Does the spring constant of rubber affect the rate at which it cools?

AI Thread Summary
The discussion centers on whether the spring constant of rubber influences its cooling rate, specifically the cooling constant 'k'. Participants express skepticism about a direct relationship, noting that both properties may correlate with rubber's density but lack a physical connection. Hysteresis is mentioned as a potential factor, though its relevance remains unclear. The spring constant is defined only within a limited range, complicating any potential relationship with thermal properties. Overall, there is a consensus that no established link exists between the spring constant and cooling rate, suggesting further investigation may not be fruitful.
namitakn
Messages
4
Reaction score
0
Does the spring constant of a piece of rubber affect its rate of cooling constant 'k' ?

1) Is there any formula, any proven relationship?

2) If there isn't, would investigating it make any sense? Or would it end up being a waste of time as the two variables are totally unrelated?
Is there a possibility of a relationship between the two.

3)I am not sure, but when I asked this to someone, I was told to read something about hysteresis (don't know if I spelt it right!). What is it? I couldn't understand anything I read about it? And could it possibly answer my question?
 
Physics news on Phys.org
I don't see any reason to expect any dependence. Both might be correlated with the density of rubber, but that is like asking for a correlation with the market price... sure, you will find some, but not because of a physical relation. In addition, spring constants for rubber are not well-defined, and a material property, whereas k depends on the shape and size of your material.
 
Okay, so ...

yes, I know that the spring constant for a piece rubber is defined only over a very small range, it reaches elastic limit very soon. Therefore, the original s.constant will be considered.

Is there any other variable that has a closer connection with spring constant that I could investigate?
 
The k factor in the Newton cooling equation and k factor denoting spring constant are coincidentally using the same letter for representation. As far as I know, there is no relationship between the two properties. Like most formulas, one can use different letters to represent different quantities, as long as it is understood which quantity goes with which letter.
 
It seems a plausible speculation (at least to somebody like me who knows a bit about continuum mechanics and heat transfer but nothing much about chemistry) that the elastic behavior and thermal properties of rubber would both be related to its behaviour at the molecular level, i.e. the way the long chain molecules "uncoil", and/or are intertwined with each other.

But I have no idea whether this has already been studied, or what the conclusions were if it has been studied.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top