- #1
theBEAST
- 361
- 0
Homework Statement
Why is the angular velocity of link BD the same as the angular velocity of ICD?
Shouldn't it be ωICD=vD/rD/IC?
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SUMMARY
The discussion centers on the relationship between the angular velocity of link BD and the instantaneous center of zero velocity (ICD). It is established that the angular velocity of link BD is indeed the same as that of ICD due to the rigid body nature of BD, which ensures all points share the same angular velocity. The calculation of angular velocity for point D is highlighted as the most straightforward approach, reinforcing the conclusion that if ICD were fixed to BD, their angular velocities would align.
PREREQUISITES- Understanding of rigid body dynamics
- Familiarity with angular velocity concepts
- Knowledge of instantaneous centers in kinematics
- Basic principles of linkages in mechanical systems
- Study the principles of rigid body motion in mechanical systems
- Learn about the calculation of angular velocity in kinematic chains
- Explore the concept of instantaneous centers of rotation in depth
- Investigate the application of these concepts in robotic arm design
Mechanical engineers, students studying kinematics, and professionals involved in robotics and linkage design will benefit from this discussion.
Why is the angular velocity of link BD the same as the angular velocity of ICD?
Shouldn't it be ωICD=vD/rD/IC?
- #2
voko
- 6,053
- 391
BD is a rigid body. All of its points have the same angular velocity. The angular velocity of point D is easiest to compute. If ICD were a rigid body firmly attached to BD, then it would also have the same angular velocity.
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