Equation of motion for the translation of a single rod

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Discussion Overview

The discussion revolves around the equation of motion for a rod system involving a mass of 50 kg and its interaction with forces and moments at various points. Participants explore the implications of mass, torque, and centripetal forces in the context of the system's motion, focusing on kinematic and free body diagram (FBD) analyses.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question why the mass of 50 kg and forces at points A and B can be ignored in certain analyses, suggesting there are rules governing this.
  • Others argue that ignoring the mass leads to infinite acceleration in a massless system, which is not applicable here since the system includes a 50 kg mass.
  • One participant explains that links AB and CD do not have mass and serve to guide the mass in a semicircular motion, while also discussing the role of torque at pivot C.
  • There is confusion about the centripetal component of acceleration and whether the mass should be considered if the lengths of links AB and CD differ.
  • Some participants assert that the mass should always be considered due to the centripetal acceleration it experiences when rotating about a point.
  • Concerns are raised about the mass 'disappearing' in the sum of moments around point C, questioning how velocity affects the moment.
  • Participants discuss the relationship between the applied moment and the direction of the mass's movement, suggesting that the mass moves in opposition to the moment's direction.
  • Free body diagrams (FBDs) are referenced, with discussions on the forces acting on point D, including the upward force from the mass and the downward gravitational force.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of the mass and forces in the system. There is no consensus on whether the mass can be ignored or how it interacts with the applied moments and forces.

Contextual Notes

Participants highlight the complexity of the system, including the dependence on the lengths of the links and the conditions under which the mass should be considered in the analysis. Unresolved mathematical steps and assumptions about the system's behavior are noted.

Who May Find This Useful

This discussion may be of interest to those studying dynamics, mechanical systems, or engineering principles related to motion and forces, particularly in the context of free body diagrams and kinematic analysis.

Mech_LS24
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Homework Statement
The angular velocity of rod CD is at the moment shown in the figure 6 rad/s. A moment is applied at rod CD of M = 450 Nm. Determine the force in rod AB, the horizontal and vertical force at pin D and the angular acceleration in rod CD given at the moment shown in the figure. The block has a mass of 50 kg and his center of mass is at point G. Neglect the masses of the rods.
Relevant Equations
(ag)n = w^2 * r
(ag)t = alpha * r
Hello,

Given the statement a described above. To find the forces at point D I drawn a kinematic scheme and FBD of rod CD. But why am I allowed to ignore the mass of 50 kg, the forces at point B and point A? I know the are some rules about this, but I just can't remember them anymore.. The figure of the situation can be found below.

See my sketch with calculations:
1625426213515.png

1625426297936.png
 
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Why ignoring the mass of 50 Kg?
Acceleration of a massless system, under the action of a force, should be infinite.
 
Lnewqban said:
Acceleration of a massless system, under the action of a force, should be infinite
But the system isn't massless? It moves/accelerates the 50 kg isn't it?

And the forces in A/B should only be considered in the total FBD?
 
Links AB and CD do not have a mass value; their only function is to force point G, which represents a mass of 50 kg, to describe a semicircular movement that is pretty close to the trajectories of points B and D.
Link CD is under the effect of a torque or moment about pivot C, which induces the acceleration of those 50 Kg.
The centripetal component of that acceleration, combined with that mass, induces centripetal forces that are equally shared by links AB and CD of similar leghts (both points, B and D have same radius of rotation).
 
Last edited:
I am still a bit confused, what do you mean with centripetal component?

Lnewqban said:
induces centripetal forces that are equally shared by links AB and CD of similar leghts
Do you mean that when the links AB and CD weren't the same length, the mass should be considered?

It is hard to imagine that point D only needs to overcome the torque while there is a mass pushing downwards with 50 kg on it.
 
The mass should always be considered.
If the center of mass of that mass is rotating about a point away from it, there is a centripetal acceleration forcing it to do so.
 
Lnewqban said:
The mass should always be considered.
But why does it kind of 'disappear' in this Sum of moment around point C? The force is pointed towards point C, but is still has a velocity which affects the moment around C?
 
Lnewqban said:
which induces the acceleration of those 50 Kg.
I think this makes sense to me, after re-reading the topic for multiple timeso0).

The mass should have been considered if there wasn't applied a moment around C. So if I erase that moment the mass should be considered? Otherwise the mass would drop downwards.
 
Mech_LS24 said:
I think this makes sense to me, after re-reading the topic for multiple timeso0).

The mass should have been considered if there wasn't applied a moment around C. So if I erase that moment the mass should be considered? Otherwise the mass would drop downwards.
In what direction the mass will move depends on how strong the applied moment is.
 
  • #10
Lnewqban said:
In what direction the mass will move depends on how strong the applied moment is.
The mass direction is opposite to the moment, so downwards, right?
 
  • #11
The moment becomes a force at point D.
The mass, combined with gravity accelerataion, becomes another force.
 
  • #12
Is this what you mean?
1625770520618.png
 
  • #13
Are those the FBD's for this structure?

1625772906636.png
 
  • #14
We have two vertical forces acting on D: M/0.6(upwards) and 50g (downwards).
Therefore, there is a resultant force pushing and accelerating G up at the instant represented in the diagram.
 
  • #15
That makes the following (below) FBD?

1625813789119.png
 
  • #16
Make that last FBD sense to you @Lnewqban ?

Like to hear :)
 
  • #17
Anyone who could help?

Thanks
 

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