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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)
6 comments:
On problem two, once the Newton Euler equations are found, the next step is kinematics. The solutions states that B is the IC for the disk. I am not sure completely how they got B as the instant center and wanted to see if someone could help me out.
Thanks
All points on the right side of the cable not wrapped around the disk are stationary (the upper end of the cable is attached to a fixed point, the cable does not stretch and it remains vertical at all times). Since the disk does not slip on the cable, then point B on the disk is also stationary (and becomes the IC for the disk).
Does this help? There is a video solution for the problem on the blog (Problem 6/140) that might also help in explaining.
In the solution it says that h_0 is 2*delta. Why is this? I thought that delta would be equal to h_0.
The above question is about problem 2. Sorry for any confusion.
The disk does not slip on the cable (given information). Point B on the disk is in contact with a stationary point on the cable. Therefore, the velocity of B is zero, and is the instant center for the disk. Point A is twice as far from the IC as is point O. Therefore, the speed of A is always twice the speed of O. As a result, A moves twice as far as O. A moves through a distance of delta. Therefore, O moves through a distance of delta/2 (h = delta/2).
An alternate explanation can come from the usual cable-pulley equation. If you write the length of the cable in terms of the distances traveled by A and O, you will find the the speed of A is always twice that of O. The rest of the above explanation then applies.
Does this help?
Yes it does, thanks.
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