None Reading_07

Table of Contents

Challenge

In the next two notebook assignments, our overall goal will be to produce a physics-based model of the truck coast-down experiment you saw in Reading 06.

In Reading 6, you created an empirical model for the velocity of the blue toy truck coasting to a stop. In Reading 5, you did the same for a tank of water emptying through a valve. Data from these two experiments (which was given to you in each relevant reading) are shown in the cell below:

In [1]:
tankdata = load('reading5data.txt');
truckdata = load('reading6data.txt');
figure(1, 'position',[0,0,900,300]);
subplot(1,2,1)
plot(tankdata(:,1),tankdata(:,2),'k')
xlabel('Time (s)')
ylabel('Water Height (m)')
subplot(1,2,2)
plot(truckdata(:,1),truckdata(:,2),'k')
xlabel('Time (s)')
ylabel('Truck Speed (m/s)')

Curiously, while the y-scale and time scale on each plot is wildly different, the shapes of the two plots are the same. This course, which is designed to teach you the systems approach to engineering modeling, should help you understand these two problems by seeing the physical similarity between them. This takes time, and it requires that we have a physical understanding of what is happening in each problem.

For the truck problem, we do have some tools we can use to understand the physics, so understanding the physics behind the truck's coasting behavior is a good first step towards understanding the physics behind both problems.

The empirical model from Reading 6 is great, but it does little to illuminate the physics. Further, an empirical model doesn't tell us how (or whether) the truck would slow down differently if we added weight to its bed, or changed the size of its wheels. Physical or physics-based models are superior to purely empirical ones because they allow us to predict the effects of design changes on the behavior of our system.

Put simply, a physics-based model is better because it generalizes to different model configurations better than an empirical one does.

While Newton's laws are a good starting place for developing a physics-based model of the truck, there are phenomena governing the truck's motion that we have not seen before. Your challenge today is to build the tools you will need to understand the physical phenomena that make the truck's behavior resemble the tank height behavior. We will look at the truck example together, and then you will be asked to develop a model of a new, similar system in today's assignment.

Challenge: Scoping

We will begin as we always do: by "scoping" a model for the truck system, showing the system's physical boundary along with notation indicating what crosses into and out of the system boundary, along with any relevant quantities that are stored. Here, we note that the truck is moving so it stores kinetic energy, and this energy leaves the system as the truck slows down in the form of losses due to friction.

image-3.png

Challenge: Model Construction

Treating the model of the truck this way, we might create an FBD of the truck, realizing that the only external force acting on it that causes any change in momentum is friction. Normal force and gravity balance one another out, and there is no change of momentum (no acceleration) in the y-direction.

image.png

Then, we might write Newton's second law as:

$$\sum \vec{F} = m\vec{\dot{v}} = N\hat{j} - mg\hat{j} - F_f \hat{i}$$

Recognizing that the normal force and gravity will cancel one another, and that $\vec{v}=v\hat{i}$, we can write the x-component of Newton's second law to obtain:

$$m\dot{v} = -F_f$$

Our only model for friction so far is $F_f = \mu N = \mu mg$, and if we made that substitution and solved for the truck's acceleration, $\dot{v}$, we would get:

$$\dot{v} = -\mu g $$

This equation says that acceleration is a constant! This equation, unfortunately, will not describe the behavior of the truck that we measured, because we have observed that the acceleration of the truck is not a constant, and that its magnitude decreases with decreasing truck speed. However, we know there must be some sort of frictional force acting on the truck, or else it would keep rolling forever at a constant velocity, via Newton's 1st law.

To understand what is going on here, it is necessary to ask ourselves: "where is the friction coming from?"

