- #1
Rigid body dynamics (kinematics of a point)
- Thread starter xzibition8612
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Discussion Overview
The discussion revolves around a homework problem related to the kinematics of a rigid body, specifically focusing on the dynamics of a point within that system. Participants are attempting to resolve discrepancies in their calculations and interpretations of the problem, which involves the application of the law of cosines and vector analysis.
Discussion Character
- Homework-related
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant reports obtaining a result of 0.92, while the textbook states the answer is 1.73, indicating a potential error in their calculations.
- Another participant questions the choice of making xA negative, suggesting it may affect the outcome.
- There are claims of sign errors in the quadratic equation and the initial application of the law of cosines, with suggestions to reassess the axis placement.
- A participant acknowledges a mistake in their calculations regarding the length of q and agrees with another's assertion about the value of xA.
- One participant expresses frustration over the problem, noting that setting the origin at point A leads to a negative answer.
- Another participant proposes an alternative approach using vector analysis and the kinematics of rigid bodies, suggesting that the negative sign does not indicate direction but rather relates to the deceleration of point B.
Areas of Agreement / Disagreement
Participants express differing views on the correctness of their calculations and interpretations of signs in their results. There is no consensus on the correct approach or final answer, as multiple competing perspectives are presented.
Contextual Notes
Participants reference attachments for problem statements and their work, indicating that some assumptions may depend on these external materials. There are unresolved mathematical steps and potential dependencies on definitions that are not fully clarified in the discussion.
Who May Find This Useful
Students working on rigid body dynamics, particularly those grappling with kinematic problems involving vector analysis and the law of cosines.
- #2
tiny-tim
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Homework Helper
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hi xzibition8612! 
why have you made xA negative?
why have you made xA negative?
- #3
xzibition8612
- 137
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xA is negative because if you look at the attachment "question", I set point C as the origin. Left of C is in the negative direction. Thanks for responding.
- #4
RTW69
- 374
- 1
I think you have made some sign errors. Check your quadratic equation, I don't think q is correct.
- #5
RTW69
- 374
- 1
You made Xa positive the first time you used the law of cosines. Move the axis to A and see if you get a better result.
- #6
RTW69
- 374
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My error, after deciding to do the math, your length of q is 13.9 m but I agree with tiny-tim, I think Xa= 10 m. The curious thing is that dKb/dt is -1.73. Right magnitude, wrong direction
- #7
xzibition8612
- 137
- 0
goddamn this problem is driving me insane, i set origin at A and the answer turns out to be negative.
- #8
RTW69
- 374
- 1
I think the negative sign means that point B is decelerating. It is does not mean that Vb is pointing toward c.
There is another way to do this problem just using vectors and the kinematics of rigid bodies. The plane motion of the rigid bar can replaced by a translation defined by the motion of an arbitrary reference point and by a rotation about that point.
The vector equation is VB= VB/A + VA
You know the magnitude and direction of VA, the direction of VB. VB/A is the vector perpendicular to the rigid bar at Pt. B causing the rigid bar to rotate CCW. You can determine the angle of VB/A from the horizontal by geometry. Through vector addition you can determine the magnitude of VB. You will see that it's magnitude is 1.73 m/s away from B. So forget the negative sign, it is not a direction sign.
There is another way to do this problem just using vectors and the kinematics of rigid bodies. The plane motion of the rigid bar can replaced by a translation defined by the motion of an arbitrary reference point and by a rotation about that point.
The vector equation is VB= VB/A + VA
You know the magnitude and direction of VA, the direction of VB. VB/A is the vector perpendicular to the rigid bar at Pt. B causing the rigid bar to rotate CCW. You can determine the angle of VB/A from the horizontal by geometry. Through vector addition you can determine the magnitude of VB. You will see that it's magnitude is 1.73 m/s away from B. So forget the negative sign, it is not a direction sign.
- #9
xzibition8612
- 137
- 0
alright man thanks a lot
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