Sample Exam Problem 28

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Aug 24, 2008

Sample Exam Problem 28

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Please add your questions and comments below.

8 comments:

BoilerBrian said...: Did you know that a_Bx was equal to a_O because they were the two contact points of the cabel?; November 28, 2008 at 11:23 AM
Phoenix said...: Yes, because the length between the two is constant, the v and a of each would be the same as long as the rope is taught.; November 28, 2008 at 12:41 PM
Phoenix said...: Or, I guess that would be x_dot and x_dot_dot, which comes out to the same thing in this problem.; November 28, 2008 at 12:49 PM
CMK said...: Yes, all points on the part of the cable not wrapped around the disk have the same acceleration. Point B (where the cable comes off the disk) is not on the part of the cable that is wrapped around the disk. Therefore, the x-component of the acceleration of B has the same acceleration as point O.; November 28, 2008 at 4:48 PM
Fady said...: In the kinematics part, can we say that the angular speed (omega) is 0 since the system was released from rest?; November 30, 2008 at 12:45 PM
CMK said...: Yes, if the system is released from rest this means that the omegas are zero.

Note that the answer found is independent of the value of omega. That is, we did not need to use any numerical value for the omegas to solve.; November 30, 2008 at 4:08 PM
Unknown said...: I am not recalling where the mr^2 comes from in I_c=I_g+mr^2, I think it has something to do with the parallel axis thereom?; November 30, 2008 at 4:11 PM
CMK said...: Yes, that equation is an expression of the parallel axis theorem. The distance between C and G is R; therefore, IC = IG + m*R^2.; November 30, 2008 at 4:51 PM