Breaking string

This is a tricky problem, because we have to change the system dynamics half way through. But we don't know that yet. To start with, let's assume that the bob remains on a circular trajectory, and see what the implications are. The relevant equations for the angle dynamics, speed, and tension are (where $\theta$ is the angle with the vertical):

$\large{\ddot{\theta} = \frac{-g \, sin \theta}{L} \\ v^2 = L^2 \, \dot{\theta}^2 \\ T = \frac{m v^2}{L} + m g \, cos \theta}$

Numerically integrating and plotting the tension vs time, we get the following. There are two things of note:

1) The tension never gets up to 35 Newtons
2) The tension goes negative, which is non-physical for a string

Therefore, we know that the bob leaves the circular trajectory at the time corresponding to the red circle in the plot. It then free-falls until the string again becomes taut, and presumably breaks due to the jarring motion. I will not actually prove that the tension exceeds 35 Newtons at that point because it's too complicated. The free-fall dynamics are:

$\large{\ddot{x} = 0 \\ \ddot{y} = -g}$

Numerically integrating again, we find that the string becomes taught when $(x,y) = (0,-L)$ . This occurs at approximately $t = 2.96$ seconds.

Note that more of this work could be done by hand, if desired.

Code:

import math

# Parameters and initialization

m = 0.5
L = 5.6
v0 = 14.0

g = 10.0

Tmax = 35.0

theta = 0.0
theta_d = v0 / L
theta_dd = -g * math.sin(theta) / L

T = 0.0

t = 0.0
dt = 10.0**(-6.0)

count = 0

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

# Numerical integration for circular motion portion (before tension goes negative)

while (T >= 0.0):

    theta = theta + theta_d * dt
    theta_d = theta_d + theta_dd * dt

    theta_dd = -g * math.sin(theta) / L

    v = L * theta_d

    T = (m/L)*(v**2.0) + m*g*math.cos(theta)

    #if count % 1000 == 0:      # optional print statement for plotting tension vs time
        #print t,theta,T


    t = t + dt
    count = count + 1

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

# intermediate results at end of circular portion

#print t
#print theta
#print ""
#print ""

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

# Initialization before free-fall portion

x = L * math.sin(theta)
y = -L * math.cos(theta)

xd = L * math.cos(theta) * theta_d
yd = L * math.sin(theta) * theta_d

xdd = 0.0
ydd = -g

r = math.hypot(x,y)

flag = 0

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

# Numerical integration for free-fall portion

while flag == 0:

    x = x + xd * dt
    y = y + yd * dt

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

    xdd = 0.0
    ydd = -g

    r = math.hypot(x,y)

    if (r > 1.0001 * L):
        flag = 1

    t = t + dt

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

# final prints

print "final time"
print t
print "final x"
print x
print "final y"
print y


# Results

final time
2.96279200006
final x
-0.000134522275623
final y
-5.60057124989

While the string is taut, we have conservation of energy $\tfrac12mL^2 \dot{\theta}^2 +mgl(1 -\cos\theta) \; = \; \tfrac12mv^2$ and the equation for central motion $T - mg\cos\theta \; = \; mL\dot{\theta}^2$ where $T$ is the tension in the string and $\theta$ is the angle made by the string with the downward vertical. With the given parameter values, we have $\dot{\theta}^2 \; = \; \tfrac{25}{28}(3 + 4\cos\theta) \hspace{2cm} T \;= \; \tfrac{15}{2}(1 + 2\cos\theta)$ Thus the string does not break, but remains taut, while $0 \le \theta \le \tfrac{2\pi}{3}$ . The time taken until $\theta = \tfrac{2\pi}{3}$ is $t_1 \; = \; \frac{\sqrt{28}}{5}\int_0^{\frac23\pi} \frac{d\theta}{\sqrt{3 + 4\cos\theta}}$ Once the string is slack, the bob moves as a projectile. From the moment the string becomes slack, the position vector of the bob is $\mathbf{r} \; = \; \left(\begin{array}{c} 2.8\sqrt{3} \\ 2.8\end{array} \right) + \sqrt{7}\left(\begin{array}{c} -1 \\ \sqrt{3}\end{array}\right)t + \left(\begin{array}{c} 0 \\ -5 \end{array} \right)t^2$ where $\mathbf{r}$ is measured relative to the point of suspension. The string will break when the bob has to be brought to instantaneous rest when $|\mathbf{r}| = 5.6$ , which occurs when $t = \tfrac{2}{5}\sqrt{21}$ . Thus the total time, in seconds, until the string breaks is $t_1 + \tfrac{2}{5}\sqrt{21} \; = \; \boxed{2.96275}$