Friday, November 30, 2012

Portals in Ocean and Sky

Portals in Ocean and Sky

The video game Portals is centered around the use of the Aperture Science Handheld Portal Device, or portal gun. The gun can fire projectiles that upon impact on a flat surface expand into either an orange or a blue portal. By entering one portal, the player can exit from the other.

For this blog post, I am taking inspiration from two other weeks. In this post, I will be looking at the flow rate of water if the portal gun created a portal at the bottom of the Mariana Trench and then, taking a flight up to the stratosphere like Felix Baumgartner did with Red Bull, fired another portal onto the balloon.

Intuitively, we know that the pressure at that height is very low whereas the pressure at that depth is very large. Thus, the water should flow from the trench portal out of the stratosphere portal.
To understand this, we will use Poiseuille's equation.

Since the portals make it so that a path from the ocean to the sky is not a straight line from a to b the idea of height, y, is made strange. One idea is that the portal is worm-hole-like. The two portals may be connected by an infinitesimally small tunnel of near zero length L.  So, the flow rate would be infinite. For the sake of argument, let us assume the tunnel is 10^-6 m long.

The portal can accommodate a young woman and have about 2 feet in height left. The average height of a 20+ American femaile is 5 ft 4 in according to Wikipedia. We will assume the portal can be approximated as a circle (it's really an ellipse) and use 1.9 meters as the diameter. The radius shall be 0.95 meters. The viscosity of water near 0 degrees Celsius will be 1.8*10^-3 Pa*s.

The Mariana Trench has a pressure of about 1086 bars, or 108,600,000 Pa according to Wikipedia. The pressure at that height is very low and we shall assume it is zero.Setting up the equation,

The resultant answer is 6.8 x 10^16 m/s . This is overly large due to the portal's small length and our assumption that the flow is not turbulent. The volume flow rate is about 1.9 x 10^16 m^3. 

Can a horseback riding incident cause a herniated disc?

