2D Rigid Body Dynamics - Newtonian Equaitons of Motion

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Discussion Overview

The discussion revolves around the application of Newtonian equations of motion to a 2D rigid body dynamics problem involving a bar that is both translating and rotating. Participants explore the calculation of forces acting on the bar, particularly focusing on the moment equation about a pivot point.

Discussion Character

  • Homework-related
  • Technical explanation

Main Points Raised

  • The initial post presents a scenario involving a 500 kg bar with a specified acceleration and angular velocity, seeking assistance in understanding the acceleration component in the moment equation.
  • One participant clarifies that the "mad" term in the moment equation represents the product of linear acceleration of the center of mass and the distance from the pivot to the center of mass.
  • Another participant confirms the solution has been found, indicating that the initial query was resolved.

Areas of Agreement / Disagreement

The discussion appears to have reached a resolution with the original poster confirming that their question was solved, suggesting agreement on the approach taken.

Contextual Notes

Participants did not elaborate on the specific calculations or assumptions made in deriving the solution, leaving some details of the problem-solving process unspecified.

mm391
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Homework Statement


In the diagram attached a bar weighs 500 kg and is a 2D plane rigid body with mass centre at G. At the instant shown, the bar is moving horizontally at but reducing speed, causing horizontal deceleration as shown by vector a=-7i m/s2. At the same time, the bar is being raised rapidly by anti-clockwise rotation of the support structure with angular velocity of magnitude ωr=2rad/s. All other kinematic components can be ignored. Assuming the forces at B can only be applied in the normal direction and that the support at point O is a frictionless pin joint construct Newtonian equations of motion, and use just the moment equation about Point O to calculate the instantaneous force in the hydraulic ram at point B.


Homework Equations


∑Fx = m*ax

∑Fy = m*ay

∑Mo = I*\alpha+m*ad

The Attempt at a Solution



I can see using the equations below that I\alpha=0 as there is no angular acceleration. I am struggling to work out what the acceleration component is in the "m*a*d" part of the equation. Can anyone please have a go at explaining (not solving) how I can find it?

Thanks

Mark
 

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The rod is rotating anticlockwise, at a constant angular speed, about one end as that end moves to the right and decelerates.

Presumably there is an unbalanced centripetal force on the rod com that points at O.

Anyway:
Can you state in words what the "mad" part of the moment sum represents?
You may need to consult your notes.

Looks like "a" is the linear acceleration of the center of mass and "d" is the perpendicular-to-a distance from the pivot to the center of mass.
 
Solved. Thanks
 
Well done :)
 

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