Wednesday, November 28, 2012

Physics of Nalgene Bottles


By Meghan Eisold

Many students at Colgate own Nalgene water bottles. I have been told that they are indestructible and that one would not even break if a car were to run over it. I decided to use physics to determine if this statement is true.  To begin, I calculated the cross-sectional area of the water bottle and the force that the car would exert on it. I used these to determine the stress on the water bottle.

Stress=F/A
A=cross-sectional area of Nalgene
L=20 cm (0.20 m)
W=9 cm (0.09 m)
A = L x W (contact area)
A=  0.018 m2

Force of average-sized car=ma
F=(1500kg)(9.8 m/s2)
F= 14,700 N

Stress =F/A
Stress= 14,700 N/0.018 m2
Stress= 816, 666 N/m2 = 816,666 Pa (0.8166 MPa)

Nalgene bottles are made of a material called Lenax (polycarbonate), which has a compressive strength of 12,500 psi. To put this in perspective, the compressive strength of a hard brick is about 12,000 psi. I compared this compressive strength to the magnitude of stress on the water bottle to determine if the bottle could withstand the force of the car.  

Compressive strength of Lexan: 12,500 psi.
1 Psi = 6 894.75729 Pascals

12,500 Psi x (6894.757 Pa)/(1 Psi) = 861, 844, 662. 5 Pa  
Compressive strength of Lexan = 861.4 MPa
861.4 MPa > 0.8166 MPa

Based on these calculations, the stress exerted on the Nalgene bottle by an average sized car is much smaller than the compressive strength of Lexan (the material from which Nalgene bottles are made). This means that the Nalgene bottle would not break under the force of an average sized vehicle. The actual results may vary somewhat due to the arched shape of the Nalgene bottle. 

Brownian Motion in the Brain: Looking at the Sodium Flow During an Action Potential


By: Chelsea Gottschalk
In class, we touched on Brownian motion, and how the flow of diffusion can be modeled by the equation:

Q = Dm/Dt = DA ((C2-C1)/L), where

Q= flow rate (Kg/m3)
D= coefficient of diffusion (m2/s)
A= cross-sectional area of substance (m2)
C2= concentration of substance on the outside of diffusion barrier (Kg/ m3)
C1= concentration of substance on inside of diffusion barrier (Kg/ m3)

            I thought it would be interesting to determine the mass of sodium that moves into a neuron during one action potential. It is no doubt a very small number, but it still represents an action that keeps us alive.

            Nerve tissue in the brain is one of the few excitable tissues in the body, a property that lends itself to fast communication within the vast network of neurons in the CNS as well as structures in the PNS. Signals are propagated throughout this network via action potentials, which start when a signal binds to a receptor on the dendritic surface, allowing some positive ions to flow into the neuron. This brings the negative resting membrane potential (-70 mV) up to its threshold of about -55 mV, which causes voltage-sensitive Na channels to open, leading to a massive influx of sodium (moving down its electrochemical gradient) into the neuron. This influx reaches a maximum of about 35 mV before potassium ion channels open, and positive charge moves out of the cell, repolarizing the neuron and preparing the cell for more signals.


      When sodium moves down its gradient, a large amount of ions flow into the neuron very rapidly. Since this is passive facilitated diffusion (no input from ATP), we can use the equation above to calculate the flow of Na into the cell. Interestingly enough, I found the diffusion coefficient for sodium in a rat brain, which is probably very close to that in humans. This value is D= 1.15 mm2/ms. The average thickness of a neuronal plasma membrane is 7 nm, and the atomic radius of Na is 180 pm. Additionally, the extracellular concentration of Na is 14mM, the intracellular concentration is 12mM, and action potentials last only about 1 ms. We can assume that extracellular sodium is right next to the plasma membrane, so the only distance that it must diffuse is across the membrane. First, let’s convert our units into SI units.

D= 1.15 mm2/ms : 1.0 x 10-12/mm2 : 1000 ms/1 sec = 1.15 x 10-9 m2/s

L= 7 nm : 1.0 x 10-9 m/ 1 nm = 7 x 10-9 m

R= 180 pm : 1.0 x 10-12/ 1 pm : 1.8 x 10-10 m

C: 1 mM : 1 M/1000mM : moles/ L : 1 L/ .001 m3 : 22.99 g/ 1 mol Na : 1 Kg/1000 g

C extracellular= 140 mM = 3.22 Kg/ m3

C intracellular= 12 mM = 0.276 Kg/ m3

Now for the calculation:

Q= (1.15 x 10-9 m2/s)(p(1.8 x 10-10 m)2)(( 3.22 Kg/ m3 - 0.276 Kg/ m3)/ 7 x 10-9 m)

Q= 4.9 x 10-20 Kg/s

Since Q = Dm/Dt, and we know that an action potential lasts approximately 1 ms, or .001 s, then we can solve for the mass of sodium that moves across the dendritic membrane during an action potential.