The answer, as you might have guessed, is that the frictional force is a result of the interaction between the truck's axle and the truck's body. In order for the axle to turn, the body of the truck has a bearing installed for each. Bearings come in many shapes and sizes, but nearly all are lubricated interfaces between two parts that either slide or rotate relative to one another. Two types of bearing are shown below: the plain bearing and the ball bearing.

image.png

In our first pass at the model, it seemed that the friction slowing the truck must have come from the ground. There is, indeed, friction between the wheels and the ground, but it is difficult to know what it is, since the wheels are not sliding. They are rolling without slip, and the frictional force applied to the wheels by the ground is just enough at any moment to keep the wheels from sliding. Friction, as it turns out, is only well-modeled by a constant when two dry objects are sliding significantly relative to one another.

In order to understand the truck's bearings and how they produce friction, we need to update our tool set for model construction using Newton's laws to include rotating systems. This is the topic of the next few sections.

Disciplined Process Model Construction Using Newton's Laws on Rotating Systems

Torque or Moment

A torque (also called a 'moment') is applied to an object any time a force is applied at some distance from a reference point. An object to which an imbalanced torque is applied will tend to rotate.

If the distance from the reference point to the application point of the force is defined by a position vector $\vec{r}$, as shown below:

image.png

Then the magnitude of the torque applied with respect to the reference point is the product of the position vector's magnitude and the component of the force vector that is perpendicular to the position vector:

$$T = r\cdot F_\perp$$

Think about the picture above and imagine that you squeezed the object shown with your thumb and forefinger at the reference point. If the force $\vec{F}$ was pointed directly perpendicular to the position vector $\vec{r}$, it would be hard to keep the object from rotating... and harder yet if the position vector got larger! But if the force vector was pointed either directly towards or directly away from your grip, the object would not "want" to rotate at all! This is the intuitive definition of the mathematical equation above, relating $F_\perp$ and $r$.

The definition of the component $F_\perp$ depends on the specific geometry of a particular problem, but if the angle $\theta$ is defined as in the figure above, then we can compute that $F_\perp=F\sin\theta$, where $F$ is the magnitude of the force vector $\vec{F}$.

The direction of the torque is said to be perpendicular to the plane defined by the position and force vectors. Specifically, its direction can be surmised using the "right hand rule," in which one curls the fingers of the right hand from the position vector into the direction of the applied force. The direction of the torque is shown by the direction in which the thumb points. Note that this means that the applied torque will always point into or out of the page when considering planar motion.

image.png

If you've never heard of the right-hand rule, or you need a refresher, check out this video!

This rule for computing the magnitude and direction of a torque vector $\vec{T}$ is a description of what is called the vector cross product. In terms of the vector cross product, torque is written:

$$\vec{T} = \vec{r}\times \vec{F}$$

Many times, the method of computing the direction and magnitude of torque explained above, as the "distance times the perpendicular component of force," is the easiest way. However, in cases where oblique forces are applied at odd distances from a reference point, more formal methods for computing the cross product can be more useful. In ES103, we will stick with the simpler definition.

"Couples" or "Pure Torques (moments)"

We said that a force at some perpendicular distance from a reference point is said to produce a torque about that reference point. However, an object with a single resultant force applied to it at some offset from its center of gravity will tend to both translate and rotate. It is possible to have a situation in which a Pure Torque, also called a couple), is applied. A couple occurs when an object is subjected to two or more forces that sum to zero via Newton's second law, but together produce a net torque. An example is shown below, in which a cylindrical body is subjected to two forces tangent to the cylinder (and thus perpendicular to the radius $r$).

image.png

If the two forces $\vec{F}_1$ and $\vec{F}_2$ have a common magnitude $F$, we could see that their vector sum would be $0$, meaning that the cylinder will not have any translational acceleration. However, these two forces each produce a TORQUE of $F\cdot r$ that, according to the right-hand rule, will be facing into the page and are clockwise torques, meaning that the net torque applied to the cylinder will be $T = 2\cdot Fr$.