Analyzing the cause of a herniated disc:  What incident to blame?
By Erin Krysinski
                  My mother recently underwent surgery to correct a herniated disc that had ruptured in her neck.  Though she has had neck problems for years, my father, as per usual, blamed the rupture on a particular incident that occurred while my mother was riding our trusty thoroughbred.  My father has taken a liking over the years to blaming most things on the horse, money, injuries, etc, probably because horseback riding is a rather expensive and dangerous hobby.  The event occurred as follows:  My mother was cantering into a fence, and misjudged the distance.  While she continued forward, the horse decided to add in another stride for safety reasons, and in doing so, significantly reduced his speed within a very short amount of time (deceleration).  My mother as a result, felt a sort of whip lash as she quickly was forced to decelerate with the horse beneath her (see these other examples: ).  Of course, various other incidents occurred that same week, including lifting heavy dogs (my mother is a veterinarian), and moving wheelbarrows at the farm, so it was difficult to pinpoint which incident was the herniation perpetrator.  My mother and even more so, my father, seem convinced that that horse jumping incident was the cause.  I on the other hand, would like to prove otherwise.  I hypothesize that a single incident of whiplash in this case, is not capable of rupturing an intervertebral disc.
                  In order to determine whether the horseback event was capable of causing the rupture, I first looked at the force on my mother when the horse quickly decelerated.  To find this force, I used the impulse equation,  ∑Fx=∆p/∆tà (m(v2-v1))/∆t=∑Fx , to find the sum of the forces on my mother in the x direction.  I estimated that as the horse added an extra stride, and thus almost stopped, decelerated from v1= 6.94 m/s (an average canter speed) to v2=1.39 m/s (an average walk speed) in t=0.25 sec.  The mass of my mother + the horse= 1115 lbs, or 505 kg.  If we plug this into the equation, (505kg(6.94m/s-1.39m/s)/.25sec =∑𝐹.  In this case, ∑F=11,211 N in the x direction. 
                  To find whether this force is enough to herniate her disc, we need to first find the ultimate shear strength of a intervertebral disc, and then find the minimum force that would cause a disc to rupture.  From “Spine Biomechanics” by Rapoff, we estimated that the ultimate shear strength of an intervertebral disc is about 17.6 N/1mm or 1.76x107N/m2 .  To find the minimum force that would break a disc, we can use the equation: ultimate shear strength=F/A.  To find the area of the disc, we use the average disc radius (.02m, estimated from and then use the circle area equation (A=2πr2) and then get that A=0.00126 m2.  From here, we can plug in our ultimate shear strength and our area to the above equation to find the minimum force applied for rupture.  Therefore, 1.76x107N/m2=F/0.00126.  F=22,116.81 N.  So therefore, the shear force would have to be greater than 22,116.81 N to herniate the disc.  Because our F is only 11,211N, it’s not likely that the horse event alone ruptured the disc, rather a series of events over many years broke down it’s shear strength, until a smaller event could
                  In reality things are a bit more complicated than this.  We would need to break the force on the rider down further and find what force is actually applied to the disc itself.  We can do this by looking at the neck as a rotating rod with a sphere on top, and can view the deceleration as a torque applied to the seat of the rider, since that’s where the deceleration occurred.  Because we know the deceleration of the horse, we could then use atan=rxalpha to find radial acceleration, and then use kinematics to find radial velocity as well as theta.   We could also find omega via w=v2/r and then use kinematics to find theta.   Regardless we can then use the conservation of energy equation, Wnc=∆KE+∆PE, or in this case, ∆1/2 I w 2 +∆1/2mv 2 + ∆mgh= (torque)(theta)  to find the torque applied.  With torque, we can then use Torque=rFsin(theta) to  find the force applied.  Another way to find the force applied to the disc, would again be the conservation of momentum theorem, this time rotational.  If we calculated I for the person on the horse as a rod with a sphere on the top and consider the axis of rotation their seat, after finding w1 and 2 as illustrated above, we can then use Iw1=Iw2 to find the change in rotational momentum, ∆L.  Because ∑torque=∆L/∆t, we can then find the sum of the torques, which in this case is only one at the seat, and then as illustrated above, find the force applied via torque=rFsin(theta).  We’ll save these calculations for another time, but since the total force of the system wasn’t enough to rupture the disc, we’ll also assume that the torque on the body alone won’t be enough either.   

The physics of a subway tunnel plug

The physics of a subway tunnel plug: an analysis for the skeptic
By Erin Krysinski

Following the flooding that threatened NYC subway systems after hurricane Sandy, scientists are now looking for new ways to block flood waters from entering the city’s transit systems in case of emergency.  One option is a new inflatable 16 ft radius “plug”, which may be used to block flood waters within the subway tubes themselves.  Like most physicists, I’m skeptical if the plug could actually work, and if so, how much pressure it would have to be able to withstand from the incoming water flow.  To find the pressure in the subway tunnels from the water, we must use Bernoulli’s equation to find the difference in pressure between the water entering the stairwells(P1) and the water flowing through the tunnel itself(P2).  Because the plug will be in the tunnel, we will be searching for P2.  However, because there are several factors we don’t know, specifically the speed of water flowing through the tunnel, we must first make some estimations about the speed of the water entering the stairwell(v1), and then use the equation of continuity to determine the speed through the tunnel (v2).  If the water is entering the stairwell at an average of 4.02 m/s (v1), and the radius of the stairwell is 1.83m, the Area of the stairwell(A1)=πr2 or A1=10.52 m2.  We can use the same equation to find the Area of the subway tunnel(A2), which has a radius(r2) of 4.88m, and thus A2=74.82 m2.  Now, with the equation of continuity (A1v1=A2v2), we can find v2, or the speed of the water in the tunnel.  10.52m2x4.02m/s=74.82xv2, v2=0.57 m/s.  We now know v1, v2, as well as that the stairwell is about 1m below the surface (y1), the tunnel is about 9.14 m below the surface(y2), and the pressure of the stairwell(P1) is 1013x105 N/m2  since it’s open to the atmosphere.  Using Bernoulli’s equation (P1+1/2pv12+pgy1=P2+1/2pv22+pgy2à1.013x105N/m2+1/2(1x103kg/m3)(4.02m/s)2+(1x103kg/m3)(9.8)(1m)=P2+1/2(1x103kg/m3)(0.57m/s)2+(1x103kg/m3)(9.8)(9.14m).  Therefore P2=29,445.75 N/m2.  Though the plug would have to with stand this pressure, in reality, the pressure of the water held back may be far greater, since water is probably entering from various other areas other than the stairwell.  Generally, there would be a lot of water to hold back, so if the plug were to break, the force of water behind could be quite dangerous.  Despite the fact that flood walls or other types of barriers may be expensive to build, they may be a safer option in the long run.  

