# The "propagation" of torque

1. Sep 6, 2015

### 24forChromium

See image. If T_1 and L is known, what is T_2? (T_2 is caused by T_1)

2. Sep 6, 2015

### Staff: Mentor

Can you draw a free body diagram (FBD) for the horizontal rod?

And if the rod does not move, what can you say about the sum of the forces and moments on the rod?

What changes if the horizontal rod can deform due to the applied torque?

3. Sep 6, 2015

### andrewkirk

I think there may be a missing piece of information, which is the length of the arm on the right of the diagram.

According to my calculations, $T_2$ depends on $\frac{h}{L}$ where $h$ is the length of the arm.

But I'm not brilliant with torques, so that may be wrong.

4. Sep 6, 2015

### 24forChromium

I think it is irrelevant, because the arm length on the right will only effect the torque T_1 if T_1 is generated by forces acting on those arms, but T_1 could be generated in many other ways, and I don't think the way T_1 is generated matters to T_2.

5. Sep 6, 2015

### 24forChromium

I don't think I can, becuase I don't know what kind of forces will be acting on it if those forces are generated by a torque on the other end. The rod should move, but if you are asking about the moment in time where is hasn't begining moving yet, then I would say the sum of the forces acting on the whole system is equal to the torque acting on the appendage, at least that would be what's logical to me. About deformation, I think that would just make things more complicated and I don't think I need that, because I am investigating a real case of such a system, I think the effect of the deformation would be unmeasureably small.

6. Sep 6, 2015

### Staff: Mentor

Is only a torque applied to the left end of the horizontal arm? How is the torque applied?

7. Sep 6, 2015

### 24forChromium

My current theory goes like this:

Rotation is relative, seen from the left pivot, a torque is accelerating the appendage in the counterclockwise (C.C.) direction; seen from the fixed joint, a torque is acting around the fixed joint and accelerating the left pivot as well as the horizontal rod in the C.C. direction, even though this is not true when the fact that the left pivot is fixed on the ground when the ground is taken into consideration. However, the angualr acceleration perceived by both points should be the same in magnitude.

From the perspective of the left pivot, a torque is accelerating the system with an angualr inertia that is the sum of the angular inertia of the pivot itself and the angular inertia of the appendage at a radius of L away. Or: α = (I_pivot + I_appendage_L) /T

From the perspective of the appendage, a torque is accelerating the system with an angualr inertia that is the sum of the angular inertia of the appendage itself and the angular inertia of the body at a radius of L away. Or: α = (I_pivot_L + I_appendage) /T

Because α is the same for both, (I_pivot + I_appendage_L) /T = (I_pivot_L + I_appendage) /T

8. Sep 6, 2015

### andrewkirk

I think it does matter, because a torque needs to arise from a force applied to some body.

Consider the cases where the arm has length L from tip to square joint and where it has length 2L, and the torque $T_1$ arises from a force applied at one tip of the arm. According to my calcs we get different answers for $T_2$ in the two cases.

9. Sep 6, 2015

### 24forChromium

yes, only a torque is applied on the left and it is applied by gyroscopic reaction in the case that I am investigating.

10. Sep 6, 2015

### 24forChromium

That sounds like it may be true, let's say the arm length on the appendage is D, can you show me how T_2 can be calculated?

11. Sep 6, 2015

### andrewkirk

This is the way I'm thinking about it: Since the joint is rigid, it doesn't matter whether the arm points down as shown, or sideways as a continuation of the horizontal strut. The latter is easier to work with, so let's consider that. The torque T1 is equivalent to a force T1/D at the end of the arm. We then use the length D+L to convert that to a torque around the pivot.

12. Sep 6, 2015

### 24forChromium

so, you are saying that T_2 = (T_1 / D) * (D + L) am I right?

13. Sep 6, 2015

### 24forChromium

I don't believe that's right, if the direction of the appendage arm does not matter and the force is always tangential to the right joint, then it should be equally correct to have the arm pointing to the left, overlapping parts of the body, and have a force point down from the arm and accelerating the whole system clockwise, which is the opposite of what happens.

14. Sep 6, 2015

### andrewkirk

Good point. I hadn't thought of that. Oh well, there goes that theory.

This nudges me towards the view that the problem is underspecified. We need to know how or where (or both) the torque T1 is applied.

15. Sep 6, 2015

### 24forChromium

There are a lot of possibilities, in my specific investigation, the torque T1 is produced by a phenomenon known as (at least I call it) Gyroscopic couple reaction. It's super complicated, by essentially it says that if a gyroscope is spinning and precessing at the same time, the gyroscope will generate a torque in a direction that is perpendicular to both of those rotations, the equation is T = ω*L. I still don't believe that the way a torque is applied matters, I still think that the variable arm length can be compensated by variable tangential force, but if you like, we can put numbers into all of those variables.
Here is the numbers in my experiment:
Body length - L: 0.6m
arm length - D (in this case the radius of the disk): 0.3m
torque - T1 (at one point in time): 75N*m

16. Sep 6, 2015

### Staff: Mentor

It is not possible to calculate given the information provided. You need to know the motion of the structure (especially angular acceleration, but maybe also angular velocity). Then you can solve for the unknown torque.

Also, are you assuming the rod is massless and rigid?

Last edited: Sep 6, 2015
17. Sep 6, 2015

### 24forChromium

You can add one such variable if you wish, then express T2 in terms of the other variables.