# Tension particle between two strings

1. Jul 30, 2014

### WannabeNewton

1. The problem statement, all variables and given/known data

While going over the exercises in my mechanics tutoring sheet I found that one of the problems' stated answer didn't make sense to me (it's apparently from a textbook):

A weight of mass $m$ is fixed to the middle point of a string of length $l$ as shown in the figure [see attachment] and rotates about an axis joining the ends of the string. The system is in contact with its environment at a temperature $T$. Calculate the tension $X$ acting between the ends of the string in terms of its dependence upon the distance $x$ between the ends.

3. The attempt at a solution

Let $\theta$ be the angle either string makes with $x$. By symmetry $\theta$ has to be the same for both strings and so too must the tension. We have $\cos\theta = \frac{x}{l}$ and $\sin\theta = \frac{2 r}{l}$ where $r$ is the radius of the circular trajectory. Then $2X \sin\theta = \frac{mv^2}{r} = \frac{2mv^2}{l \sin\theta}$ so $X = \frac{mv^2}{l (1 - \cos^2\theta)} = \frac{mv^2 l}{l^2 - x^2} = 2K \frac{l}{l^2 - x^2}$ where $K$ is the kinetic energy. Using the equipartition theorem we then have that the average tension in either string is $\langle X \rangle = k_B T \frac{l}{l^2 - x^2}$.

The answer however is apparently supposed to be $\langle X \rangle = k_B T \frac{x}{l^2 - x^2}$. But how can this be? This would imply that for $x = 0$ there is no tension in either string which is obviously not true. The tension in either string for $x = 0$ should be $\frac{mv^2}{l}$ as this is exactly half of the centripetal force $\frac{mv^2}{l/2}$.

My other issue is, the problem statement says "calculate the tension $X$ acting between the ends of the string...". How can someone possibly infer from this that the problem wants the thermal average $\langle X \rangle$ for an ensemble of such systems and not just the actual tension $X$ for any one copy of the system? Is one to infer it from the lack of specification of the angular velocity $\omega$ of the particle together with the specification of the temperature $T$ of the heat bath?

#### Attached Files:

• ###### problem statement string.png
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2. Jul 31, 2014

### Matterwave

Your answer looks right to me, as well as producing the correct limits. The book's answer can't be right given your argument. And yes, the book should have asked for an ensemble average since there is no one unique tension, given the variables given in the problem.

3. Jul 31, 2014

### jackwhirl

My reading of this is that it is asking for an axial tension.

Here you find the radial tension.

For x = 0 there would be no axial tension, only radial tension.
Also, $\frac{mv^2}{l}$is exactly double the centripetal force $\frac{mv^2}{l/2}$.

4. Jul 31, 2014

### WannabeNewton

What is axial vs. radial tension?

No it isn't. $\frac{mv^2}{l/2} = \frac{2mv^2}{l}$ is double $\frac{mv^2}{l}$ not the other way around.

5. Jul 31, 2014

### jackwhirl

Ah. Missed the division symbol there. Sorry.

Anyway, axial tension would be the component of the tension in line with x. The radial tension would be the component of the tension in line with the radius, perpendicular to the axis.

6. Jul 31, 2014

### WannabeNewton

I see, thanks. Well if the problem wants the axial tension then that would certainly give the required result. But how does a person reading the problem go about interpreting the statement "find the tension $X$ acting between the ends of the string" as "find the component of the tension in either segment of the string along $x$" as opposed to "find the resultant tension in either segment of the string" which is what I calculated above, or even "find the net tension in the string"?

By "segment" I mean one of the two halves of the string in the depicted diagram since the problem considers the two joint half segments to be the entire string.

7. Jul 31, 2014

### Matterwave

The only consistent definition of tension is along the direction of the string itself. If the problem wanted you to project this vector along the x-axis, it would have to explicitly say that. I do not agree that the problem is asking for this.

8. Jul 31, 2014

### WannabeNewton

This is how I read the problem as well so I went about calculating the tension along either half of the string.

Thank you for your earlier reply by the way.

If it was ambiguous to me then it might be ambiguous to the students as well so I will definitely be more clear when giving them the problem.

9. Aug 1, 2014

### Matterwave

It also physically makes no sense to me why you want to search for the x-projection.