Physics of Tom Petty’s Free Fall

 By Jessica Mondon

In the song “Free Fallin’” Tom Petty repeats the words “I’m free falling” a lot. I
decided to look at the physics of this free fall. I want to see how far he would free
fall if he fell for the whole time that he said he was free falling. Then I want to see if
he produces enough power with this fall to light the Christmas tree at Rockefeller
Center. He says that he’s free falling for two minutes and 28 seconds (148 s) in the
song. I am assuming he starts at rest and since he is free falling, the only force acting
on him is the force of gravity. Also, since he is free falling the acceleration is just the
acceleration due to gravity, or 9.8 m/s2. Using kinematics we can find the distance
(Δy) that he falls:
Δy = vot + ½ at2
Δy= ½ (-9.8 m/s2) (148 s)2
Δy = -107330 m

I looked up Tom Petty’s weight, and he is approximately 75 kg. This means the force
with which he is falling is:
F = mg
F = (75 kg) (9.8 m/s2)
F = 735 N

I can then calculate the work done by gravity, which is the total work for the system
because gravity is the only force. We now know the force of gravity, and the distance
he free falls so:
W = FdcosΘ
W= (735 N) (-107330 m) cos(270)
W = 7.9 x 107 J

Now that we found work, and we know time we can calculate the power with which
he falls:
P = W/t
P = (7.9 x 107 J)/(148 s)
P = 533022 W

Now that I found the power with which he falls I need to find the power it takes to
light the tree in Rockefeller Center. I found that there are roughly 30,000 bulbs on
the tree, and each bulb is 7.5 W, so the power needed to light the tree is 225,000 W.
The power with which Tom Petty fell is over twice this, so he fell with enough power
to light the tree in Rockefeller Center twice.

The Physics of the Adhesive Duck Deficiency

By Beth Shore

I love the show the Big Bang Theory, and in it there is a lot of physics both
shown and talked about. In one episode, Penny falls in her shower and dislocates
her shoulder. This, according to Sheldon, was due to the lack of adhesive ducks on
the tub.

So I wondered, just how important are adhesive ducks (or any other stickers) in preventing shower injuries? By how much do they increase the friction coefficient, providing a safer shower experience? The ASTM F-462 (American Society of Testing and Materials) requires that tubs have a static coefficient of friction of at least 0.04 1. Assuming Penny weighs 55.0 kg, we can calculate the angle at which her force due to gravity.

The vertical components, since she is not accelerating in this direction, sum to 0. 
Fn - Fg cosθ = 0
Fn = Fg cosθ

And, since we don’t want her to slip in the x direction either, we set this
acceleration equal to 0 as well. This is a situation known as static equilibrium.
Fgsinθ - Ffriction = 0
Fgsinθ = Ffriction

Ffriction = μ Fn
Fgsinθ = μ Fg cosθ
sinθ = μ cosθ

sinθ / cosθ = μ
μ = tan θ
tan θ = 0.04
θ = 2.29o

This isn’t a very large angle, thereby making the shower a very dangerous place without anything to increase the coefficient of friction. There are no specifications for the amount that adhesive stickers, such as the whimsical ducks Sheldon talks about, increase the static coefficient of friction. So, what would the coefficient need to be so that Penny could take a large step in the shower, say 30o, without falling? Using what we found before, tan (30) = 0.58. This is a significant increase from the static coefficient in a tub without such protective measures.

