# Shear: What would be the equation for ''pure'' torque?

#### kamhogo

1. Homework Statement
We had this elementary experiment in class. An aluminium rod is fixed at one end and at the other end you have a disk. You hand increasingly heavy weights on the disk and measure how much the rod shears (twists). We were asked to make a graph of the shear torque vs. angle of torsion. It came out quadratic. I'm trying to figure out why, physically and mathematically.

2. Homework Equations
Shear Torque = F x d; Equation = aΘ² + bΘ + c (specifically: 6888.1Θ² + 87.951Θ - 1.0003) - Quadratic
Oscillatory Torque = -k*Θ (specifically: - 0.3670Θ) - Linear
ω = 3.27051 rad/sec
k = I*ω² = 0.3670 Nm
I = 0.034315 kgm² (Given)
J = (πD^4)/32 = 6.1359*10^-11
T = 1.92/sec
L= 0.57m
3. The Attempt at a Solution
I figure out that the rod was trying to resist the forced torque with an oscillatory torque in the opposite sense. I looked on the web and found the equation (see above). My data seem valid since I could calculate the G (Coulomb's modulus or modulus of rigidity or shear modulus) of the material with a percent difference of 10, which I read was acceptable in physics. I think my shear torque equation gives me the *net* torque, i.e. it takes into account the oscillatory torque. Now I'd like to know how to find what ''pure'' torque, i.e. what the torque would be without the oscillatory torque. Do I add up, substract, multiply or divide the two equations? My gut feeling tells me nothing.....Any hint please? Thanks !

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#### haruspex

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Not sure I understand the set up. Is the rod horizontal? Is the plane of the disc normal to the rod, with the rod attached at its centre? Are weights attached at the rim of the disc, at the same horizontal height as its centre?
Where do the oscillations come in? The problem description makes it sound static.

#### kamhogo

Sorry about that. The rod is horizontal and normal to the plane of the disc, attached at it's centre. Weights are attached at the rim of the disc but not at the same height as its centre, lower rather (measure not taken). When you hang a weight, the disc rotates (its rim is graduated in mm) but then it oscillates back and forth (counter-rotation and back) then it stops. That's when you read the displacement in mm and convert them to rads.

I thought some more about it. Coulomb's modulus (G) cannot possibly be found using the quadratic equation alone because in that equation, it keeps changing, which is normal. As the rod undergoes more and more shear, it is no longer the same rod. So for every weight you add (every bit of additional shearing), G must get smaller and smaller (hence the hyperbola), until it reaches 0 and then the torque skyrockets (quadratic becomes delta Dirac?).

Do I make sense?

#### haruspex

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then it stops. That's when you read the displacement
So, it is effectively static. The oscillations are not interesting, right?
Coulomb's modulus (G) cannot possibly be found using the quadratic equation alone because in that equation, it keeps changing, which is normal. As the rod undergoes more and more shear, it is no longer the same rod.
I did not understand this part at all. How is it a different rod?
Which equation do you have in mind?
Perhaps you mean that as the rod is twisted, the torque changes, because the weight does not have the same horizontal distance from the rod. If that is what you mean, knowing exactly where the weights are attached in relation to the disc's centre will be crucial. If a bit above the horizontal, the torque will increase initially.

#### kamhogo

Oh I get it. The two graphs represent a static situation (quadratic) and a dynamic one (linear). Senseless to try and combine them.

I meant that as the rod gets twisted, the particles that make it up are no longer arranged in the same way so it is not really the same rod at the location of the twist. At that location, it becomes easier and easier to twist it: G becomes smaller and smaller.....

The weights were attached below the horizontal.

#### kamhogo #### kamhogo #### haruspex

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the weight does not have the same horizontal distance from the rod.
Seeing the diagram, it looks like the weights are attached to a band which wraps around the top of the disc. If so, the point of application of the weight is always on the level of the disc centre.
at the location of the twist. At that location, it becomes easier and easier to twist it:
The twist is distributed all along the rod. It behaves like a spring, the greater the twist the greater the torque.

Right. Thanks!

#### kamhogo

Note to everybody reading this post: please disregard all the nonsense I wrote above. :)

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