\cos{\theta} = \frac{mg \cos^2{\theta}}{qB}$ $v_{oy} = -v\sin{\theta} = -\frac{mg \sin{\theta}\cos{\theta}}{qB}$

Now, looking at the instant when the particle loses contact, we have:

$mg \cos{\theta} = qvB$ $\implies mg \cos{\theta} = qBg \sin{\theta} t_o$ $\implies t_o = \frac{m \cot{\theta}}{qB}$

$\implies v = g \sin{\theta} t$

Displacement along the incline:

$s = \frac{g \sin{\theta} t^2}{2}$

At the instant of loss of contact:

$L = \frac{g \sin{\theta} t_o^2}{2}$ $L = \frac{m^2g \cos^2{\theta}}{2q^2B^2\sin{\theta}}$

The X and Y coordinates of the particle at this instant are:

$x_o = L \cos{\theta}$ $y_o = L \sin{\theta}$

Now that the particle has lost contact with the incline, this instant from now is considered to have occurred at time $t=0$ . When the particle loses contact with the incline, Newton's second law can be applied to find the equations of motion:

$m \vec{a} = q(\vec{v} \times \vec{B}) - mg \ \hat{j}$

Simplifying these equations gives the following differential equations:

$m \ddot{x} = -qB\dot{y}$ $m \ddot{y} = qB \dot{x} - mg$ $x(0) =x_o \ ; \ y(0) = y_o$ $\dot{x}(0) = v_{ox} \ ; \ \dot{y}(0)= v_{oy}$

Consider the equation:

$m \ddot{x} = -qB\dot{y}$

Integrating both sides:

$m \dot{x} = -qBy+ C$

Where $C$ is an integration constant. Applying initial conditions and solving for $C$ gives:

$v_x = -\frac{qBy}{m} + \frac{mg \cos^2{\theta}}{2qB}$

Plugging this result into the differential equation for $\ddot{y}$ gives:

$\ddot{y} = -\frac{q^2B^2}{m^2}y + g\left(\frac{\cos^2{\theta}}{2}-1\right)$

Let:

$\omega = \frac{qB}{m}$ $K = g\left(\frac{\cos^2{\theta}}{2}-1\right)$ The general solution is:

$y = A_o \sin{\omega t} + B_o \cos{\omega t} + \frac{K}{\omega^2}$

The unknown constants $A_o$ and $B_o$ can be found by applying the initial conditions and solving gives the following:

$A_o = -\frac{m^2g\sin{\theta}\cos{\theta}}{q^2B^2}$ $B_o = \frac{m^2g\sin^2{\theta}}{q^2B^2}$

Now:

$y = A_o \sin{\omega t} + B_o \cos{\omega t} + \frac{K}{\omega^2}$ $y = \sqrt{A_o^2+B_o^2}\sin(\omega t + \phi) + \frac{K}{\omega^2}$

The amplitude:

$\sqrt{A_o^2+B_o^2} = 2L\tan^2{\theta}$

Looking at the diagram, one can see that $h$ is the vertical distance between the maximum and minimum of the cycloid. This distance is basically twice this amplitude. This implies that:

$h = 2\sqrt{A_o^2+B_o^2} = 4L\tan^2{\theta}$ $\implies \boxed{L = \frac{h \cot^2{\theta}}{4}}$

Commented code solution below, along with a plot of the motion:

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import math

# Constants

dt = 10.0**(-6.0)

theta = math.pi/3.0
B = 5.0
m = 1.0
g = 10.0
q = 1.0

cot = math.cos(theta)/math.sin(theta)

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

# When particle leaves ramp, q*v*B = m*g*math.cos(theta)

v = m*g*math.cos(theta)/(q*B)

# Use conservation of energy to determine L
# Magnetic field does no work

# 0.5*m*(v**2.0) = m*g*L*math.sin(theta)

L = 0.5*m*(v**2.0)/(m*g*math.sin(theta))

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

ux = math.cos(theta)   # Unit vector in direction of initial velocity
uy = -math.sin(theta)

ymin = 999999999.0
ymax = -99999999.0

# Initialize time, position, velocity

t = 0.0
count = 0

x = 0.0
y = 0.0

xd = v*ux
yd = v*uy

xdd = 0.0
ydd = 0.0

while t <= 5.0:

    x = x + xd*dt         # Numerical integration
    y = y + yd*dt

    xd = xd + xdd*dt
    yd = yd + ydd*dt

    FBx = q*(-yd*B)     # Magnetic force = q*(v cross B)
    FBy = q*(xd*B)

    Fgx = 0.0        # gravity force
    Fgy = -m*g

    Fx = FBx + Fgx   # Total force
    Fy = FBy + Fgy

    xdd = Fx/m        # acceleration
    ydd = Fy/m

    if y > ymax:     # store min and max y values
        ymax = y

    if y < ymin:
        ymin = y

    t = t + dt
    count = count + 1

    if count % 1000 == 0:
        print t,x,y

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

h = ymax - ymin    # h is difference between min and max y values

#print h

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

# Find the integers corresponding to alpha and beta

print ""
print ""

A_store = 0
B_store = 0

minres = 999999999999.0

for A in range(1,20):
    for B in range(1,20):

        a = float(A)
        b = float(B)

        res = math.fabs(h*(cot**a)/b - L)

        if res < minres:
            minres = res
            A_store = A
            B_store = B

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

print minres
print A_store
print B_store

#3.32539514421e-06
#2
#4
#>>>