Next, I decided to look at the force that was exerted on Penny’s shoulder to
dislocate it. We know that the moment of intertia of a long uniform rod rotating about the end is 1/3 ML2. Assuming Penny is 5’6’’ and weighs 120 pounds, this comes out to be 1/3 (55 kg) (1.7m)2 = I
I = 53 kg m2
T= I α sin θ

Because her initial angular velocity was 0, and we can assume the fall took approximately 1 second, and she rotated 90 degrees, we can calculate the angular acceleration using the equation Δθ=ωot + (1/2) α t2
Π - π/2 = (0 rad/s)(1s) + (1/2) α (1 s)2
π/2 * 2 = α
α = π rad/s2

Using our torque equation, T= I α sin θ
T= (53 kg m2) (π rad/s2) sin (π/2)
T = 167 Nm

To figure out force, we have to refer to the equation
T = r F sin θ
167 Nm = (.85 m) F
F = 196 N

Which is a very large force. So, had Penny added adhesive ducks to her hower floor; she could have saved her shoulder a lot of force from the fall.

Physics of a syringe

By Tue Nguyen

I want to look at how much force the liquid is entering the body with from the tip of the syringe when a nurse injects some drugs into someone’s vein.

I assume that the force applied is 5 N, which is a good estimate of the force applied onto the liquid.
I assume that the radius of the base of the syringe is 0.5cm = 0.005 m. The radius of the needle tip is
0.5mm = 5x10^-4m.

According to Pascal principle, F1/A1=F2/A2 F2=F1*A2/A1= 5*π*(5x10^-4)^2/(π*(5x10^-3)^2)=0.05 N. This is a very small amount of force so it ensures that the liquid coming in does not do any damages to the cells circulating in the blood.

Vasovagal Reaction

By Danielle Scheer

In the summer of 2008 I was diagnosed with neurocardiogenic syncope, a condition
described as being “a temporary loss of consciousness associated with a drop in
arterial blood pressure, quickly followed by a slowed heart rate (1). I wanted to know
the physics of what is going on:

Because we are considering a fluid flow, and assuming non-viscous liquid:

Q= πR4 (P1-P2)/8ηL
In a vasovagal response, a person’s veins dilate when they should constrict, so
assuming Q stays constant, ΔP will decrease as radius increases.

P1 + ρgy1 + ½ ρv12 = P2 + ρgy2 + ½ ρv22

If the pressure is decreased, the blood won’t be able to get as high up, which is why
people with vasovagal conditions lose consciousness (lack of blood and oxygen to
the brain).

(1) Grubb, B.P. & McMann, M.C. (2001) The fainting phernomenon: Understanding why
people faint and what can be done about it. New York: Futura Publishing Company,
Inc. p. 133

The Collapse of the Silver Bridge (A Not-So-Happy Physics Tale)

By Madison Daly

On December 15, 1967 around 5 pm, The Silver Bridge (which connected Ohio and West
Virginia) collapsed. The bridge was suspended by eyebar chains (an eyebar is a long
steal plate with a large circular end with a hole through which a pin is used to connect to
other eyebars and make a chain). The collapse was due to a fracture in an eyebar induced
at least partially from the cold. What I want to figure out is what was the change in
temperature and the change in volume the steel bridged experienced, causing it to crack
(assuming the crack was purely temperature induced and not from corrosion/stress from
car weight).
The Silver Bridge eyebars were around 13.7- 16.8 m long; 300mm wide, varied in
thickness (but average thickness of typical eyebar was around 25-50 mm, so we’ll
approximate around 35 mm). Assuming the eyebar is a perfect rectangle, its volume is
approximately 14.5mm*300mm*35mm= 0.0145m*0.300m*0.035m= 1.52 x 10-4 m3


