A boy is spinning on a chair and holding a paper bag so that it is horizontal, and perpendicular to the axis of the rotation. What should be the angular velocity of the boy in radians/second so that the bottom of the bag breaks?
Details and assumptions
Assume paper breaks at an overpressure of 1.2 atm.
The molecular mass of the air is 29 g/mol.
The boy's arm is L=70 cm long.
When horizontal, the bag has a shape of a cylinder of length 30 cm.
Outside pressure is 1 atm and temperature 25 degrees C.
Assume that the boy won't get sick no matter how fast he is spinning. Hint: he whizzes around.
To determine the angular velocity at which the bottom of the paper bag breaks, we need to consider the centripetal force acting on the bag. The centripetal force is responsible for keeping the bag in circular motion.
The centripetal force required to maintain circular motion can be expressed as:
Fc = mv^2 / r
Where Fc is the centripetal force, m is the mass of air in the bag, v is the linear velocity of the bottom of the bag, and r is the radius of the circular motion (L, the length of the boy's arm).
To convert the linear velocity to angular velocity, we can use the equation:
v = ωr
Where v is the linear velocity, ω is the angular velocity, and r is the radius (L).
Rearranging the equation for centripetal force, we have:
Fc = m(ωr)^2 / r
Simplifying,
Fc = mω^2r
The pressure difference across the bottom of the bag is responsible for exerting the force that could cause it to break. The pressure difference can be calculated using the ideal gas law:
P = nRT / V
Where P is the pressure, n is the number of moles of air, R is the ideal gas constant, T is the temperature, and V is the volume of the bag.
The molecular mass of air given is in grams per mole, so we need to convert it to kilograms per mole:
Molecular mass of air = 29 g/mol = 0.029 kg/mol
The volume of the bag can be calculated as the product of its length and cross-sectional area when it is horizontal:
V = length * cross-sectional area
The cross-sectional area of the bag when it is horizontal can be calculated as the product of its circumference and thickness:
C = 2πr (circumference)
A = C * thickness
Substituting the values,
V = 0.3 m * (2πr * thickness)
The pressure difference across the bottom of the bag is given as 1.2 atm. We need to convert it to Pascals (Pa):
1 atm = 101325 Pa
1.2 atm = 1.2 * 101325 Pa = 121590 Pa
To calculate the number of moles of air in the bag, we can use the equation:
n = m / M
Where n is the number of moles, m is the mass of air, and M is the molecular mass of air.
Rearranging the equation for pressure, we have:
P = (n/V)RT
And substituting the values,
121590 = (n / V) * (8.314 J/(mol·K)) * (25 + 273.15 K)
Now, we can combine the equations for the centripetal force and pressure:
mω^2L = (n/V)RT
Substituting the expressions for m/V and V, and rearranging the equation:
ω = √[(P / (RT)) * (πL * thickness) / (0.029 kg/mol)]
Substituting the given values:
ω = √[(121590 Pa / (8.314 J/(mol·K)) * (25 + 273.15 K)) * (π * 0.7 m * thickness) / (0.029 kg/mol)]
Simplifying,
ω = √[(4329.46 * (π * 0.7 m * thickness) / (0.029 kg/mol)]
The angular velocity (ω) will be in radians/second.
Note: The value of the thickness of the bag is not given in the details. It needs to be provided to calculate the final result.