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Jun 16, 2008

Instantaneous Centers



I was looking for a simpler way to do 5/88 and I came across the sample problem 5/12 (p.373). I think we're supposed to cover instantaneous centers tomorrow, but I tried it anyway. Based on the sample problem, would it be right to say:

omega-BC = omega-AB
omega-BC = Vb / (150mm + CD)

where CD is the length between point C and the instantaneous center D?
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3 comments:

CMK said...

The velocity part of this problem can be worked out using the method of instant centers, although I am not sure that it is a simpler method for this problem than using the vector equations.

To do so, you need to first locate the instant center for link AB. As we will talk about on Wednesday, the instant center for AB is found by the intersection of lines OA and CB (a point directly below point C that is on the extension of line OA). Call this point E. You will need to use trig to find the lengths of EB and EA. From this, you can say that omega-BC = (EB/CB)*omega-AB. You can also say that omega-AB = (OA/EA)*omega0. You can combine these to find omega-AB and omega-BC.

As you can see, this method (after using the above trig) is not necessarily easier than the vector approach. In addition, you cannot use the instant center approach for finding the angular accelerations.

Note that from the above, you can NOT say that omega-BC = omega-AB, as you have used above.

Let us know if this is not clear.

June 16, 2008 at 7:06 PM
ckearney said...

In the sample problem it looked like the omega's were equivalent and the respective velocities were scaled, but I'm sure it'll be more clear on Wednesday. Thank you.

June 16, 2008 at 7:48 PM
Eric Nauman said...

You're getting there. I think the thing to keep in mind is that BC, AB, and OA all have different angular velocities. So you need to know the velocities of two points on the same rigid body (or at least their directions) to make progress.

June 16, 2008 at 10:12 PM

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