βsteel= 33.0
αsteel= 11.0
Ysteel= 200 x 109 Pa
Ultimate strength of steel (max)= 550,000,000 Pa

550,000,000 = (200 x 109) (11.0) ΔT
ΔT= 2.5 X 10-4 K

ΔV = (33.0) (1.52 x 10-4) (2.5 X 10-4) =1.3 x 10-6 m3.

Why are the geese still here?!

By Jon Ranieri

Every winter I am here at Colgate I am dumbfounded by how long the geese stay here. They swim in the freezing lake, and waddle around on the ice when Taylor Lake has frozen over, constantly pooping on our sidewalks. Well, I wanted to look into how these geese don't get frostbite and die (even though I wish they did).

Geese are warm blooded, but they are generally fine in really cold temperatures because their feathers provide amazing insulation. They have also adapted similar 'shivering' mechanisms to keep their body temperature high. More shivering means higher metabolic rate, which, unfortunately means more eating... and pooping. Ok so fine, their core can stay warm, but what about those little feet? They have no insulation, but geese seem to not mind standing on the ice for hours at a time, staring down their next grass plot to attack.

Lets take a step back. Why do we get frostbite? Because our body determines that sending blood to our extremities (which are poorly insulated and COLD), is not worth the cooling effect that blood will have on our core. Since we need to maintain a high body temperature to carry out normal bodily functions, our body chooses life over fingers and toes.  

How do geese do it? They have something called countercurrent heat exchange. In their legs, geese have arteries and veins positioned very close to each other. Arteries bringing warm blood to the feet, and veins bringing cold blood to the heart. BUT, we are trying to avoid the whole cold blood to the heart part. SO, the cold blood from the feet is warmed by the warm blood going to the feet, so the returning blood is actually somewhat warmer, so the goose doesn't mind sending blood down there. And the warm blood going to the feet is cooled by the cold blood coming from the feet, but thats OK, because the feet are very exposed and are going to be cold anyway. Usually, geese feet are just above freezing, while the core is around 100 F. Its all thanks to thermo equilibrium that allows the geese to stand on ice and snow.

Thanksgiving Balloons

By Mike Jones

After watching the Macy’s Thanksgiving Day parade I started thinking about the relative physics involved with keeping balloons from flying away.  I noticed that each balloon has many strings held down by many bodies, and was curious as to the approximate force needed by each individual to keep the balloon from flying away.
I found this video and used the volume of this new balloon (14,000 ft^3= 396.44m^3) (  I approximated 30 strings coming off the balloon, ρair=1.2041kg/m^3, and ρHe=0.1786kg/m^3.  Solving for the buoyant force:
FB=ρVg=(ρair - ρHe)Vg= (1.2041kg/m^3 - 0.1786kg/m^3)( 396.44m^3)(9.8m/s^2)
FB= 3,984.18N or approximately 4,000 N.
Assuming the buoyant force is equally distributed over the 30 attached strings then the approximate force each person exerts on each string is 4,000 N/30= approximately 133 N per string. 

Hybrid Cars

           By Andrew Long

           When going down a hill, you have to hit the breaks in order to not quickly accelerate (this creates energy lost as heat, keeping kinetic energy constant while decreasing the potential energy of the car). When you go back up a hill, you have to put a significant amount of energy to keep the kinetic energy constant while increasing the gravitational potential energy. I wanted to see if there was a way that you could reduce the amount of energy needed to go up and down a hill.
It turns out that, using a flywheel; it is ideally possible to spend no energy going up and down two equivalent hills. Using a flywheel in a car, as you go down a hill and apply the “breaks” to the car, you slow down the car (as potential energy turns to kinetic energy) by applying torque to the flywheel. As the flywheel spins, the wheel can create energy in the form of electricity. This is what they use in hybrids like the Toyota Prius.

