ME 203 Exam


Spring 2010

United States
Questions, questions, post them here...
How many questions should we expect on the exam, is there any particular emphasis on certain materials that you have gone over in class, and is homework #4 cover all that you would expect us to know for the test?
United States
The exam will have 3-4 questions covering Cartesian dynamics. So anything that can be described with Cartesian coordinates is fair game, such as projectiles or blocks. This eliminates rotation from the topics on the test. In homework 04, only the last three problems are within the scope of the exam.
Thanks.
Wednesday you said you were going to try to go to a different room for the exam if you could. I have looked through my email and forums and have not found any postings on the subject, so am I to then assume that the test is in the usually room?
Greetings was curious, if you would post class averages and such stats on this website for us?
What day is the next exam? I forgot to write it down.
Next exam is Tuesday immediately following break during the problem session.
Question: My plan for this week and spring break is to study and watch the video solutions under the planar heading. Is there anything else I should be mindful of? I know in class Dr. Quinn spoke of posting examples, but when and where are they posted ?

Thanks
I'm working on the exam solutions for exam 1 problem 2... and am confused on how I can use the second point x(final)=d and y(final)=0 any suggestions?
Is radius of gyration going to be covered on the exam? I don't remember going over it in class.
i have a question regarding question 5 from the review questions that you posted for exam 2

in the angular momentum balance why is "(fr − F)*l/2" written the way it is?

i would have written it as "(F-fr)*l/2"

is it because the the object would initially resist a change and swing to the left (negative k by r-hand rule) "(F-fr)*l/2 *-k" which is equal to your "(fr − F)*l/2"

thank you
United States
The radius of gyration is simply a different way to measure the moment of inertia. The moment of inertia I depends on both the mass of an object and the way that mass is distributed. The radius of gyration just describes how the mass is distributed so that I = m k^2 with m the mass of the object. However, on the exam I will give you moment of inertia directly, rather than giving it to you in terms of the radius of gyration.
The moment was written the way it was because, well, that's what it is. In a little more detail, the moment M from a force F is defined as M = r x F, with the vector r from the point about which moments are taken to the point at which the force is applied. In this case r = l/2 j. Working through the cross product with these forces gives the result in the solutions. Ultimately, this has nothing really to do with which way the bar moves, and everything to do with how the forces are applied.
For anybody who is interested, there will be a study group this Monday at 4:00 in Bulger residence hall. We can work through the practice exam, as well as the homework.
How can we tell just from looking at it, if the ωx(ωxr) term in the acceleration equation is zero?
>Just wondering..will there be anything on this coming test about Moving Frames of Reference?
where are these review problems
On question 1 (Exam 2 Spring 2009), what exactly is the contact force? I took it as a normal force on the top of the slot. Since theta is 0, the slot is horizontal to the ground, meaning the normal and gravitational forces cancel out, leaving only Fc acting down, which means it is equal to 0.

On question 2 (Exam 2 Spring 2009), how do you relate the acceleration of point A, to the center of the big pulley without knowing the radii of the pulleys?
United States
Question 1: In this question there is no gravity (I think I told the students to assume no gravity during the exam), The contact force is the normal force in this case and vanishes only because at this instant the acceleration normal to the direction of the slot vanishes, so that the net force must vanish.

Question 2: The radius of the pulley is not necessary to find out how the velocities at either edge relate to the velocity of the center. The radius is only necessary to find the angular velocity of the pulleys, which was not requested in this problem.
Question 2, in order to find the tension in the cable, we need to know the acceleration of the center of the bigger pulley. The known variable that we have is the acceleration of A, which we need to relate to the center point to use F=ma. We can get the forces from the free-body diagram and the mass is given, but without the acceleration, cannot find the tension. I am not sure how to relate the accelerations without the radii of the pulleys.
I don't know if anyone else is having trouble, but I cannot get the video solutions to work for the review homework assignment. It won't open a media file labeled as a .pdf
Did anyone else have this trouble?
Page: 1/2  [Next]
1  2