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\title{Twix Challenge - Chocolate Orange}
\author{Andrew Cooper}
\maketitle
	A chocolate orange of delicious, smooth, uniform chocolate of mass \textit{M}, radius \textit{a}, is sliced into segments by planes through a fixed axis. It stands horizontally on a table, with this axis vertical, and is held together by a narrow ribbon around its equator.\medskip 

	
	Show the tension in the ribbon is $ \dfrac{3 Mg}{32} $ .
	You may assume that the centre of mass of a segment of angle $2\theta$ is at distance  $\dfrac{3\pi asin\theta}{16\theta} $ from the axis.
	
\medskip 
	
	By looking at the chocolate orange from above, and drawing a triangle of hypotenuse \textit{ a}, and resolving the force of tension so that it passes through the centre of the segment, we can see that the component of tension toward the axis for a segment of any value of $\theta$ is:
	
	$$F_T = 2Tsin\theta$$
	\textit{ Where T = Tension}\\
\medskip	

\includegraphics[scale=0.35]{../TwixChallenge4.png}	
	
\medskip 

The fraction of the sphere of chocolate that any segment is can be written as:\\
 
	$\dfrac{2\theta}{2\pi}  =   \frac{\theta}{\pi}$\\
	\smallskip
	
Therefore the mass represented by a segment is:\\

	$$M = M_{total}\dfrac{\theta}{\pi}$$
	
	\textit{Where $M_{total}$ is the mass of the whole chocolate orange. }\\


Therefore the force due to gravity is:

	$$F_g = Mg\dfrac{\theta}{\pi}$$
	
\medskip 

The moment of a force is:\\

	$$Moment = \overrightarrow{F} * \overrightarrow{r}$$
\smallskip
Where r represents the distance of the centre of mass from the point at which the chocolate orange touches the surface of the table. $\dfrac{3\pi asin\theta}{16\theta}$ 	 
\\
\\
The sphere and its segments are stationary, so the force due to gravity and the tension provided by the ribbon must be equal.
\smallskip
	$$Mg\dfrac{\theta}{\pi}  \dfrac{3\pi asin \theta}{16\theta} = 2aTsin\theta 
$$ 	
Cancels to:
	$$Mg\frac{3 }{16} = 2T$$
    Therefore: $$T = Mg\frac{3}{32} $$

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