ME274 - Basic Mechanics II: Sample Exam Problem 36

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Aug 24, 2008

Sample Exam Problem 36

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8 comments:

gbaker said...: For this problem, when you sum the moments about point C you say it's a counter clockwise moment. I agree. Why does this make the terms be negative then?

Garrett; November 23, 2008 at 4:41 PM
CMK said...: Note that the angle theta is defined as being positive in the CW direction (see problem figure). This sets the sign convention for both theta AND moments (since you eventually set summation of moments = I*theta_dot_dot).

If theta had been defined positive in the CCW direction, then the moments would also be defined positive in the CCW direction.

Good question.; November 23, 2008 at 4:46 PM
Fady said...: When drawing the FBDs for this problem, how do we know that the spring force acts on the wheel and not on the bar?

Fady; November 30, 2008 at 12:04 PM
CMK said...: Actually, you can put the spring force on either the wheel or the bar. The result is the same.

In truth, the spring is attached to pin G that connects the wheel to the bar. I chose to include this pin as part of the wheel. You can include the pin as part of the bar.

In summary, put the spring on either the wheel or the bar, NOT both.

Good question.; November 30, 2008 at 4:11 PM
Anonymous said...: Why isn't the weight included in the FBD for this problem?; December 1, 2008 at 11:30 PM
CMK said...: The weight forces do not act in the x-direction and do not contribute to the moment equation for the disk. Therefore, they do not appear in the EOM.; December 1, 2008 at 11:56 PM
Dale Szul said...: When faced with a EOM problem such as this one, would it be wise to attempt a sum of forces and then a sum of moments equation, or go directly to the sum of moments equation (such as for the disk)?; December 2, 2008 at 9:01 AM
CMK said...: I suppose that this comes down to personal choice.

To be safe, I usually recommend summation of forces along with a moment about the center of mass. This is safe because we know that the moment (Euler) equation about G always gets rid of the second term on the RHS.

To choose a point other than G (or a fixed point) requires that you deal with that second term on the RHS of Euler's equation. In this problem it drops out because the r-vector is parallel with the a-vector. This is a more advanced move.

In summary, to be safe you should always choose your moment equation about G.; December 2, 2008 at 5:07 PM

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