Dm/.001s= 4.9 x 10-20 Kg/s

Dm= 4.9 x10-23 Kg of Na moved during one action potential!

This value may seem incredibly small, and it is. It only takes a very small percentage of the total extracellular concentration of sodium to flow into the neuron for the membrane potential to reach threshold. That is why a neuron could fire for several hundred or a thousand times (without a working Na/K pump) before a noticeable change in concentration would occur.

Although Q= A x v, this is not a closed system, so I do not think we can accurately determine the area over which this diffusion occurs (especially since the surface area of dendrites is so great), meaning that we can not determine the velocity of the sodium ions.

Reference:

Goodman, J. A., Kroenke, C. D., Bretthorst, G. L., Ackerman, J. J. H. and Neil, J. J. (2005),                   Sodium ion apparent diffusion coefficient in living rat brain. Magn Reson Med, 53: 1040–1045. doi: 10.1002/mrm.20444

Physics News – Milk Protest in Brussels


By Kate Allaway

On Monday, November 26th, thousands of dairy farmers coordinating with the European Milk Board (EMB) came to Brussels, Belgium to protest low milk prices in the European Union. EU milk is often sold below the production price, forcing small farmers out of business. In their demonstrations, which are ongoing, protestors are spraying fresh milk out of large hoses onto policemen and the European Parliament building, preventing many EU employees from getting to work.

I wanted to determine the approximate volume rate of flow out of these hoses. This is most easily accomplished using Poiseuille’s equation (Q = (πR4ΔP)/(8ηL)).

I made the following assumptions about the hose based on the pictures:
R=5 cm (0.05 m)
Δheight between ground and person holding the hose=1.5 m
L=15 m

I found the following values for milk:
ρ= 1030 kg/m3 (for whole milk, from http://physics.info/density/)
η= 3 mPas = 0.003 Pas (from http://physics.info/viscosity/)

First I found the ΔP:
ΔP = ρgΔh = (1030 kg/m3)(9.8m/s2)(1.5 m) = 15,000 Pa
Then plugged in to find Q:
Q = (πR4ΔP)/(8ηL) = (π(0.05 m)4(15,000 Pa))/(8(0.003 Pas)(15 m ))
Q = 0.8 m3/s

With that volume rate of flow and the number of protestors spraying milk with hoses, it is clear why many EU employees are struggling to get past the mayhem to their offices.

OLD FAITHFUL


By Rachel Walsh


Old Faithful is a geyser in Yellowstone National Park that is powered by an instrusive pluton that is believed to be at a depth of 2.4 km to 4.8 km below the surface. The water is heated, and pumped through a plumbing system up through many geysers, one of which is old faithful. The height the water can reach is up to 56m. The duration of each eruption is 1.5 to 5 min. and the volume of water can reach up to 32 m3 . I am going to investigate the rate, velocity of water at the hole and the possible pressure by the pluton.

Assumptions:
-                    the plumbing is one vertical pipe of equal diameter through all 3.6 km (average) of the earth to the pluton.
-                    The eruption is at a constant rate.
-                    The change of temperature is irrelevant
-                    The pressure at the opening of the hole is atmospheric

Radius of opening: 0.2m
Possible viscosity of brine:1.3 x 10-3 Pa *s (Francke et al)
Density of brine: 1090 kg/ m3

Rate:
            Q= v/t  to 32 m3 / 300s = 0.11 m3/s


Velocity of water:
            pgh = ½ pv2  to (9.8 m/s)(56m) = ½ v2
                       
            v = 23.4 m/s

Change in pressure from pluton to surface:

            Q = pR4(deltaP)/8hL à   0.11 m3/s = p(0.2m)4(deltaP)/8(1.3x10-3 Pa*s)(3600m)

            deltaP = 1100 Pa

Sources:
 Francke, Henning. "Density and Viscosity of Brine: An Overview from a Process Engineers Perspective." Sci Verse. Cheme Der Erde - Geochemistry, n.d. Web.

Monday, November 26, 2012

How Cats Always Fall on their Feet


By Danielle LaPaglia
Whenever cats fall, they are almost always able to orient themselves so that they land on their feet. They are able to do this because they have a very flexible spine and can rotate the front half of their body separately from the back half. First, the cat tucks in its front legs to rotate the front half of its body quickly. This works because it decreases its moment of inertia and causes angular velocity to increase. Then, the cat stretches out its front legs and tucks in its back legs so that the back half of its body can follow suit. The cat is then in position to land feet first on the ground. The conservation of angular momentum (L=Iω) allows the relationship between moment of inertia and angular velocity to be true. No external forces are acting on the cat when it is falling, therefore its momentum will be constant.
Fd
 
Cats can also land safely because it has been said that they do not have a deadly terminal velocity. To figure out what the terminal velocity of a cat is, all the forces acting on the cat must be equal to zero.




