# How to find the Angular Velocity?

1. Jul 8, 2013

### Rococo

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$pV = nRT$

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

1 atm = 101325 N/m²
1.2 atm = 121590 N/m²
R = 8.31 Jmol-¹K-¹

3. 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):

$pV = nRT$
(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:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
T2 = (T1 x P2)/P1
T2 = (298 x 1.2)/1
T2 = 357.6

$pV = nRT$
(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

$mrω² = PA$
(0.356A)(0.7+0.3)ω² = (121590)A
0.356ω² = 121590
ω² = 3.42x10⁵

However this is incorrect, so I need some ideas as to how to proceed or where I have gone wrong.

Last edited: Jul 8, 2013
2. Jul 8, 2013

### voko

The gas in the bag experiences centrifugal force, not unlike the gas in atmosphere experiences the force of gravity.

3. Jul 8, 2013

### Rococo

Thank you for the response

So, is it then correct to say:

F = mrω²

Where
F = Force experienced by gas in the bag at point of breaking
m = mass of gas in the bag at point of breaking
r = distance from bottom of bag to the center of the circular motion
ω = angular velocity of the bag at point of breaking

4. Jul 8, 2013

### voko

This will work only as a rough approximation. Observe that in our atmosphere, pressure changes with altitude. You should have the same effect here, except: it should change not vertically, but horizontally, along the axis of the bag; it should change stronger, because gravity is approximately constant in atmosphere, while the centrifugal force changes.

5. Jul 8, 2013

### Rococo

I see, so is there an equation I need to use that shows how the pressure of the air in the bag changes with distance along the axis?

6. Jul 8, 2013

### voko

I am afraid you are supposed to derive the equation here. Are you familiar with the barometric formula? It should be quite similar.

7. Jul 8, 2013

### Rococo

I'm not familiar with it although I have looked it up: $P_h=P_oe^\frac{-mgh}{kT}$

I have tried to come up with an equation relating the pressure, P, to the distance along the cylinder, r. But it does not look similar to the Barometric Formula so I think I must be missing something.

Pressure at a point in the cylinder = Force at that point / Cross sectional area of cylinder

$P = \frac{m(0.7+r)w^2}{A}$

Where m is the mass of the gas from the opening of the cylinder to that point.
m = Vρ
m = Arρ

$P = \frac{(Arρ)(0.7+r)ω^2}{A}$
$P = rρ(0.7+r)ω^2$

The density of air, ρ can be calculated. Now seeing as the question wants me to find ω when r = 0.3m and P = 1.2atm, can this equation not be used?

8. Jul 8, 2013

9. Jul 8, 2013

### Rococo

I could follow it, although I haven't encountered the equations $P=ρgh$ or $dP = -ρgdz$ before.

Would this be the correct adaptation:

$dP = \frac{dF}{A}$

$dP = \frac{mω^2dr}{A}$

And as $P = \frac{ρRT}{M}$
(where M = molar mass)

Then dividing both sides by P to form a differential equation:

$\frac{dP}{P} = \frac{Mmω^2}{AρRT}dr$

10. Jul 8, 2013

### voko

Not exactly. It is correct that you changed the sign. Now, what should you have instead of g? Observe that g is the acceleration due to gravity.

11. Jul 8, 2013

### Rococo

The gas is experiencing an acceleration due it its circular motion, equal to rω^2, as opposed to g, the acceleration due to gravity.

So is the correct equation just:

$dP = ρrω^2 dr$

12. Jul 8, 2013

### voko

Yes, this is correct. Now continue the way it was done with the barometric formula.

13. Jul 8, 2013

### Rococo

Since $P = \frac{ρRT}{M}$

Then dividing both sides by P:

$\frac{dP}{P} = \frac{ρrω^2M}{ρRT} dr$

$\frac{dP}{P} = \frac{rω^2M}{RT} dr$

$\int_{P_0}^{P} \frac{dP}{P} = \int_0^{r} \frac{rω^2M}{RT} dr$

Where:
P = Pressure at a distance r from open end of the bag
Po = Pressure at the open end of the bag (1 atm?)
r = Distance from open end of the bag
M = Molar mass of air

Working this out I get:

$P = P_0 e^\frac{Mω^2r^2}{2RT}$

Is this correct?

14. Jul 8, 2013

### voko

Good, but a quibble. $\omega^2 r$ implies that $r$ is the distance from the center of rotation, not from the open end of the bag. So your right-hand integral must be from $r_0$, not from $0$.

15. Jul 8, 2013

### Rococo

Thanks, so is this now correct:

$P = P_0 e^\frac{Mω^2(r^2 - r_0^2)}{2RT}$

And so would $r_0$ be equal to 0.7m, the length of the girl's arm?

16. Jul 9, 2013

### voko

Looking good!

17. Jul 9, 2013

### Rococo

Thanks, I've solved the problem now and found the correct value of ω using the equation.