- #1
Rococo
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Homework Statement
A girl 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 the girl's angular velocity be in radians per second, so that the bottom of the bag breaks?
The girl's arm is 0.7m long.
When horizontal, the bag is a cylinder of length 0.3m.
The paper will break at an over-pressure of 1.2 atmospheres.
A mole of air has a mass of 0.029 kg.
The outside pressure and temperature is 1 atmosphere and 25°C
Homework Equations
[itex]pV = nRT[/itex]
[itex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/itex]
1 atm = 101325 N/m²
1.2 atm = 121590 N/m²
R = 8.31 Jmol-¹K-¹
The Attempt at a Solution
Let the cross sectional area of the cylinder = A m²
Volume of cylinder = 0.3A m³
Volume of air in cylinder = 0.3A m³
Before the rotation, the air in the cylinder is at 1 atm (101325 N/m² ) and 25°C (298K):
[itex]pV = nRT[/itex]
(101325)(0.3A) = n(8.31)(298)
n = 12.275A moles of gas
mass of air in cylinder = number of moles X mass of one mole
m = 12.275A x 0.029
m = 0.356A kg
During the rotation, if the air in the cylinder is exerting a pressure of 1.2 atm (121590 N/m²) on the bottom of the bag then:
[itex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/itex]
T2 = (T1 x P2)/P1
T2 = (298 x 1.2)/1
T2 = 357.6
[itex]pV = nRT[/itex]
(121590)(0.3A) = n(8.31)(357.6)
n = 12.275A moles of gas
mass of air in cylinder = number of moles X mass of one mole
m = 12.275A x 0.029
m = 0.356A kg
I'm not sure of the next steps however I tried using the equation for centripetal force:
Centripetal force = force on bottom of bag
[itex]mrω² = PA[/itex]
(0.356A)(0.7+0.3)ω² = (121590)A
0.356ω² = 121590
ω² = 3.42x10⁵
ω = 584 radians/second
However this is incorrect, so I need some ideas as to how to proceed or where I have gone wrong.
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