One thing's for sure. After this, we're all going to be a lot thinner!

To calculate the coefficient of friction, we will examine the balloon in two states: its initial state, without any contact with the ball; and its final state, when its contact with the ball makes it stop slipping.

Initially, the volume, $V$ of the ball is the volume of a sphere: $\frac {4\pi}{3}r^3 = .0041888 m^3$ . The initial given pressure, $P$ is $2 ATM$ , or $202650 Pa$ , and the initial given temperature, $T$ is $20^\circ C$ , or $293.15K$ .

Using the Ideal Gas Law, $PV = nRT$ with $R = 8.314 \frac {J}{K·mol}$ , we may find that in the balloon, there are $.346$ moles of gas. Given that the air has a molar mass of $0.029 kg / mol$ we may calculate the mass of the air in the balloon: $0.010034 kg$ . The total mass of the balloon and the air inside is $0.1100034 kg$ .

In its final state, when the balloon is squished to the walls, many values stay constant: the temperature T, the number of moles in the balloon, and the mass of the balloon.

What we notice first that changes is the volume of the balloon. We may calculate the new volume of the balloon through subtraction. The volume of a spherical segment that is "squished out" by the wall may be calculated by the formula: $V_{spherical segment} = \frac {\pi}{6}h(3a^2 + h^2)$ , in which $h$ is the height of the segment and $a$ is the radius of the base. Because the walls are $18cm$ apart, while the balloon has an initial radius of $10cm$ , $h = 10cm - 18cm/2 = 1cm$ , and $a = \sqrt{(10cm)^2 - (9 cm)^2} = \sqrt{19} cm$ . The volume of each spherical segment is thus $30.3687 cm^3$ or $3.03687*10^{-5} m^3$ . The volume of the balloon is $V_{initial} - 2*V_{spherical segment} = .0041281m^3$ .

We must still observe the Ideal Gas Law, $PV = nRT$ . Because the volume changed between the initial and final states, while both the temperature and number of moles stayed constant, the pressure of the balloon also must have changed. For this, we may use Boyle's Law, that $P_{1}V_{1} = P_{2}V_{2}$ . This gives us a new pressure of $205631 Pa$ .

Pressure is the force per unit of area; to calculate the normal force exerted by the balloon on the surface of the wall, we must calculate the area it has in contact with the wall. We already calculated the radius of the base of the spherical segment: $a = \sqrt{19} cm$ . The area, $\pi a^2$ is thus $\pi19cm^2$ for each spherical segment. Because there are two spherical segments cut out (the balloon is squished by walls on both sides), we multiply this surface area by 2. The total normal force is thus $205631 Pa * 2\pi19cm^2 = 2454.83 N$ .

The gravitational force on the balloon is $MA = g*mass_{total}$ . We calculated the mass earlier to be $0.1100034 kg$ . The net gravitational force is $1.078 N$ . In order for the balloon to stop slipping down, the frictional force is of equal magnitude of this gravitational force, and in opposite direction.

The frictional force is equal to the normal force times the coefficient of friction. We solve the equation, $\mu * 2454.83 N = 1.078 N$ to get $\mu = 0.000439$ .

By the ideal gas law: $n=\dfrac{P_0 V_0}{RT}$ Where:

$P_0=2\ atm=202,650\ Pa$

$V_0=\dfrac{4\cdot 0.1^3}{3}\pi \ m^3$

$R=8.3144\ J/(mol\cdot K)$

$T=20\ ^oC=293.15\ K$

And the mass of the ball in kg is $m_b=0.29\cdot n+0.1$ .

When the ball stops slipping down, its volume is $V_f=V_0-2V_c$ , where $V_c$ is the volume of a spherical cap with height $h=0.01\ m$ and radius of sphere $r=0.1\ m$ , obtaining: $V_c=\dfrac{\pi h^2}{3}(3r-h)=\dfrac{2.9\cdot 10^{-5}}{3}\pi\ m^3$ As the temperature remains constant, being $P_f$ the final pressure of the gas: $P_f=\dfrac{P_0V_0}{V_f}$ The area of the ball in contact with each wall when the ball stops slipping down is $S=\pi(r^2-(r-h)^2)=1.9\cdot 10^{-3} \pi\ m^2$ .

The normal force between the ball and each wall is $F_N=P_f\cdot S$ .

The vertical force between the ball and each wall is half of the ball's weight., i.e., $F_V=\dfrac{9.8\cdot m_s}{2}$

When the ball stops, being $\mu$ the coefficient of friction between the surface of the bal and the wall:

$\mu F_N=F_V$ $\mu=\dfrac{F_V}{F_N}$

Solving, one can obtain, with three significant digits:

$\mu=4.39\cdot 10^{-4}$

Given: r = 10 cm = 0.1 m d = 1 cm = 0.01 m t = 20° C = 293 K p = 2 atm = 202,650 Pa m₁ = 100 g = 0.1 kg g = 9.8 m/s² M = 29 g/mol R = 8.3144621 J/mol*kg

Solution: mg = μ 2A p' => μ = mg/2Ap' , where p' is the internal pressure of the ball after it squishes; m is the total mass of the ball and the air inside it; A is the area of the contact surface between the squished ball and one of the walls. According to the Boyle's law pV = p'V' , where V = 4πr³/3 and V' = 2π(2r³/3 - rd² + d³/3) Let m₂ be the mass of the air inside the ball => m₂ = n M , where n is the amount of the air inside the ball According to the ideal gas law pV = nRT => n = pV/RT => m₂ = pVM/RT = 0.0101 kg m = m₁ + m₂ = 0.1101 kg p' = pV/V' = 205,631.659 Pa A = π(r² - (r - d)²) = π 0.0019 m² μ = mg/2Ap' = 4.3953E-4

area of contact of the deformed ball with the wall(when the separation is 18cm ) is equal to the area cut off by a plane intersecting at a distance of 9 cm from the centre ,an approximate circle whose radius can be determined using Pythagoras Theorem.(coefficient of friction) (pressure) (area)*2=mg.[no of moles can be found out by ideal gas equation and accordingly the weight of the gas inside].

We first need to find the pressure when the ball is squeezed. Since the ball is squeezed slowly, the process is isothermal and so we have

$\frac{p_f}{p_i}=\frac{V_i}{V_f}$ .

The initial volume is $V_i=\frac{4}{3}\pi r^3=0.00419~\mbox{m}^3$ . As we squeeze the ball, we remove the volume of two spherical caps on either side, each of which has volume $V_{cap}=\frac{\pi h^2}{3}(3r-h)$ where $h$ is the amount squeezed. Hence the final volume is

$V_f=0.00419-0.00006=0.00413~~\mbox{m}^3$ .

The final pressure is therefore $p_f=2 \times 101325 \times 0.00419/0.00413=205631~\mbox{Pa}$ . The surface area in contact with one of the walls is $A=\pi (r^2-(r-h)^2)=0.00597~\mbox{m}^2$ . Hence the force of the ball on the wall (and hence the normal force of the wall on the ball) is $N=0.00597 \times 205631=1227~\mbox{N}$ .

The total mass of air inside the ball is $0.029 \times p_i V_i/(RT)=0.0101~\mbox{kg}$ . The total gravitational force on the ball is therefore $F_g=(0.1+0.01)g=1.078~\mbox{N}$ . This force must be balanced by the friction between the ball and the wall, so $2 \mu N=F_g$ which yields $\mu = 0.000439$ .