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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)
5 comments:
In the solution you have the mass moment of inertia of point C written as I(c)=I(g) + mR^2, where you have I(g)=0.5mR^2. I remember at some point doing a wheel problem and using just mR^2 for the center of mass moment of inertia, or am I confusing that with something else?
The centroidal mass moment of inertia for a SOLID homogeneous disk is given by: IG = (1/2)*m*R^2.
The MMI for a RING is given by: IG = m*R^2. (A ring is a thin circular rigid body with all of its mass distributed at a distance of R from the center G).
You might be thinking about the problem we did earlier in the semester that dealt with a ring rather than a disk.
For the exam, you will be given the MMI for a solid disk and for a thin bar.
Let us know if this does not help.
When I relate theta_dot_dot to phi_dot_dot, I used: a_E = a_C + phi_dot_dot x R_E/C - omega*R^2, where a_E is equal to -theta_dot_dot*R (based on relating theta_dot_dot to x_dot_dot).
my final relation ended up being phi_dot_dot = theta_dot_dot/2. Is this correct? What intuition I have tells me that this is wrong and they should be equal because they have equal circumferences and don't slip, and therefore would rotate the same degree for a given change in position of the bar. Thanks for any insight you can give me.
--to john--
Your analysis is correct as well as your final result: phi_dot_dot = theta_dot_dot/2.
* The wheel of the left rotates about point A, and the wheel on the right rotates about point C.
* The distance from C to point E is twice that of the distance from A to point D.
* Since points D and E on the bar have the same acceleration, then the angular acceleration of the wheel on the right is half that of the angular acceleration of the wheel on the right. And they have the same sign.
Does this make sense?
Yes, this makes sense. I think my problem was with the rotation of the right wheel being about point C, instead of being about the center of the wheel. Thanks.
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