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.


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Sep 7, 2008

Problem 5/25

I can't seem to get the right solution to problem 5/25 for tomorrow. I'm not quite sure what problem states about the 0.6 number, and there are no units. Do I need to convert the angular velocity to rad/sec? Any ideas?
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4 comments:

kyle brooks said...

I agree with you. The question statement says something about hte tangent, but then if you take the tangent of 0.6 that is 120 degrees approximately and does not align with the schematic at all.

September 7, 2008 at 10:45 PM
CMK said...

The problem statement gives us that tan(theta) = 0.6, where theta is the angle between the acceleration of B and the radial line from O to B. Note that the tangent function gives a non-dimensional number, hence no units. From this we know that theta = 30.96 degrees.

As suggested in an earlier post (see it further down on the page), you need to find the two components of the acceleration of point B (these components will be in terms of alpha and omega^2). With trig, you can relate the ratio of these components to tan(theta).

September 7, 2008 at 11:06 PM
nour said...

this makes stuff a bit clearer. thank you

September 8, 2008 at 2:14 AM
ClayDT3 said...

Using a_n = v^2/r, you get the normal acceleration of point A.
Using the inverse tangent function to find theta, you take 90-theta to find the angle between the a_n and a_t vectors.
Then you can use trig and relate the ratios to find a_t.
Finally, using a_t = alpha*r, you find alpha.

September 8, 2008 at 11:28 AM

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