Assumptions:
m=4.5 kg
Cd=0.4
r=.125 m

Fd= ½ Cd ρ A v2
Fg=mg
mg=½ Cd ρ A vmax2
(4.5 kg)(9.8 m/s2)=1/2 (0.4)(π*(.125m)2)(1.2754 kg/m3)(v2)
v=60 m/s

The moment of inertia of a cat with its legs extended would be (assumed to be a cylinder):
I=1/2mr2
I=1/2(4.5 kg)(.0625m)2
I=.0087 kg m/s

As opposed to having it’s legs tucked in (r increases):
I=1/2(4.5 kg)(.125 m)2
I=.035 kg m/s

This shows how if angular momentum is kept constant, this change in moment of inertia can have a big effect on angular speed.

Sunday, November 25, 2012

The Physics of a Trapeze


Trapeze artists gained popularity by performing at the circus, swinging high above the ground.  The trapeze and trapeze artist can be modeled as a simple pendulum because the artist is basically a mass at the end of a string, which in this case is the trapeze made of a horizontal bar hanging by two cables.  I assumed that the trapeze artist doesn’t move his body, but hangs straight in line with the middle of the bar.  I wanted to determine the maximum speed that the trapeze artist moves.  Air resistance was ignored.

I estimated the mass of the trapeze artist to be 65 kg and the mass of the horizontal bar to be 4 kg (the mass of the cables is negligent).  Therefore, the mass of the trapeze system is 69 kg.  The cables connecting the horizontal bar to the ceiling are 4.2 m long and the trapeze artist is 1.6 m tall.  The trapeze artist starts by standing on a platform.  I assumed that the platform was 4.5 meters above the floor.  The man is modeled as a point mass and his center of mass is assumed to be half of his body length.  Therefore, the distance of the trapeze system is 4.5 m + (1.6 m/2) = 5.3 m from the floor.

The trapeze artist starts by standing on a platform at the highest point of his swing.  Therefore, while standing on the platform, at rest, the trapeze artist has his maximum potential energy.

PE = mgh = (69 kg)(9.8 m/s2)(5.3 m) = 3.6 x 103 J

When the trapeze artist swings to the lowest height in his swing, he is at his maximum speed and his maximum kinetic energy.  The trapeze artist is at his lowest potential energy here.  I assumed that there are no nonconservative forces.  I also assumed that the trapeze system was now at a height of 4.6 m from the floor.











One trick performed on a swinging trapeze, which is a trapeze in which the trapeze artist starts from rest and swings the trapeze himself, instead of creating the swing by walking off the high platform (this is a flying trapeze), is the stand seats-off.  In this trick, the trapeze artist first stands upright on the bar, swings, and eventually finishes hanging upside down from the bar with his feet.  I assumed the angle of the trapeze artist’s swing to be 50°.  To determine the maximum speed at which the trapeze artist swings, I used the same formula as above.  This time the height is the height of the “pendulum” or the trapeze system, which is the length of the cables plus half of the performer’s height (4.2 m + (1.6 m/2) = 5.0 m.  The performer starts from rest.




















Now I want to find the maximum velocity of the trapeze artist after he starts falling.  I estimated that the performer fell 2.2 m.  First, I have to find the potential energy before the performer falls.  I ignored the mass of the bar.
PE = mgh = (65 kg)(9.8 m/s2)(5.0m – (5.0 m * cos(50°))) = 1137 J
The PE at the bottom of the fall is:
PE = mgh = (65 kg)(9.8 m/s2)(-2.2 m) =  -1401 J
The PE is negative because the man is falling down.










The trapeze artist has a large increase in velocity when he performs this trick.


Sources:
http://newyork.trapezeschool.com/resources/inv1.php
http://www.flying-trapeze.com/The-Physics-of-Flying-Trapeze/the-swinging-trapeze.html

The force exerted on the rugby ball by Francois Steyn (The longest rugby conversion kick)


Tinashe Nyanhete

Steyn’s record 66m kick.  http://www.youtube.com/watch?v=muI0ULBSWmw
Horizontal distance (x) = 68m
Time of flight of ball (t) = 4s
Vxo = x/t = 68m/4s =17m/s
Vy0 = gt = 9.8ms-2 * 4s = 39.2m/s
V = 43m/s
The approximate change in velocity of the ball after Francois struck the ball was 43m/s, which is higher than the normal reported value of 30m/s, since the ball travelled the highest ever recorded horizontal distance for a conversion kick.
Now to determine the force on the ball due to the kick (Use Newton’s 2nd and 3rd laws):
F = ma, and action = -reaction (Force exerted on ball = force exerted on the foot by the ball)
F = mv/t
Ft = mv, where t – time of contact between the foot and the ball; v – change in velocity of ball; m – mass of ball
Time of contact with the ball = 0.013s
Mass of ball = 0.44kg
Change in velocity of ball = 43m/s
Therefore, F = mv/t = 1455N
This is larger than the normal estimated force of 1015N, which explains why Francois Steyn’s ball travelled a record distance.
Reference: http://www.coachesinfo.com/index.php?option=com_content&id=228&Itemid=147