Exam No. 3 - posted solution and statistics

Click here for the solution of Exam No. 3. 

Click here for the individual statistics of the three problems on Exam No. 3.

Homework Problem 8/48

For this problem and forced vibrations in the vertical direction in general, should we still include the gravitational component? I wondering this because I don't see how I can group the cosine and sine terms since mg doesn't have either associated with it.

About the Exam 3 Solution

Professor, I noticed on the solution for the first problem on the exam that there was no force due to gravity shown acting on the bar in the Free Body Diagram. Is there supposed to be? I am unclear as to why there wouldn't be one somewhere on there. The way it's shown, it looks like the only force acting in the y direction is the Normal force at A pushing the bar upward, which I assume would imply that it's being lifted off the ground.

Homework Problem No. 8/46

I am having troubles setting up 8/46. My main issues deal with the effect of the arm with the dashpot on the system.

Lecture Example P. 75

For the problem on page 75 in our lecture notes, what is the numerical value for delta s that is used when computing the work for the system? I think its either the ten inches the drum drops or its the amount of cord that gets wrapped around the drum as it drops. Thanks

Solution Video 6/121

Proff.

On this video, the answer that was put down for w doesnt correspond to the numbers used.
I tried to do the calculations, i think there s a little mistake with Ic, actually two.
The way i see it Ic= Mr^2+Mr^2+4( (mr^2/12)+m(r/2)^2+mr^2)
Ic= 2Mr^2+(16/3)mr^2
And this would give the recorded answer for w..w=13.2 rad/s

Please let me know if this is not true.
Thanks

Exam No. 3 - last minute hint

During the help session this evening, I shared a hint related to Exam No. 3 with those present. I would like to share this with the whole class.

The grading of this exam will focus on setting up the problem (FBD, kinetics equations and kinematics). The algebra in solving will not be heavily weighted.

• For Problems 1 and 2, Steps 1-3 will be weighted at 18/20 points. The algebra and arithmetic in solving for the final answer will be weighted at 2/20 points.
• For Problem 3 (vibrations), Steps 1-4 (FBD, Newton, kinematics and EOM) will be weighted at 18/20. Since natural frequency is simple algebra from the EOM, this part will be weighted at 2/20 points.

In order to receive the full 18 points for setting up the problem, you will need to include ALL work. Only algebra/arithmetic should remain to be done in order to receive the 18 points.

Please use this hint in preparing for the exam.

If you have any questions on this, please ask me.

Soln Video 6/8

Why cant you sum the moments about A and set that equal to I(about A) times alpha? That was my first instinct when solving this problem. It is moving but not relative to the bar, which is our FBD. Do we need to look at the whole system when solving the moment equation instead of our FBD?? I think Im a little confused.

VIB 3

In VIB 3, I am not able to follow a certain step. We have the disk moment equation: -k(R*theta)*(R) + T*R = Io*omega_double dot. We then have from the sum of forces in Y for the block that T = m*g - m*y_double dot. combining the two, -k*R^2*theta + (mg - m*y_double dot)*R = 1/2*mR^2*theta_double dot. If you expand ..."(mg - m*y_double dot)*R" it would be "mgR - m*y_double dot*R" correct? Later, in the video solution, I can't find out what happened to the R in the mgR term. That would still carry through to the EOM and make it have an R^2 term on the left and an R term on the right, that could cancel. If you could clear that up for me, I would appreciate it.

Exam No. 3 - Common Access Point for Videos

Feel free to use the following links to get to the video solutions in preparation for the exam. Or, just use the usual links. If you want to download the videos onto your computer and do not know how, just leave a comment.

Sample Exam Solutions: Use scrollbar to locate video; click on image to view video.

Homework Solutions: Use scrollbar to locate video; click on image to view video.

Solution Video Modules: Use mouse to turn pages; click on image to view video.

Exam No. 3 details

Exam No. 3 is scheduled for 6:30-7:30 PM on Tuesday, December 2 in SMTH 108 (same room as for Exams 1 and 2). You can start 10 minutes earlier (at 6:20) if you want extra time for the exam.

The exam will cover Lectures 26-37. There will be three full-length questions (no multiple choice or short calculations questions on this exam) on:

1. Newton-Euler equations
2. Work-energy or impulse-momentum equations
3. Vibrations: EOM and natural frequency

Sample exam questions 28-37 are related to the material covered on this exam. Video solutions for these sample exam questions are provided through the links on the left side of the blog. Each video solution is presented on a blog page on which you can directly view and/or add comments. Please use this to provide assistance to and receive assistance from others in the course.

A review video is provided to assist you in your studying for this exam. Click here for a copy of the equation page for Exam No. 3.

A help session is scheduled for 7PM on Monday, December 1 in Room ME 161. This review session will be run a little differently than for the earlier exams. For this help session, you are asked to work through the sample exam questions on your own prior to the help session. Use the solution videos provided to guide you with problems on which you have difficulty. During the help session, I will try to assist you with problems that you are having in general and/or with concepts that might not be clear in the solution videos for the sample exam questions.

Let me know if you have any questions related to Exam No. 3.

