Part 1 -- Equations of motion       Due Monday, April 12, 1999

We will be investigating the motion of the system described in problem 10.32 in the textbook. The first part of this assignment is to complete problem 10.32. Following are some hints to help you complete part 1:
 

• Because of the reaction forces and the fairly complex kinematics of the problem, it is much easier to find the equation of motion given in part (a) using conservation of energy rather than Newton's second law. The main idea is to write an expression for the kinetic and potential energies in terms of the angle theta, then write out the equation, T + V = constant, and take the derivative of that equation with respect to time. The resulting expression will be the equation of motion given in part (a). See example 10.2 for a similar problem.
• For the expression of T, the kinetic energy, remember that you will have rotational and translational energy for both the bar and the disk.
• For the expression of V, the potential energy, you should have the height of the center of mass of the bar, and the energy stored in the spring. Notice that the spring is unstretched when theta is zero. Confirm to yourself that the stretch of the spring is given by L(1-cos(theta)).
• Set up your T + V = constant equation. So far, this expression should involve theta, theta_dot, velocity of the center of mass of the bar, velocity of the center of mass of the disk, and theta_dot of the disk. We want to relate these last three quantities to theta_dot itself, so that the entire expression is in terms of theta and its derivatives. The easiest way to do this is using instant centers.
• Locate the instant center of the bar. Now locate the instant center of the disk. What is the distance between the instant center of the bar and the center of mass of the bar (hint: it is a very simple expression which does not depend on theta)? Write an expression for the velocity of the center of mass of the bar involving that distance and theta_dot (of the bar). Now write an expression for the velocity of the end of the bar which is attached to the disk in terms of the instant center of the bar, and in terms of the instant center of the disk. Equate these two expressions. This gives you the relation between theta_dot of the bar and theta_dot of the disk.
• Substitute your kinematic relations into your conservation of energy equation. Now it should only involve theta, theta_dot, and some constants. Now take the derivative of the equation with respect to time. Don't forget to use the chain rule. This should result in the expression given in part (a).
• Now to tackle part (b): start by finding theta_e, the equilibrium angle of the system. (Remember how to do this? Set all of the derivatives of theta to zero, then solve for theta_e.)
• Now define a new variable theta_tilde = theta - theta_e. This is the deviation of the system from its equilibrium position. We want to write the equation of motion in part (a) in terms of the new variable theta_tilde. Do this by straight substitution; i.e., wherever you see theta, replace it with theta_tilde + theta_e.
• Now you will need to use trig identities to expand terms like cos(theta_tilde + theta_e). For example,

• After you have expanded all of the trig expresssions, use the fact that we are looking at small deviations of the angle from equilibrium. In other words, theta_tilde << 1, which means sin(theta_tilde) --> theta_tilde, and cos(theta_tilde) --> 1. Also, because theta_tilde is small, any nonlinear terms in theta_tilde can be neglected. Nonlinear terms are any terms which multiply theta_tilde with itself or any of its derivatives. So , for example, a term like theta_tilde^2 can be neglected. The process described above is called linearization of the equation of motion. We do it because it greatly simplifies the equation of motion and is valid for a range of angles around the equilibrium angle.
• After all of these simplifications have been made, you should end up with the expression in part (b) of problem 10.32.

You should have already derived the two equations of motion for the bar-disk-spring system in homework problem 10.32. We will be using MATLAB to compare the solution to part (a) (the exact solution), and the solution to part (b) (the approximate solution at angles close to the equilibrium angle).

To start, show that the equilibrium angle theta_e is 0.5358 rad or 30.70^o for the following parameter values: m = 2 kg, m_d = 4 kg, l = 1 m, k = 70 N/m.

Exact solution

Using MATLAB, write a program which will plot theta and theta_dot as a function of time. You should write a function file and a main program as you did in CA#1 and use ODE45. Since this is only a one-degree-of-freedom problem, assign the state variables as:

u(1) = theta
u(2) = theta_dot

Thus, your resulting u matrix will have only 2 columns instead of 4 columns as it did in CA#1. Use the parameter values given above.

Approximate solution

The approximate solution, given in part (b) of problem 10.32, is valid for angles close to the equilibrium angle. This equation of motion is simply a single-degree-of-freedom oscillator with no damping and no forcing. It is given in terms of theta_tilde, the difference between theta and the equilibrium angle theta_e. Write the expression of theta as a function of time if the initial conditions are

theta(0) = theta_0
theta_dot(0) = 0

(Remember, you should use the general solution

theta_tilde = A*sin(wn*t) + B*cos(wn*t)

and solve for A and B using the initial conditions).

Remember,

theta_tilde = theta - theta_e

so

theta = theta_tilde + theta_e

Add a few lines to your MATLAB code which will plot the approximate solution and its derivative with respect to time (i.e., theta and theta_dot). Do not use ODE45 to do this. You already have an expression for theta and you can find one for theta_dot  -- just use MATLAB to plot them.

Plots

Use your MATLAB program to plot the exact and approximate solutions of theta and theta_dot from 0 to 10 seconds in 0.05 second intervals. Overlay the two solutions of theta on the same plot. Do the same for the theta_dot solutions. Print three sets of plots--one for each of the following initial conditions:

1. theta(0) = 15^o (don't forget to convert to radians), theta_dot(0) = 0
2. theta(0) = 30^o, theta_dot(0) = 0
3. theta(0) = 45^o, theta_dot(0) = 0

1. For which set(s) of initial conditions do the exact and approximate solutions match closely? Why?
2. For which set(s) of initial conditions do the exact and approximate solutions differ significantly? Why?
3. What are the advantages and disadvantages of the exact solution?
4. What are the advantages and disadvantages of the approximate solution?
5. As an engineer, you will be asked to make judgements about the range of validity of approximate solutions. Suppose your boss wants you to use the approximate solution to make calculations for a physical system. For what range of initial angles would you say the approximate solution is valid? Justify your reasoning (simply exercising your program by trying a few different initial angles is fine).