How Do You Calculate Angular Speed and Acceleration for a Rotating Robot Arm?

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SUMMARY

The discussion focuses on calculating the angular speed and acceleration of a rotating robot arm, specifically link AB, when θ1 is 30 degrees and θ2 is 60 degrees. The constant angular speed of link OA is established at 0.8 rad/s. The calculations yield an angular speed of link AB (Omegaa/b) at 2.13 rad/s and an acceleration (Ab) of 4.56 rad/s² directed downwards. The equations utilized include the cross product for velocity and acceleration, demonstrating the relationship between the links in motion.

PREREQUISITES
  • Understanding of angular kinematics, specifically angular speed and acceleration.
  • Familiarity with vector operations, particularly the cross product.
  • Knowledge of trigonometric functions, especially sine and cosine.
  • Proficiency in applying equations of motion for rotating bodies.
NEXT STEPS
  • Study the application of the cross product in rotational dynamics.
  • Learn about the dynamics of robotic arms and their kinematic equations.
  • Explore advanced topics in angular motion, such as torque and moment of inertia.
  • Investigate simulation tools for robotic motion analysis, such as MATLAB or Simulink.
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Mechanical engineers, robotics students, and professionals involved in robotic arm design and analysis will benefit from this discussion, particularly those focusing on kinematics and dynamics of rotating systems.

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Homework Statement



The link OA of a robots arm is rotating clockwise with a constant angular speed of 0.8 rad/s. At the same time the link AB is also rotating clockwise so that the end B moves down the vertical dashed line. Determine the angular speed and the acceleration of the link AB when θ1 = 30 degrees. θ2 = 60 degrees


Homework Equations


cp = crossproduct
Va = Omega(oa)R(oa)
Vb = Va + (Omegaa/b)cp(Ra/b)
Ab = Aa + ALPHAa/b cp R(b/a) - ((Omegaa/b)^2)* Ra/b


The Attempt at a Solution


Va = Omega xRoa = 0.8x0.8 = 0.64m/s
Vb = Va + (Omegaa/b)cp(Ra/b), Va = 0.64i, Vb = -Vbj, Ra/b = (0.6cos30i -0.6sin30j), Omegaa/b = Omegaa/bk

Equate i's to get Omegaa/b = 2.13rad/s
Equate j's to get Vb = 1.109m/s down

For Aa, it is in the i direction only so Aa = (Omega(oa)^2)R(oa) = (0.8x0.8)x0.8 = 0.512 rad/s^2

so Ab = Aa + ALPHAa/b cp R(b/a) - ((Omegaa/b)^2)* Ra/b
equate i's
0 = 0/512 - ALPHAa/b x 0.6sin30 - (2.13^2)x 0.6cos 30
ALPHAa/b = 6.15 rad/s^2
equate j's
i get Ab = 4.56 rad/s^2 down
Could anyone verify my calculations. i would be very greatful please see the attachment for the diagram
 
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this is the diagram theta 1 is between OA and theta 2 between AB. OA = 0.8m and OB 0.6m
 

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