In situations like these, we will often just represent the cylinder as being exposed to a known Pure Torque, which is often shown using an arc-like symbol like the one in the figure below that indicates its direction. In planar problems, counter-clockwise torques are positive, and clockwise torques are negative. This is also due to the right-hand rule, which can be used to show us the directions of positive $x,y,z$ axes as shown below.

image-2.png

Newton's Second Law applied to angular momentum for a rigid body rotating about its center of mass in a plane

Newton's second law, a statement about the conservation of momentum, applies to angular as well as linear momentum. For a rigid body moving in a plane, angular momentum is defined as:

$$\vec{H} = J \omega \hat{k}$$

Where $\omega$ is the magnitude of the rigid body's angular velocity (radians/second), and $J$ is the mass moment of inertia of the rigid body. Think of the mass moment of inertia like a "mass for rotation." The larger this number is, the harder it is to accelerate an object in rotation. Notice that the angular velocity $\omega$ points in the $\hat{k}$ direction for an object moving in a plane. This is because of the right-hand rule. When we curl our right-hand fingers in the direction of rotation, our thumb points in the direction of the rotation vector. Some more details about angular velocity can be found here if desired.

Mass moment of inertia is a scalar quantity that is related to how mass is distributed in an object. A list of mass moments of inertia is available here, and the formal definition is given as:

$$J = \int r^2dm$$

Where "dm" represents an infinitesimal piece of mass that makes up the rigid body. It is rare that one needs to calculate a mass moment of inertia using this formula-- tables of mass moments of inertia for most common shapes are widely available, and most modern computer-aided design software can compute mass moments of inertia for parts you design. However, looking at the equation is helpful for understanding what moment of inertia really is. Imagine an object with all of its mass concentrated at a point. We could calculate that mass's moment of inertia with respect to some arbitrary axis as $J = mr^2$. By contrast, a wheel or a cube would have a different moment of inertia, since the mass of those objects is distributed in space.

Angular velocity, which has units of $\frac{rad}{s}$ in the SI system, has a direction either into or out of the plane of motion, according to the right-hand rule. With a standard coordinate system defined on a page as $x$ right and $y$ up, counterclockwise rotation is positive (out of the page), and clockwise rotation is negative (into the page). Newton's second law deals with the derivative of momentum. For a rotational system in a non-accelerating (also called "Newtonian") reference frame, we can write:

$$\sum \vec{T} =\frac{d\vec{H}}{dt}$$

where The angular momentum is computed about a rigid body's center of mass. If moments (torques) are to be summed about an arbitrary point, this formula does not, in general, apply. There are special exceptions to this rule, but we will not treat them here.

As with a translational system, if the rotational inertia of a rigid body is constant, the right-hand side of the above equation can be written as $\frac{d\vec{H}}{dt} = J \dot{\vec{\omega}}$. In making this simplification, and considering only motion in one plane, we can write Newton's second law for a rigid body rotating about its center of mass as:

$$\sum \vec{T} = J \vec{\dot{\omega}}$$

You will often see this equation written as a scalar equation for objects that rotate only in one plane (such as a wheel functioning normally. If you imagine a wheel flying off of a car, wobbling and bouncing down the road, you can see that it might have angular velocity about more than one axis of rotation). This scalar version of Newton's second law comes from the fact that if a body rotates only about one axis, its angular velocity (and thus, the torques that produce it) must all be going either into or out of the plane.

In summary, summing the torques about an object's center of mass to see how its rotational velocity changes over time is analagous to summing the forces applied to an object to see how its velocity will change over time.

Example: Newton's Laws for a rotating plain bearing

You may recall that we needed to develop the concept of torque and angular momentum in order to understand how the bearings in the truck's axles contribute to the truck's overall behavior. Let's analyze a "plain bearing" using Newton's laws for rotating systems. We can assume that there is a "known" pure torque $T_2$ acting on the inner race of the bearing, and another "known" pure torque $T_1$ acting on the outer race. The inner and outer races of the bearing move at different velocities, and they are separated by a thin film of lubricant.

image-4.png

Scoping

Scoping this system is straightforward. The bearing itself will define our system boundary, and we will assume that the bearing can be modeled by two elements: the inner and outer races. Kinetic enery of the inner and outer races is our stored quantity. The known torques on these inner and outer races originate from outside the system boundary, and have the potential to accelerate the bearing's inner or outer races, increasing stored kinetic energy (especially since both are drawn positive). So they are "sources." The friction in the lubrication between the two races always tends to slow the races down, decreasing the system's total kinetic energy, so it is a "sink."

image-2.png

Model Construction

  • Our independent variables will be:
    • the known torques $T_2$ and $T_1$ applied to the inner and outer races, respectively.
    • the moments of inertia $J_1$ and $J_2$ of the outer and inner races, respectively
    • properties of the friction between the inner and outer races (as yet unknown)
  • Our dependent variables will be:
    • $\omega_1$, the angular velocity of the outer race (and its derivatives, if needed)
    • $\omega_2$, the angular velocity of the inner race (and its derivatives, if needed)
  • To relate the independent and dependent variables, we will use Newton's Laws.

To proceed with model construction, we need to know how the lubricated interface between the two races develops friction. Often, we will use a linear model of friction to represent the fact that a pure torque develops on each of the races that is roughly proportional to the difference between the angular velocity of each race:

$T_f = b(\omega_2 - \omega_1)$

This frictional force, like dry Coulomb friction, always opposes the relative motion of a body with respect to the surface that is providing the friction, so this equation assumes that the torque will be positive if $\omega_1>\omega_2$. Depending on how you define your torques, you may need to flip the signs. The "damping constant" $b$ is a property of the bearing's lubricated interface that must be added to our independent variable list.

Our process for relating our independent and dependent variables is to write the Free-Body-Diagram (FBD) for each component, and apply Newton's laws. To do this, we will assume that the outer race is moving faster than the inner race. This will allow us to assume a direction for the frictional torque $T_f$ on each body. If we are wrong, the signs should reverse themselves as long as we're consistent.

image.png

Notice that the frictional torque $T_f$ was drawn in opposite directions on each component. This is because of Newton's third law; if the frictional torque on the outer race is due to the inner race, then Newton's third law states that the inner race must be subjected to an equal and opposite torque by the outer race!!

Applying Newton's Second Law (N2L) for rotation about the outer race's center of mass, assuming the bearing's motion is in plane and therefore all torques and angular velocities are in the $\hat{k}$-direction (out of + or into - the page), we get:

$$\sum T_{outer} = J_{1} \dot{\omega_1} = T_1-T_f$$

Substituting in our linear friction model, we can solve for the outer race's angular acceleration.

$$\dot{\omega_1} = \frac{1}{J_{1}} \left( T_1 - b( \omega_1 - \omega_2)\right)$$

Note that this model now only contains independent and dependent variables, and so it is complete. Examining the units of each term, we find that all are angular accelerations (units of $\frac{rad}{s^2}$). We can also confirm that if $\omega_1>\omega_2$, the friction will be negative as drawn, and the outer race will tend to slow down.

Repeating this process for the inner race yields:

$$\sum T_{2} = J_{2} \dot{\omega_2} = T_2+T_f$$

Substituting in our linear friction model, we can solve for the inner race's angular acceleration.

$$\dot{\omega_2} = \frac{1}{J_{2}} \left(T_2 + b(\omega_1 - \omega_2)\right)$$

Again, all terms have the correct units, and we can see that if $\omega_1>\omega_2$, the outer race is tending to accelerate the inner race. If $\omega_2>\omega_1$, the inner race will tend to slow down as expected.

We also see that what we have developed is a model that is actually TWO differential equations that are coupled to one another; the motion of the inner race depends on the motion of the outer race, and vice versa. At this point in the course, we do not know how to solve a system of two coupled differential equations. Therefore, we cannot perform model evaluation for this example.

