Very fast spinning

Consider an area $A$ of infinitesimally small thickness $dr$ of the bag. We use Newton's Second Law to look at the forces of the two sides of the area. By doing so, we obtain an equation $dF = (p + dp)A - pA \ .$ By spinning, the pressure on the area closer to the opening of the bag will be greater than the pressure on the other side, as there will be more air particles pushing down. With Newton's Second Law, we have $dF = a\, dm \ ,$ where $a = \omega^2 r$ . Thus subbing in $dF$ and $dm = \rho Adr$ gives $dp = \omega^2 \rho r dr \ .$ Because the density throughout the bag changes, we use the ideal gas law to write the density as a function of pressure, $pV = nRT\ .$ Subbing in $n = \frac{m_{air}}{M_{air}}$ , where $M_{air}$ is the molecular mass, we can rearrange the ideal gas law to yield $\rho = \frac{m_{air}}{V} = \frac{M_{air}p}{RT} \ .$ With this, we obtain the equation $dp = \omega^2 \frac{M_{air}P}{RT} rdr \ .$ This is simply a separable first-order DE, so we can rearrange the equation and integrate, $\int_{p_0}^P \frac{dp}{p} = \omega^2 \frac{M_{air}}{RT}\int_{l_0}^l rdr \ ,$ where $p_0$ is the outside pressure, $P$ is the pressure at the bottom of the bag, $l_0$ is the length of the boy's arm, and $l$ is the length of the boy's arm and the length of the cylinder. Thus by integrating and rearranging, we obtain $\omega = \sqrt{\frac{2RT}{M(l^2 - l_0^2)}ln\left(\frac{P}{p_0}\right)} \ .$ Subbing in the numbers, we get a final answer $\omega = \sqrt{\frac{2(8.314)(25+273.15)}{0.029(1-0.07^2)} ln\left(\frac{2.2}{1}\right)} = 514 \ .$

Since the centrifugal force acts on the air in the bag, the air creates an additional pressure on the bottom of the bag. Once that pressure reaches $2.2~\mbox{atm}$ , the bag will break ( $2.2~\mbox{atm}$ since the outside pressure is $1~\mbox{atm}$ and the pressure difference has to be $1.2~\mbox{atm}$ ).

If we take infinitesimally small cylindrical intersection of the width $dr$ of the air in the bag, we can calculate the pressure it exerts on the rest of the air. Its mass is $dm = \rho A dr$ , so the force it exerts is $dF = dm \omega^2 r$ and the pressure $dp = \rho \omega^2 r dr$ .

Since from the ideal gas law we have that $\rho = \frac{pM}{RT}$ , we can plug it into the above formula, and we get $\frac{dp}{p} = \frac{M \omega^2 r dr}{RT}$ .

To obtain the formula for $p(x)$ , we should integrate both sides, $p$ from $p_0$ to $p$ , and $r$ from $L$ to $L+x$ . We get that $\ln(\frac{p}{p_0}) = \frac{M\omega^2}{2RT} x(2L+x)$ . Plugging in $x = 30~\mbox{cm}$ and $p = 2.2~\mbox{atm}$ we get $\omega = 513.95 ~\mbox{rad/s}$ .