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Cranking it Up


The purpose of this lab is to give you an idea of how derivatives can be used to help analyze mechanical systems. A simplified model of the slider crank mechanism used in piston engines is investigated.


Before we talk about the slider crank, we need to review uniform circular motion. By uniform circular motion, we mean the motion of a point $P$ on a circle when the circle is revolving about its center at a constant angular speed. If we assume a circle of radius $r$ is centered at the origin and $\theta$ is the angle between the positive $x$ axis and the radius from the center of the circle to $P$, then the coordinates of $P$ are $x = r \cos(\theta)$ and $y= r \sin(\theta)$. Uniform circular motion means that $\theta$ satisfies $\theta(t) = \theta_0 + \omega t$, where $\theta_0$ is the angle at $t=0$. The parameter $\omega$ is called the angular velocity, because its value determines how fast the circle is revolving. The period $T$ is the length of time it takes the point $P$ to go around once, and can be obtained from the relation below.

\begin{displaymath}T = \frac{2 \pi}{\omega} \end{displaymath}

Another way to measure how fast the circle is revolving is the frequency $\nu$ or number of revolutions per unit time. Typical units are revolutions per minute or revolutions per second. The equation below shows how the frequency is related to $T$ and $\omega$.

\begin{displaymath}\nu = \frac{1}{T} = \frac{\omega}{2 \pi} \end{displaymath}

As described in Prof. Norton's text Design of Machinery, the slider crank mechanism is used in most internal combustion engines. A schematic diagram of this mechanism appears below. The circle represents the crankshaft and the line from point $P$ to point $Q$ is the connecting rod which links the crankshaft to the piston. Burning of fuel in the combustion chamber above the piston forces the piston down, and the vertical movement of the connecting rod is converted to rotational motion of the crankshaft.


In class, we derived an equation that relates the $y$ coordinate of the point $Q$ to the angle $\theta$ that a radius through the point $P$ makes with the positive $x$ axis. This equation appears below.

\begin{displaymath}y = r \sin(\theta) + \sqrt{L^2-r^2 \cos^2(\theta)} \end{displaymath}

If the crank is rotating at an angular speed $\omega$ then this equation gives the $y$ coordinate of the point $Q$ as

\begin{displaymath}y(t) = r \sin(\omega t) + \sqrt{L^2-r^2 \cos^2(\omega t)} \end{displaymath}

The first and second derivatives of this function then give the velocity and acceleration of the piston.

In engine design, the speed and acceleration of the pistons are two of the factors which determine the stress on certain parts of the engine. For example, a larger crankshaft radius can increase the torque the engine produces, but can also lead to more stress on the crankshaft, piston and the connecting rod. In the exercises, you will use Maple to help you investigate how changes in the parameters $r$, $L$, and $\omega$ affect the piston velocity and acceleration.

The command below shows how to define the function $y(t)$ in Maple.

> y := t -> r/12*sin(omega*t) + sqrt(L^2-r^2*cos(omega*t)^2)/12;
Here the units of $r$ and $L$ are in inches and $t$ is in seconds. The mysterious factor of $12$ just converts the units for $y$ into feet. We will want to plot this function and its derivatives for several parameter values and the following commands show a convenient way to do this, by defining sets of parameter values, and using the subs command. When you want to use more than one set of parameter values, this is a good way to do it because it ensures that the correct values are used.
> par_set1 := {r=1,omega=80*Pi,L=5};
> plot(subs(par_set1,y(t)),t=0..0.1);
> plot(subs(par_set1,diff(y(t),t)),t=0..0.1);
> plot(subs(par_set1,diff(y(t),t,t)),t=0..0.1);
From the plots, it is easy to approximate the minimum and maximum values of the position, velocity, and acceleration by using the mouse. If you don't know how to do this, ask your IA to explain.


  1. Suppose that a circle of radius 6 inches is revolving about its center at a frequency of $300$ revolutions per minute. Find the value in seconds of the period $T$ and the angular frequency $\omega$ in radians per second. Then plot graphs of the position in feet, velocity in feet per second, and acceleration in feet per second squared of the $x$ coordinate of a point $P$ on the circle that starts at $\theta =
0$ at $t=0$.
  2. In the example in the background, $\omega = 80 \pi$ radians per second. Convert this value to revolutions per minute. Is this a reasonable value for an internal combustion engine?
  3. Find approximate values for the maximum and minimum values for the position, velocity, and acceleration of point $Q$ for the example in the background. The easiest way to do this is to use the mouse.
  4. The maximum and minimum values of the velocity of the point $P$ on the crankshaft are $\pm r \omega$ and the maximum and minimum values of the acceleration of the point $P$ are $\pm r
\omega^2$. Compute these values for the example in units of feet per second for the velocity and feet per second squared for the acceleration. Compare them to the approximate values you obtained in the previous exercise.
  5. Plot the position, velocity, and acceleration of the piston for the parameter set $r=2$, $\omega = 80 \pi$ and $L=5$. Compare the maximum and minimum values for position, velocity, and acceleration to the ones in the example in the background. Is the motion of the piston simple harmonic motion? Why or why not?

next up previous
Next: About this document ... Up: lab_template Previous: lab_template
William W. Farr