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 *
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]:
In [5]:
eq2 = Eq(T - m * g, m * a)
Out[5]:
In [6]:
eq3 = Eq(T * r, I * alpha)
Out[6]:
In [7]:
soln = solve([eq1, eq2, eq3], [T, a, alpha])
Out[7]:
In [8]:
soln[T]
Out[8]:
In [9]:
soln[a]
Out[9]:
In [10]:
soln[alpha]
Out[10]:
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]:
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]:
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]:
In [15]:
system.init
Out[15]:
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]:
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]:
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]:
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]:
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 [ ]: