yoyo_soln


Modeling and Simulation in Python

Case study

Copyright 2017 Allen Downey

License: Creative Commons Attribution 4.0 International


In [1]:
# Configure Jupyter so figures appear in the notebook
%matplotlib inline

# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'

# import functions from the modsim.py module
from modsim import *

Yo-yo

Suppose you are holding a yo-yo with a length of string wound around its axle, and you drop it while holding the end of the string stationary. As gravity accelerates the yo-yo downward, tension in the string exerts a force upward. Since this force acts on a point offset from the center of mass, it exerts a torque that causes the yo-yo to spin.

This figure shows the forces on the yo-yo and the resulting torque. The outer shaded area shows the body of the yo-yo. The inner shaded area shows the rolled up string, the radius of which changes as the yo-yo unrolls.

In this model, we can't figure out the linear and angular acceleration independently; we have to solve a system of equations:

$\sum F = m a $

$\sum \tau = I \alpha$

where the summations indicate that we are adding up forces and torques.

As in the previous examples, linear and angular velocity are related because of the way the string unrolls:

$\frac{dy}{dt} = -r \frac{d \theta}{dt} $

In this example, the linear and angular accelerations have opposite sign. As the yo-yo rotates counter-clockwise, $\theta$ increases and $y$, which is the length of the rolled part of the string, decreases.

Taking the derivative of both sides yields a similar relationship between linear and angular acceleration:

$\frac{d^2 y}{dt^2} = -r \frac{d^2 \theta}{dt^2} $

Which we can write more concisely:

$ a = -r \alpha $

This relationship is not a general law of nature; it is specific to scenarios like this where there is rolling without stretching or slipping.

Because of the way we've set up the problem, $y$ actually has two meanings: it represents the length of the rolled string and the height of the yo-yo, which decreases as the yo-yo falls. Similarly, $a$ represents acceleration in the length of the rolled string and the height of the yo-yo.

We can compute the acceleration of the yo-yo by adding up the linear forces:

$\sum F = T - mg = ma $

Where $T$ is positive because the tension force points up, and $mg$ is negative because gravity points down.

Because gravity acts on the center of mass, it creates no torque, so the only torque is due to tension:

$\sum \tau = T r = I \alpha $

Positive (upward) tension yields positive (counter-clockwise) angular acceleration.

Now we have three equations in three unknowns, $T$, $a$, and $\alpha$, with $I$, $m$, $g$, and $r$ as known parameters. It is simple enough to solve these equations by hand, but we can also get SymPy to do it for us.


In [2]:
from sympy import init_printing, symbols, Eq, solve

init_printing()

In [3]:
T, a, alpha, I, m, g, r = symbols('T a alpha I m g r')

In [4]:
eq1 = Eq(a, -r * alpha)


Out[4]:
$\displaystyle a = - \alpha r$

In [5]:
eq2 = Eq(T - m * g, m * a)


Out[5]:
$\displaystyle T - g m = a m$

In [6]:
eq3 = Eq(T * r, I * alpha)


Out[6]:
$\displaystyle T r = I \alpha$

In [7]:
soln = solve([eq1, eq2, eq3], [T, a, alpha])


Out[7]:
$\displaystyle \left\{ T : \frac{I g m}{I + m r^{2}}, \ a : - \frac{g m r^{2}}{I + m r^{2}}, \ \alpha : \frac{g m r}{I + m r^{2}}\right\}$

In [8]:
soln[T]


Out[8]:
$\displaystyle \frac{I g m}{I + m r^{2}}$

In [9]:
soln[a]


Out[9]:
$\displaystyle - \frac{g m r^{2}}{I + m r^{2}}$

In [10]:
soln[alpha]


Out[10]:
$\displaystyle \frac{g m r}{I + m r^{2}}$

The results are

$T = m g I / I^* $

$a = -m g r^2 / I^* $

$\alpha = m g r / I^* $

where $I^*$ is the augmented moment of inertia, $I + m r^2$.

You can also see the derivation of these equations in this video.

To simulate the system, we don't really need $T$; we can plug $a$ and $\alpha$ directly into the slope function.


In [11]:
radian = UNITS.radian
m = UNITS.meter
s = UNITS.second
kg = UNITS.kilogram
N = UNITS.newton


Out[11]:
newton

Exercise: Simulate the descent of a yo-yo. How long does it take to reach the end of the string?

I provide a Params object with the system parameters:

  • Rmin is the radius of the axle. Rmax is the radius of the axle plus rolled string.

  • Rout is the radius of the yo-yo body. mass is the total mass of the yo-yo, ignoring the string.

  • L is the length of the string.

  • g is the acceleration of gravity.


In [12]:
params = Params(Rmin = 8e-3 * m,
                Rmax = 16e-3 * m,
                Rout = 35e-3 * m,
                mass = 50e-3 * kg,
                L = 1 * m,
                g = 9.8 * m / s**2,
                t_end = 1 * s)


Out[12]:
values
Rmin 0.008 meter
Rmax 0.016 meter
Rout 0.035 meter
mass 0.05 kilogram
L 1 meter
g 9.8 meter / second ** 2
t_end 1 second

Here's a make_system function that computes I and k based on the system parameters.

