Equation of motion for the translation of a single rod

In summary, the mass should always be considered when there is an applied moment. The direction of the mass will be opposite to the moment, so it will move downwards.
  • #1
Mech_LS24
148
16
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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  • #2
Why ignoring the mass of 50 Kg?
Acceleration of a massless system, under the action of a force, should be infinite.
 
  • #3
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?
 
  • #4
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:
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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
 

1. What is the equation of motion for the translation of a single rod?

The equation of motion for the translation of a single rod is given by F = ma, where F is the net force acting on the rod, m is the mass of the rod, and a is the acceleration of the rod.

2. How is the equation of motion derived for a single rod?

The equation of motion for a single rod is derived using Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

3. What factors affect the motion of a single rod?

The motion of a single rod is affected by factors such as the net force acting on the rod, the mass of the rod, and the surface on which the rod is moving.

4. Can the equation of motion be used for any type of motion?

Yes, the equation of motion can be used for any type of motion as long as the motion is along a straight line and the net force acting on the object is constant.

5. How can the equation of motion be applied in real-world situations?

The equation of motion can be applied in real-world situations, such as calculating the acceleration of a car or the force needed to move an object, by using the known values of mass, acceleration, and net force.

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