Further Simplification of the model

However, it is almost always true that bearings are connected to objects much larger than themselves. On the toy truck, the outer race of each axle's bearing is connected to an object that does not rotate (the chassis)! The inner race is connected to the truck's axle. This simplifies the problem a lot because for the truck, it means that $\omega_1=\dot{\omega}_1 = 0$.

In fact, it is common to further simplify the analysis of bearings when modeling their role in larger systems, because we can often assume that the inner and outer races have negligible moments of inertia $J_1$ and $J_2$.

Making this assumption for our system means that, for the outer race, N2L becomes: $$\require{cancel}$$ $$\sum{T_{outer}} = \cancel{J_1}\dot{\omega}_1 = T_1 - T_f$$ And this implies that: $$T_1 = T_f = b(\omega_1-\omega_2)$$ For the inner race, this means that N2L becomes: $$\sum{T_{inner}} = \cancel{J_2}\dot{\omega}_2 = T_2 + T_f$$ And this implies that: $$T_2 = -T_f = -b(\omega_1 -\omega_2)$$

Taken together, this means that we could really treat the bearing as a single "element" in a larger system, with the following free-body diagram that shows that the external pure torque applied to the bearing must be equal in magnitude and opposite in sign between the inner and outer races, and that this torque will always have a magnitude of $T_b = b(\omega_1-\omega_2)$: image.png

When we make these assumptions, the bearing falls into a special category of element called an "idealized rotational damper," which we formally define below.

Idealized Rotational and Translational Dampers: Force and Torque

Idealized rotational and translational dampers are fictitious single elements in a dynamic system that have insignificant translational and rotational inertia, but that produce a force or torque that resists motion proportionally to the velocity difference between each part of the damper. "True" idealized dampers do not exist, because nothing is massless, and almost nothing has a purely linear relationship between force and velocity (or torque and angular velocity). However, empirical observations have led scientists and engineers to model many "real" system elements as if they were idealized dampers.

For example, rotational bearings and lubricated sliding contacts are often modeled as idealized dampers, as are translational "shock absorbers" such as those found on a car's suspension. Dampers are often indicated on a drawing using a shorthand sketch of their free body diagrams.

image-2.png

Because idealized rotational and translational dampers do not have significant mass, the torque or force (respectively) on each end of the damper must be equal and opposite.

The magnitude of the torque on a rotational damper, as drawn above, is given by:

$$ T_b = b(\omega_1 - \omega_2)$$

Where $b$ is the damper's "damping coefficient." $\omega_1$ represents the angular velocity of one race of the damper, and $\omega_2$ represents the angular velocity of the other race. The damping coefficient $b$ is often experimentally determined, and its units are $\frac{N\cdot m \cdot s}{rad}$ in SI units so that multiplying the coefficient with an angular velocity in $\frac{rad}{s}$ yields a torque in $N\cdot m$. Note that using the right-hand rule, we can confirm that if $\omega_1$ is larger than $\omega_2$, the torque will point "outwards" from each end of the damper.

Similarly, for a translational damper (such as a shock absorber in a car), the equation for the force produced by the damper is given by:

$$ F_b = b(v_1-v_2)$$

The damping coefficient $b$ for a translational damper has units of $\frac{N\cdot s}{m}$ such that multiplying the coefficient by a velocity in $\frac{m}{s}$ produces a force in $N$.

With our new concept of Torque, Angular Momentum, and Idealized Dampers, we can update our disciplined process for model construction using Newton's Laws so that we can apply these concepts to more complex systems.