I estimated I by modeling the yo-yo as a solid cylinder with uniform density (see here).

In reality, the distribution of weight in a yo-yo is often designed to achieve desired effects. But we'll keep it simple.


In [13]:
def make_system(params):
    """Make a system object.
    
    params: Params with Rmin, Rmax, Rout, 
                              mass, L, g, t_end
    
    returns: System with init, k, Rmin, Rmax, mass,
                         I, g, ts
    """
    L, mass = params.L, params.mass
    Rout, Rmax, Rmin = params.Rout, params.Rmax, params.Rmin 
    
    init = State(theta = 0 * radian,
                 omega = 0 * radian/s,
                 y = L,
                 v = 0 * m / s)
    
    I = mass * Rout**2 / 2
    k = (Rmax**2 - Rmin**2) / 2 / L / radian    
    
    return System(params, init=init, I=I, k=k)

Testing make_system


In [14]:
system = make_system(params)


Out[14]:
values
Rmin 0.008 meter
Rmax 0.016 meter
Rout 0.035 meter
mass 0.05 kilogram
L 1 meter
g 9.8 meter / second ** 2
t_end 1 second
init theta 0 radian omega 0.0 radi...
I 3.0625000000000006e-05 kilogram * meter ** 2
k 9.6e-05 meter / radian

In [15]:
system.init


Out[15]:
values
theta 0 radian
omega 0.0 radian / second
y 1 meter
v 0.0 meter / second

Write a slope function for this system, using these results from the book:

$ r = \sqrt{2 k y + R_{min}^2} $

$ T = m g I / I^* $

$ a = -m g r^2 / I^* $

$ \alpha = m g r / I^* $

where $I^*$ is the augmented moment of inertia, $I + m r^2$.


In [16]:
# Solution

def slope_func(state, t, system):
    """Computes the derivatives of the state variables.
    
    state: State object with theta, omega, y, v
    t: time
    system: System object with Rmin, k, I, mass
    
    returns: sequence of derivatives
    """
    theta, omega, y, v = state
    g, k, Rmin = system.g, system.k, system.Rmin
    I, mass = system.I, system.mass
    
    r = sqrt(2*k*y + Rmin**2)
    alpha = mass * g * r / (I + mass * r**2)
    a = -r * alpha
        
    return omega, alpha, v, a

Test your slope function with the initial paramss.


In [17]:
# Solution

slope_func(system.init, 0*s, system)


Out[17]:
(0.0 <Unit('radian / second')>,
 180.54116292458264 <Unit('1 / radian ** 0.5 / second ** 2')>,
 0.0 <Unit('meter / second')>,
 -2.888658606793322 <Unit('meter / radian / second ** 2')>)

Write an event function that will stop the simulation when y is 0.


In [18]:
# Solution

def event_func(state, t, system):
    """Stops when y is 0.
    
    state: State object with theta, omega, y, v
    t: time
    system: System object with Rmin, k, I, mass
    
    returns: y
    """
    theta, omega, y, v = state
    return y

Test your event function:


In [19]:
# Solution

event_func(system.init, 0*s, system)


Out[19]:
1 meter

Then run the simulation.


In [20]:
# Solution

results, details = run_ode_solver(system, slope_func, events=event_func, max_step=0.05)
details


Out[20]:
values
sol None
t_events [[0.879217870162702]]
nfev 134
njev 0
nlu 0
status 1
message A termination event occurred.
success True

Check the final state. If things have gone according to plan, the final value of y should be close to 0.


In [21]:
# Solution

results.tail()


Out[21]:
theta omega y v
0.706147 44.0182 121.32 0.327291 -1.76573
0.756147 50.268 128.609 0.236982 -1.84499
0.806147 56.8724 135.496 0.142962 -1.91405
0.856147 63.8093 141.885 0.0457628 -1.97198
0.879218 67.1147 144.631 9.02056e-17 -1.99469

Plot the results.

theta should increase and accelerate.


In [22]:
def plot_theta(results):
    plot(results.theta, color='C0', label='theta')
    decorate(xlabel='Time (s)',
             ylabel='Angle (rad)')
plot_theta(results)


y should decrease and accelerate down.


In [23]:
def plot_y(results):
    plot(results.y, color='C1', label='y')

    decorate(xlabel='Time (s)',
             ylabel='Length (m)')
    
plot_y(results)


Plot velocity as a function of time; is the yo-yo accelerating?


In [24]:
# Solution

v = results.v # m / s
plot(v)
decorate(xlabel='Time (s)',
         ylabel='Velocity (m/s)')


Use gradient to estimate the derivative of v. How does the acceleration of the yo-yo compare to g?


In [25]:
# Solution

a = gradient(v)
plot(a)
decorate(xlabel='Time (s)',
         ylabel='Acceleration (m/$s^2$)')



In [ ]: