A perfectly spherical, but squishable ball of radius 10 cm is blown up to an internal pressure of 2 atm. The ball is placed between two perfectly vertical walls and the walls are slowly squeezed together. Initially the ball slips down the walls, but when the walls are 18 cm apart the ball stops slipping down. What is the coefficient of friction between the surface of the ball and the walls?
Details and assumptions:
-The ambient air temperature is 20∘C.
-Air has a molar mass of 29 g/mol.
-The mass of the ball (not including any air inside) is 100 g.
-1 atm is 101,325 Pa.
-The ball squishes as the walls close in, but you may assume the ball does not deform otherwise.
-The acceleration of gravity is −9.8 m/s2.
Hint: the answer is very small.
To solve this problem, we need to consider the forces acting on the ball and the conditions under which it stops slipping down the walls.
Let's start by analyzing the forces acting on the ball. There are two main forces to consider: the gravitational force (weight) pulling the ball downward and the normal force exerted by the walls pushing back on the ball.
1. Gravitational Force:
The weight of the ball can be calculated using the mass of the ball. In this case, the mass of the ball is given as 100 g (or 0.1 kg). Since the acceleration due to gravity is -9.8 m/s^2 (negative because it is acting opposite to the positive direction), the weight can be calculated as follows:
Weight = mass x acceleration due to gravity = 0.1 kg x (-9.8 m/s^2) = -0.98 N (negative because it is acting downward).
2. Normal Force:
The normal force is the force exerted by the walls perpendicular to their surfaces. When the ball is not moving, the normal force is equal in magnitude and opposite in direction to the gravitational force.
Now, let's consider the forces involved when the ball stops slipping down the walls. At this point, the static friction force between the ball and the walls opposes the force of gravity. Therefore, the static friction force should exert enough force to counteract the component of the weight acting downward.
The friction force acting on the ball can be calculated as:
Friction force = coefficient of friction x normal force.
To calculate the normal force, we need to determine the pressure inside the ball when it stops slipping. We can use the ideal gas law to find the pressure:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given that the internal pressure of the ball is 2 atm (101,325 Pa) and the radius is 10 cm, we can calculate the volume of the ball:
Volume = (4/3)πr^3 = (4/3)π(0.10 m)^3 = 0.004189 m^3.
Since we know the molar mass of air is 29 g/mol and the mass of the ball (excluding air) is 100 g, we can determine the number of moles:
Number of moles = mass/molar mass = 0.1 kg/ 0.029 kg/mol = 3.4483 mol.
Now, let's calculate the temperature in Kelvin. Given that the ambient air temperature is 20°C, we need to convert it to Kelvin:
Temperature (in Kelvin) = 20°C + 273.15 = 293.15 K.
Substituting the values into the ideal gas law equation, we can find the normal force.
Once we have the normal force, we can equate it to the weight of the ball and solve for the coefficient of friction:
Friction force = coefficient of friction x normal force,
Friction force = -0.98 N (weight of the ball),
Coefficient of friction = Friction force / normal force.
By calculating the coefficient of friction, we can determine the value between the surface of the ball and the walls.
To recap:
1. Calculate the volume of the ball using its radius.
2. Calculate the number of moles of air inside the ball using its mass and the molar mass of air.
3. Convert the ambient air temperature to Kelvin.
4. Use the ideal gas law to find the internal pressure of the ball.
5. Calculate the normal force by equating it to the weight of the ball.
6. Calculate the coefficient of friction by dividing the friction force by the normal force.
Please note that due to the complexity of these calculations, it's recommended to use a scientific calculator or a computer program that can handle mathematical operations with precision.