Torque required to keep a body in pure translation as it accelerates

In summary, the conversation is about solving a problem of a cantilevered gantry system by determining the torque needed to keep it moving in translation without rotation. The solution involves summing the forces and moments, and taking moments about the center of mass to find the torque. The conversation also touches on the concept of D'Alembert's principle and the use of pseudo inertial forces.
  • #1
pharcycle
5
0
Hello All,

I've been searching for days to try and solve this but I'm going round in circles so thought I'd fire it out to into the ether! I'm analysing a gantry system that is essentially cantilevered from one end but the part of the problem I'm struggling with can be simplified thus:

Imagine pushing a ruler on a desk from one end so that it has a tendency to want to rotate. To keep it moving as a pure translation you also apply a torque to the end. How do you work out what that torque it!?

I must be missing something brutally simple here but for the life of me I can't work it out. We can ignore friction for the purposes of this and since we're moving across a surface the force due to gravity is out of plane.

From the sum of forces = ma, it's pretty simple to work out what the applied force is to get it to accelerate. Summing the moments to zero (I'm assuming zero here as it's in pure translation) varies depending on where you take moments about so clearly something is wrong with my FBD here but I can't work out what. I don't think it should be sum of torques = I * alpha as it's not rotating, although I could certainly do that to get an answer.

It's doubly frustrating as 10 years ago I'd have been able to do this in a heartbeat but instead I've had to waste hours on this!

Hope someone can help and sorry my first post is a question!

Thanks,
David
 

Attachments

  • ruler on table.pdf
    29.7 KB · Views: 181
Engineering news on Phys.org
  • #2
OK.. think I've realized my problem (well, one of many!)

For pure translation the sum of the forces = mass * acceleration but the sum of the moments only equals zero about the centre of mass...

Could someone confirm this for me?

Thanks
 
  • #3
The sum of forces can be taken from any point on the body. The center of mass is usually chosen to simplify things.
 
  • #4
timthereaper said:
The sum of forces can be taken from any point on the body.

For statics, yes, but I think for an accelerating body it has to be about the centre of mass.

In my FBD, the reaction moment would change depending on where you take moments about as there are no other forces acting on it. So T - F*r = 0 and depending on where you take moments from (i.e. r) will change the value of your moment. This is clearly wrong as the reaction moment will be constant - this is what triggered my further thinking into it as I was under the same assumption as you!

I've also found in one of my old textbooks a statement to the same effect so I'm reasonably confident I'm right now... I hope!
 
  • #5
pharcycle said:
For statics, yes, but I think for an accelerating body it has to be about the centre of mass.

In my FBD, the reaction moment would change depending on where you take moments about as there are no other forces acting on it. So T - F*r = 0 and depending on where you take moments from (i.e. r) will change the value of your moment. This is clearly wrong as the reaction moment will be constant - this is what triggered my further thinking into it as I was under the same assumption as you!

I've also found in one of my old textbooks a statement to the same effect so I'm reasonably confident I'm right now... I hope!
You are correct...you must sum moments about the center of mass = 0 in this case of translational acceleration to solve for the torque required to prevent rotation of the ruler about its center of mass under the application of the end force F. You can sum moments about any point for the accelerating case only by using a pseudo inertial force (D'Alemberts Principle) applied at the center of mass, equal to F but in the opposite direction, but for now, don't worry about that, it gets a bit confusing using that concept, so just do as you did and sum moments about the center of mass to get your desired result.
 
  • Like
Likes 1 person
  • #6
thanks,

yeah I thought about using D'Alemberts principal to solve it but from my uni days I remembered that you don't really need to use it for classical mechanics problems. Plus I don't like the F -ma = 0 approach... just seems wrong to me.

Cheers for the confirmation,

Dave
 
  • #7
pharcycle said:
For statics, yes, but I think for an accelerating body it has to be about the centre of mass.

In my FBD, the reaction moment would change depending on where you take moments about as there are no other forces acting on it. So T - F*r = 0 and depending on where you take moments from (i.e. r) will change the value of your moment. This is clearly wrong as the reaction moment will be constant - this is what triggered my further thinking into it as I was under the same assumption as you!

I've also found in one of my old textbooks a statement to the same effect so I'm reasonably confident I'm right now... I hope!

So, you have one equation and 2 unknowns, F and T?

What are you going to do next?
 
  • #8
256bits said:
So, you have one equation and 2 unknowns, F and T?

What are you going to do next?

... Haha, not sure if that's sarcasm or a genuine question - if you look at my scanned picture I actually have 2 equations. But, for the sake of anyone who may come across this, the solution is thus:

Sum F = ma (upwards +ve)
so F = ma (1)

Sum T=0 about C.O.M CW +ve
So F*xg -T = 0
So T = F* xg (2)

a=5 m/s2
m=100kg (it's not actually a ruler by the way)
So F=500N

xg=600mm
So T = 0.6 * 500 = 300Nm

I just forgot that in pure translation it's only the sum of the moments about the centre of mass that equal zero.
 
  • #9
OK. I had a preconceived idea about the problem so no posts made sense.
( is there a smiley to go with that! )

I thought you were pushing the ruler from the end along the long axis and the ruler would rotate from it being unstable similar to balancing a pencil on your finger, such as an inverted pendulum.

Thanks for reply to clear things up!
 
  • #10
pharcycle said:
For statics, yes, but I think for an accelerating body it has to be about the centre of mass.

Using D'Alembert's principle, you can apply inertial forces and torques equivalent to the accelerating body at the center of mass and get an equivalent statics problem (like F-ma = 0). Then you can take the sum of forces and moments about points that make the resulting equations simpler, like the center of mass.
 

1. What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation.

2. How is torque related to acceleration?

In order to accelerate an object in a linear or straight-line motion, a certain amount of torque must be applied. This torque is directly proportional to the mass and acceleration of the object.

3. What is the role of torque in maintaining pure translation?

When an object is undergoing pure translation (moving in a straight line without rotation), the total torque acting on the object must be zero. This means that the torque required to maintain pure translation is equal to the torque caused by any external forces acting on the object.

4. How does the torque required for pure translation change as the object accelerates?

As the object accelerates, the torque required to maintain pure translation will also change. This is because the acceleration of the object will alter the forces acting on it, and therefore the torque needed to counteract these forces will also change.

5. Can the torque required for pure translation ever be completely eliminated?

In theory, yes, the torque required for pure translation can be completely eliminated if the net force acting on the object is zero. This would result in the object moving at a constant velocity without any rotation.

Similar threads

  • Mechanical Engineering
Replies
2
Views
1K
Replies
10
Views
899
  • Mechanical Engineering
Replies
1
Views
965
Replies
9
Views
5K
  • Mechanical Engineering
Replies
8
Views
3K
  • Mechanical Engineering
Replies
1
Views
2K
Replies
3
Views
2K
  • Mechanical Engineering
Replies
8
Views
1K
  • Mechanical Engineering
Replies
11
Views
2K
  • Mechanical Engineering
Replies
5
Views
3K
Back
Top