Homework Problem 8/130 - more discussion

There have been several good questions raised in the post below. Maybe I can jump in at this point to say a few things about these questions and some other important points.

• The EOM that I get for this system is [by summing moments about the no-slip contact point C -- see the lecture example on page 8 of the notes]: (3*m*r/2)*x_dot_dot + k*r*x = 0. Therefore the natural frequency, omega_n, is given by: omega_n = sqrt((2*k)/(3*m)). This natural frequency is NOT equal to sqrt(k/m)!!
• The actual IC's that you use to get a particular x_max in the response is somewhat arbitrary. That is, you can start the system out with an infinite number of sets of IC's x(0) and x_dot(0) and get the same x_max. [Can you look at the animations above to see what I used for IC's in this simulation?] For your work, I recommended using x(0) as some non-zero number and x_dot(0) = 0, as Phil B explains in a comment in the following post.
• The amplitude of oscillation for the orange wheel above is roughly twice that of the blue wheel. However, the two wheels take EXACTLY the same amount of time to make one complete cycle of oscillation -- do you know why this is true? Does this agree with your intuition?
• A friction force is required to prevent slipping as the disk rolls. Watch the friction force (FF) in the above animation as the system moves. The friction force for the orange wheel is larger than the friction force for the blue wheel -- why is this necessary? At what point in one cycle of oscillation is the friction force a maximum? Why does it occur that that time?

Homework Problem 8/130

How do you find the max amplitude if you don't know the displacement. for the equation x(t)= C cos(wn t) + S sin(wn t)?

Homework Problem VIB-3

Right now I am stuck on VIB-3. I am unable to determine how to relate the force of the spring and the force of the block on the disk in terms of theta. I feel like the point of no slip should be at a 45 degree angle above the horizontal from point O on the disk. I don't know if this is a correct assumption or if this will even help me out. Any help would be great!

Question on page 8 example

In reviewing my notes, I was wondering why it was necessary to sum the moments instead of simply summing the forces in the x direction as we had done in the previous example. With the sum of the forces in x, I get Fx = -3kx =m x(double dot) and I'm wondering why this is not a valid EOM.

The pitcher's mound and dynamics (posted for rvanklom)

Click here to see an explanation provided by Ryan for having a pitcher's mound in baseball. 

What do you think -- is this the reason for the mound?

Homework Problem No. 6/226

Please note that this problem was worked out in lecture. Because of this we will collect the other problem (6/194) on Wednesday.

Problem 6/194 - animation

Shown above are animations for two speeds of the auto: the upper frame is a speed slightly below the critical speed, and the lower frame is a speed above the critical speed. 

In both cases, the angular momentum about point O before impact is not due to rotation of the auto but is given by HO = m(rG/O x vG). 

Note that, on impact, the velocity of the center of mass changes instantaneously from directly to the right (at "1") to a direction that is perpendicular to line OG (at "2"). Having the velocity being perpendicular to OG shows that the auto is rotating about point O.

After impact, the angular momentum about O is due to the auto's rotation and is given by HO = IO*omega.

In the upper frame, the speed of G drops to zero slightly before G passes over O. In the lower frame, G has enough speed to make it over the top, and therefore the auto flips on its side.

[Note that I have modeled the curb as a small obstacle so as to not interfere with the tire as the auto tips.]

Homework Problem No. 6/194

Suggestions:

This problem is naturally broken up into two parts: from immediately before impact ("1") to immediately after impact with the curb ("2"), and from immediately after impact ("2") to the position with G directly over the contact point of the tire with curb ("3").

From 1 to 2:

The impact force of tire with curb, FC, is an impulsive force. Compared to this force we can ignore the non-impulsive forces of NL, NR and mg. Angular momentum about the contact point of the tire with the curb (call this point O) is therefore conserved. Before impact, the auto moves as a particle -- use the expression of angular momentum for a particle. After impact, the auto moves as a rigid body -- use the expression of angular momentum for a rigid  body.

From 2 to 3:

Energy is conserved. During this period of time the auto rotates as a rigid body about O.

Let us know if you have questions on this problem.

Homework Problem No. 6/185

Suggestions:

1. FBD: Use pendulum and bullet TOGETHER as your system. Draw an FBD of this system.
2. Impulse momentum equations: From your FBD, find a point about which angular momentum is conserved (no external moments acting on system about this point). [HINT: It is point O.] Recall the definitions of angular momentum for a particle (bullet) and for a rigid body (pendulum). Add these together to find the angular momentum for time 1 (before impact), HO1. Add these together to find angular momentum for time 2 (after impact), HO2. Equate: HO1 = HO2.
3. Kinematics: After impact (time 2), the bullet is stuck to pendulum. Ignore the width of the pendulum, and assume that the embedded bullet is directly under G (at a distance of h from O) immediately after impact.
4. Solve

Note that this problem is very similar to the angular impulse Problem 3/248 that we worked earlier in the semester. The difference here is that we treat the pendulum as a rigid body (HO = IO*omega) rather than as particles (HO = m*(rP/O x vP) ). See the solution of that problem on the Homework Page of the course site.

Let us know if you have any questions on this problem.