Welcome to the website of ME 274 for the Fall 2008 semester. On this site you can view blog posts, add your own blog posts and add comments to existing posts. In addition to the blog are links to course material: course information, information on solution videos, exams, quizzes, homeworks and other course-related material. Direct links to the homework solution videos are also available on the left side of this page.


The following is a reverse chronological order listing of the posts for the course blog. To add a post, click here (when adding posts, be sure to add a "label" in the box at the lower right side of the post window). To add a comment to an existing post, click on the "Comments" link below the post.


____________________________________________________

Aug 24, 2008

Exam Problem 34




Right-click here to download video onto your computer.

Please add your questions and comments below.



Posted by at
Labels:

5 comments:

nour said...

Unless i missed something, i think Vg= (L/2 - R)w .
right ?

December 1, 2008 at 5:27 PM
CMK said...

Yes, you are correct: vG = (L/2 - R)*omega. The picture clearly shows that to be true. Unfortunately, I must have not been looking at the picture correctly.

Apologies for the confusion.

December 1, 2008 at 6:16 PM
Unknown said...

There is a mention of finding the IC of the bar when coming up with the W/E eqn. How would I go about doing this? If I am not mistaken, it would reduce the terms in the T_2 eqn.

December 1, 2008 at 10:09 PM
Unknown said...
This comment has been removed by the author.
December 1, 2008 at 10:10 PM
CMK said...

Excellent question. The IC of a rigid body is the point of the body with zero velocity. What is that point for the bar?

Let's not worry about this right now as you study for the exam. I recommend just using G as the point for the KE of the bar. After the exam, I would be very open to you (or someone else) posting this answer for blog points).

December 1, 2008 at 10:49 PM

Post a Comment

Subscribe to: Post Comments (Atom)