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

1. Feb 18, 2014

### pharcycle

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

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2. Feb 18, 2014

### pharcycle

OK.. think I've realised 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. Feb 18, 2014

### timthereaper

The sum of forces can be taken from any point on the body. The center of mass is usually chosen to simplify things.

4. Feb 18, 2014

### pharcycle

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 text books a statement to the same effect so I'm reasonably confident I'm right now... I hope!

5. Feb 18, 2014

### PhanthomJay

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.

6. Feb 18, 2014

### pharcycle

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. Feb 18, 2014

### 256bits

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

What are you going to do next?

8. Feb 18, 2014

### pharcycle

... 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. Feb 18, 2014

### 256bits

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. Feb 19, 2014

### timthereaper

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.