Finding relation between velocities of wedge and particle

Applying Conservation of momentum $0 = -Mv' + m(v\sin{\theta} - v')$ $\boxed{v' = \frac{mv\sin{\theta}}{M + m}}$

Finding time

Applying Conservation of Energy

$mgR\sin{\theta} = \frac12Mv'^2 + \frac12m(v\sin{\theta} - v')^2 + \frac12m(v\cos{\theta})^2$

From the relation between $v$ and $v'$

$mgR\sin{\theta} = \frac12M\bigg(\frac{mv\sin{\theta}}{M + m}\bigg)^2 + \frac12m\bigg(\frac{Mv\sin{\theta}}{M + m}\bigg)^2 + \frac12m(v\cos{\theta})^2$

$2gR\sin{\theta} = \frac{mMv^2\sin^2{\theta}}{(M+m)^2}+ \frac{M^2v^2\sin^2{\theta}}{(M+m)^2} + v^2\cos^2{\theta}$

$v = \sqrt{\frac{2gR\sin{\theta}}{\frac{mM\sin^2{\theta}}{(M+m)^2}+ \frac{M^2\sin^2{\theta}}{(M+m)^2} + \cos^2{\theta}}}$

$\frac{Rd\theta}{dt} = \sqrt{\frac{2gR(M+m)\sin{\theta}}{M + m\cos^2{\theta}}}$

$dt = \sqrt{\frac{R}{2g}} \sqrt{\frac{M + m\cos^2{\theta}}{(M+m)\sin{\theta}}}$

$t = \sqrt{\frac{R}{2g}} \int_0^{\pi}\sqrt{\frac{M + m\cos^2{\theta}}{(M+m)\sin{\theta}}}$

Putting $g = 10m/s^2$ , $R = 80m$ , $m = 2\text{ kg and } M = 4\text{ kg}$

$t = \sqrt{\frac{80}{20}} \int_0^{\pi}\sqrt{\frac{2 + \cos^2{\theta}}{3\sin{\theta}}}$

$t \approx 9.865 \text{ sec}$

$\text{Time Period } T = 2t$

$\color{#3D99F6}{\boxed{T \approx 19.73 \text{ sec}}}$

Let the coordinates of the particle be $(x,y)$ , and let the center axis of the wedge be at position $x_0$ . The position and velocity of the particle are:

$x = x_0 + R \cos \theta \\ y = R + R \sin \theta \\ \dot{x} = \dot{x}_0 - R \sin \theta \, \dot{\theta} \\ \dot{y} = R \cos \theta \, \dot{\theta}$

Combined kinetic energy of the particle and the wedge:

$T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) + \frac{1}{2} M \dot{x}_0^2 \\ = \frac{1}{2} m (\dot{x}_0^2 - 2 \dot{x}_0 \dot{\theta} R \sin \theta + R^2 \dot{\theta}^2) + \frac{1}{2} M \dot{x}_0^2$

Gravitational potential energy of the particle:

$V = m g y = m g (R + R \sin \theta)$

Lagrangian:

$L = T - V = \frac{1}{2} m (\dot{x}_0^2 - 2 \dot{x}_0 \dot{\theta} R \sin \theta + R^2 \dot{\theta}^2) + \frac{1}{2} M \dot{x}_0^2 - m g (R + R \sin \theta)$

Equations of Motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}_0}} = \frac{\partial{L}}{\partial{x_0}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

Evaluating results in the following system of equations to be solved for the second derivatives:

$\begin{bmatrix} m+M & -mR \sin \theta \\ -mR \sin \theta & m R^2 \\ \end{bmatrix} \begin{bmatrix} \ddot{x}_0 \\ \ddot{\theta} \\ \end{bmatrix} = \begin{bmatrix} mR \cos \theta \, \dot{\theta}^2 \\ - m g R \cos \theta \\ \end{bmatrix}$

Initialize a numerical solution as follows:

$x_0 (0) = 0 \\ \theta (0) = 0 \\ \dot{x}_0 (0) = 0 \\ \dot{\theta} (0) = 0$

On each time step, solve the linear system for the double-dot terms, and numerically integrate. Plots of $x_0$ and $\theta$ are given below. The period of motion is about $19.73$ . Simulation code is also attached.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71

import math
import numpy as np
import random

# Constants

M = 4.0
m = 2.0
R = 80.0
g = 10.0

dt = 10.0**(-5.0)

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

# Initialize variables
t = 0.0
count = 0

theta = 0.0
x0 = 0.0

thetad = 0.0
x0d = 0.0

thetadd = 0.0
x0dd = 0.0

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

# Numerical integration

max = -9999999.0

while t <= 40.0:

    x0 = x0 + x0d*dt
    theta = theta + thetad*dt

    x0d = x0d + x0dd*dt
    thetad = thetad + thetadd*dt

    J11 = m + M
    J12 = -m*R*math.sin(theta)

    J21 = -m*R*math.sin(theta)
    J22 = m*(R**2.0)

    J =  np.array([[J11,J12],[J21,J22]])
    vec = np.array([m*R*math.cos(theta)*(thetad**2.0),-m*g*R*math.cos(theta)])

    Sol = np.linalg.solve(J, vec)

    x0dd = Sol[0]
    thetadd = Sol[1]

    if (t>15.0) and (t<25.0) and (theta > max):
        max = theta
        t_store = t

    t = t + dt
    count = count + 1

    #if count % 1000 == 0:
        #print t,x0,theta

print dt
print t_store

#1e-05
#19.73016999