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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
- Lectures (1)
- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
14 comments:
Prof Neuman told me that they will not be giving out the solution. We should discuss the exams on the blog.
mark
I can tell you that for problem 2 you should be getting 6 or 7 equations/unknowns depending on how you do it.
On problem 3. when looking at the moments on the pulley, do you make the angular acceleration = (x_double_dot)/r
Yes, theta_double_dot = x_double_dot/r_double_dot
I hope that helps!
Also, you may leave off gravity since you can declare it at equilibrium position. Also make sure to use 2mx** for and not mx** for the newtwon Eq's.
On #3, but what is the Newton eq. for the mass?
Currently I have
-kx + k(x-x_b) = 2m*x_dot_dot
but how does the pulley fit into that equation? Or does it even need to?
For number 3:
Is the only reason we have to take a moment about O because the pulley has mass? If the pulley had no mass it would be an ideal pulley and therefore the tension would be equal on both sides, correct?
Yes, that is correct. Sample Problem 8/5 is very similar. It says to ignore the mass and friction of the pulleys and they do not take a moment about the pulley.
on prob 4 # 7 I thought all the equations were true, I was wrong does anyone know which are not true?
I thought the same, but someone posted earlier that only b and c are true. On problem 4 #9 someone posted that the correct answer is "c", but the only answer that makes sense to me is "a". does anyone know why it is "c"? Also what is thoesen talking about when he said to use 2mx** why would that be. I tried looking in the book and it says nothing about assuming equilibrium position.
For number 4 part 9, I put “a” and got it right. I believe the correct answer is suppose to be “a” and not “c”.
For problem 4 #9 the answer is b and d. The other equations are wrong because if you look at the second part the MMI must be about the center of mass or an instant center. You do not use I_A or I_B. #9 I believe is C because if you would look at the wheel you see that the centripetal acceleration is pointing up. Then if you do sum of Forces in the y the centripetal acceleration is counteracting the mg and therefore N would be less than mg. Does this help?
mark
Problem 4 part 9:
Sum forces in y
N - mg = m*a_Gy
Therefore
N = m*a_Gy + mg
Where a_Gy is positive because G is accelerating in the positive y direction.
For problem number 1 part 2 how do you calculated the work done on the disk?
I assume you mean Problem II under Problem No. 4. For the work done by F you add the work done by the force times the distance it acts over PLUS the work done by the moment caused by F since it doesn't act at G. The equation looks like U = -F*delta_x - F*R*(delta_x/R), which simplifies to U = -F*delta_x - F*delta_x.
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