Disciplined Process: Using newton's laws to obtain a mechanical system model

Step 3 of our disciplined process for Model Construction is to relate our model's independent and dependent variables. If we choose to use Newton's Laws to perform this step, we can create a differential equation to serve as a model for a system comprised of "idealized elements" such as masses and dampers by:

  1. Drawing the FBD for each individual element in the system, indicating whether it meets the requirements for an idealized damper or an idealized mass.
  2. Writing Newton's second law for each individual element in the system, using sums of forces ($\sum \vec{F} = m\vec{a}$) and/or torques ($\sum \vec{T} = J\vec{\dot{\omega}}$) as appropriate given the motion of your system. Be sure to apply Newton's 3rd law to link forces and torques between components (free bodies) in your system.
  3. Eliminate all motion and force terms other than terms multiplied by the system's dependent variables (and their derivatives, as appropriate) using algebraic manipulation. Solving an equation for an unwanted term and substituting into another equation is a good way to eliminate the unwanted term.
  4. Check to make sure that all terms multiplied by your model's dependent variables contain only known or measurable quantities (independent variables).
  5. Check the model for internal consitency by confirming that the units of each term are correct, that the signs match your intuition, and that the equation has the correct overall form.

Assignment

Before we can apply our new concepts to the truck model, we should probably isolate the new skills we've learned to build a model of a simpler system. We will continue with the construction of the physics-based toy truck model in the next reading.

In this assignment, you will use all of the knowledge you have gained so far in this course to analyze the following system, consisting of a pulley attached to a fixed point of rotation via a ball bearing. You are interested in how the pulley's angular velocity evolves over time as it coasts down.

You know the following things for sure about the system's components:

  • The pulley (dark grey circle in the simulation above) has a radius of $0.05m$
  • The pulley is made of solid steel and is $0.01m$ thick. You may need to look up the density of mild steel to use this thickness and the pulley radius to calculate the inertial parameters of the pulley.
  • The pulley is bolted to a wall using a small, lubricated bearing (brass color in the simulation above)

You can compute other parameters you may need based on this information, but not all. In particular, you do not know the bearing's damping constant $b$.

You can use the simulator above to collect a dataset in which the pulley coasts down from a known initial angular velocity. Configure the simulation parameters appropriately using the text boxes in the simulator. When you are ready to collect a dataset, you can check the "Record Data" checkbox in the simulation, and then click "Run Experiment." You can watch the experiment happen, and when it is complete, you will see a text file download automatically from your browser.

Upload this text file to the Reading_07 folder, renaming it something appropriate and descriptive like "pulleydata_r05_3rads.txt." You can then load this datafile in any code cell.

Deliverables: System Scoping

Scope your system below, including explicit delineation of any separate elements you include. Provide your diagrams as hand-drawn sketches or as a www.draw.io diagrams.

YOUR ANSWER HERE

Deliverables: Model Construction

Using what you know, develop a physics-based model for the system's behavior. This model should be symbolic because you want to make sure that the model will generalize to similar systems that may have different mass properties or friction properties. Present your model construction steps and your final differential equation model in the markdown cell below. You can add more markdown cells to capture multiple pages of work using the cell menu, but do not delete the one I provided. Please make sure any pictures you include are right side up.

YOUR ANSWER HERE

Deliverables: Model Evaluation

The first step in evaluating your model is to produce an estimate for the pulley bearing's viscous damping coefficient $b$. Do this by setting up an Euler integration code that will predict the angular velocity of the pulley, and making a "guess" for $b$ in your code. Adjust your guess for $b$ until your model matches the data from your simulation.

In [4]:
% YOUR CODE HERE
error('No Answer Given!')

error: No Answer Given!

Deliverables: Model Evaluation

Simulate what would happen if you doubled the pulley's radius, leaving all other parameters (especially $b$) constant. Collect a new dataset from the simulator and compare your model with the new data. Does the pulley's new behavior match your intuition? Provide plots of your simulation below along with your new dataset for model validation. Provide a short discussion of the agreement between model and data as comments at the bottom of your code cell. Specifically address whether the behavior of this larger pulley matches your intuition and why.

In [5]:
% YOUR CODE HERE
error('No Answer Given!')

error: No Answer Given!

Deliverables: Model Evaluation

Explain, in your own words, why having a physics-based model for the pulley is more useful than having a simple empirical model, ala reading 6.

YOUR ANSWER HERE