Probability of string snapping proportional to tension, and polar co-ordinates for probability density function

Question

Picture the setup:

A particle of mass $m$ attached with a string of length $R$ spins in a vertical circle around a fixed peg with constant velocity $v>\sqrt{Rg}$ in the $x$-$z$ plane, acting under its weight $mg$. In this model, the probability that the string snaps when the string is at an angle $\theta\in[0,2\pi)$ is proportional to the tension in the string, $T(\theta)$.

We know the tension is given by $T(\theta) = \frac{mv^2}{R}-mg\sin\theta$.

So the tension is greatest when the particle is at the bottom of its motion (i.e. when $\theta=3\pi/2$) and the tension is least when the particle is at the top (when $\theta=\pi/2$).

If we let the random variable $\Theta$ be the value of theta at which the string snaps, then we can work out the probability density function for $\Theta$: $$f_{\Theta}(\theta) = \frac1{2\pi}-\frac{Rg\sin\theta}{2\pi v^2}.$$

The issue arrises here: the expectation of $\Theta$ is given by $$ \mathbb{E}(\Theta) = \int_0^{2\pi}\theta\; f_{\Theta}(\theta)\;d\theta $$ which evaluates to $\pi+Rg/v^2$, which is not intuitively correct. It would make sense for the expectation of $\Theta$ to be $3\pi/2$, the point at which the string is most likely to snap. Furthermore, if we instead define $\theta$ by the angle with the verticle rather than the angle with the positive horizontal, and so the pdf is in terms of $\cos\theta$ instead of $\sin\theta$, we get $\mathbb{E}(\Theta)=\pi$, which does make sense in this context.

I suspect the cause of this inconsistency is that the expectation doesn't take into account that $0=2\pi$ in this context. We need to use a different formula for expectation that doesn't bias towards $2\pi$ over $0$. I reckon transforming to polar co-ordinates would make sense, but I have no idea how expectations translate into polar, and how to make the outcome independent of the choice of co-ordinate system.

Perhaps changing the factor of $\theta$ in the integral for the expectation, i.e. work out $\mathbb{E}(g(\Theta))$ for some function $g$ that deals with the moment of an angle, or something...

Any thoughts are appreciated. The endgame is to work out $\textrm{Var}(\Theta)$ and maybe other statistics about $\Theta$, but obviously I need a good structure for the pdf and stuff first.

Answer 1

The expected value is a summation of all the positions it could break times the probability of it breaking at that particular position. The definite integral does this summation and the result is an average angle. It should be biased towards the angle of maximum tension but it is an entirely different metric to the angle of maximum tension. The expected value of a series of bets isn't the value of the most likely win.

Answer 2

Okay I think I've solved my own question here. We have the tension $T(\theta)$ is given by $$ T(\theta) = \frac{mv^2}{R} - mg\sin\theta. $$ Then to find the pdf, we normalise $T$ by dividing by the total "mass" of $T$, the $\textbf{polar}$ integral from $0$ to $2\pi$: $$ \int_0^{2\pi} \int_0^{T(\theta)} r\; \mathrm{d}r \; \mathrm{d}\theta $$ I.e. we needed to use polar co-ordinates from the very start, since the random variable $\Theta$ is inherently an angle. We can simplify this integral by writing $T$ in the form $T(\theta)=a+b\sin\theta$, where, in this case, $a=mv^2/R$ and $b=-mg$.

This gives the total "mass", $k$, of $T$ as \begin{align*} k=\int_0^{2\pi} \int_0^{T(\theta)} r\; \mathrm{d}r \; \mathrm{d}\theta &= \frac12 \int_0^{2\pi} (a+b\sin\theta)^2 \;\mathrm{d}\theta \\ &= \frac\pi2 (2a^2+b^2) \\ &= \frac\pi2\left( 2\frac{m^2v^4}{R^2} + m^2g^2 \right) \\ &=\frac{\pi m^2}{2R^2}(2v^4 +R^2g^2) \end{align*} So it makes sense to use $$ f_{\Theta}(\theta) = \frac{1}{2k}\; [T(\theta)]^2, $$ as the p.d.f. for $\Theta$, which actually ends up being rather redundant.

Now, we calculate the "centre of mass" of $T$ by finding the cartesian coordinates, $x=r\cos\theta$ and $y=r\sin\theta$, for the centre of mass, $(\bar{x},\bar{y})$ by finding the moments:

\begin{align*} \bar{x} &= \frac1k \int_0^{2\pi} \int_0^{T(\theta)} xr\; \mathrm{d}r \; \mathrm{d}\theta = \frac1k \int_0^{2\pi} \int_0^{T(\theta)} r^2 \cos\theta\; \mathrm{d}r \; \mathrm{d}\theta \\ &= \frac1k \int_0^{2\pi} \frac13 (a+b\sin\theta)^3 \cos\theta \; \mathrm{d}\theta = 0 \end{align*}

and

\begin{align*} \bar{y} &= \frac1k \int_0^{2\pi} \int_0^{T(\theta)} yr\; \mathrm{d}r \; \mathrm{d}\theta = \frac1k \int_0^{2\pi} \int_0^{T(\theta)} r^2 \sin\theta\; \mathrm{d}r \; \mathrm{d}\theta \\ &= \frac1k \int_0^{2\pi} \frac13 (a+b\sin\theta)^3 \sin\theta \; \mathrm{d}\theta = \frac1{4k}\pi b(4a^2+b^2) \\ &= \frac{b(4a^2+b^2)}{2(2a^2+b^2)} = \frac{-mg(4v^4+R^2g^2)}{2(2v^4+R^2g^2)} \end{align*}

Since $\bar{x} = 0$ and $\bar{y} < 0$, we get that (excuse sloppiness) $\bar{\theta} = \textrm{"arctan"}\left(\frac{\bar{y}}{\bar{x}}\right) = \frac{3\pi}2 = \mathbb{E}(\Theta)$, as we had anticipated.

We also obtain $\bar{r} = |\bar{y}|$. Does this equal $\mathbb{E}(T(\Theta))$? It seems plausible since the dimensional analysis works this value out to be a force.

This is a clunky solution and I don't particularly like moving back to cartesian to get back to polar, but at least it's a solution. It's also interesting that in this case, the expected value of $\Theta$ is not an integral with $f_\Theta$ but instead with $T$.

This would also be the case for the variance of $\Theta$, where we might compute $$ \left( \frac1k \int_0^{2\pi} \int_0^{T(\theta)} x^2 r \;\mathrm{d}r \; \mathrm{d}\theta, \frac1k \int_0^{2\pi} \int_0^{T(\theta)} y^2 r \;\mathrm{d}r \; \mathrm{d}\theta \right) $$ to find the second moments, and then this could give us $\mathbb{E}(\Theta^2)$. Who knows... food for thought.