Sample calculations

mgh=(1/2)mv02+electricity stored by the flywheel.
(6000kg)(9.8m/s2)(60m) turns completely into electrical energy. This energy is used to get back up the hill.
(1/2)(6000kg)(13.4m/s)2 this amount of energy is kept the same throughout the entire ride down the hill.

The Physics Behind The Turntable Stylus

The Physics Behind The Turntable Stylus

     Alright, so I'm somewhat of a nerd when it comes to music, and my preferred method of listening is vinyl on a turntable. After a couple of years of collecting albums and having to maintain my equipment, I've become curious about the kinds of forces that act upon records and needles. So, here's a look at the physical world of analog music.

     For the sake of brevity, I am going to stick to LP's, which rotate at 33 1/3 revolutions per minute and have a radius of 6 inches, or 0.1524 meters. 

     The angular velocity of an LP is calculated as follows:

velocity(angular) = (33.33333 rev/min)(2*Pi radians/revolution)(1 min/60 sec) = 3.49 radians/sec

     At the outermost point of an LP, the linear velocity is:

v = (omega)(radius)
   =(3.49 radians/sec)(0.1524 m) = 0.531876 m/s
                                              approx. 0.532 m/s

     Now, I would like to be able to calculate the kind of force the stylus exerts on a vinyl record. First, I need to calculate the radial acceleration of the LP as it plays on a turntable.

a(radial) = (omega)^2(radius)
              = (3.49 radians/sec)^2(.1524 m)
              = 1.856 m/s^2

     The force itself is a bit tricky to calculate. Typically, a turntable's tonearm is like a fulcrum, such that it is balanced so that only a portion of the tonearm's weight is applied to the record itself (this is called a tonearm's effective mass). This value varies drastically between turntables, anywhere from less than 10 grams to more than 25 grams. After some research, I've approximated my tonearm's effective mass to be roughly 16.5 grams (0.0165 kilograms). We know force is mass times acceleration, so:

Force = (0.0165 kg)(1.856 m/s^2)
          = 0.030624 Newtons

         approx. 0.031 Newtons

     This may seem like a small force, but this force is concentrated at a very small point (the stylus tip) and acts constantly on this point for an entire album (which is roughly 23 minutes per album side). So even though this seems like an insignificant force, it really begins to add up after playing albums for long periods of time. Styluses are generally made of diamond, so it may seem surprising that these tips even wear out at all. But because the area of contact between the diamond tip and the vinyl groove is so tiny, the forces that act upon the stylus are massive in comparison. Interestingly enough, there is a tremendous amount of heat generated at the contact point between the stylus and record, and the particles that are found in a record groove is most commonly silicon dioxide, a substance nearly as hard as diamond. So over time, these interactions will actually wear down the "needle" quite significantly. Whoever said "Diamonds are forever" clearly did not own a turntable!

-Ian Vannix

The Physics of Hurricane Sandy

By Elizabeth Morea

Hurricane Sandy was a very strong and powerful super storm that hit my coastal hometown in New Jersey. The destruction caused by the hurricane was unbelievable. Entire buildings and trees were ripped out of the ground due to the high wind speeds. In order to cause this much destruction I decided to find out an estimate of what the energy of the hurricane was. To do this I first had to find the  inertia of the hurricane. I estimated the hurricane to be a cylinder of air with the each area of the cylinder having the same angular velocity. According to the website “Our Amazing Planet”, the super storm had hurricane-force winds of about 120km/hr, as a level 3 hurricane, extending up to 175 miles from the center of the storm.  Based on data from NASA, the hurricane had a height of about 13km.  Because of the assumption that the hurricane is a cylinder the equation for the moment of inertia is ½ MR2. The mass of the hurricane can be found by multiplying the volume of the cylinder times the density of the air in the hurricane.

Density of Air= 1.3kg/m3

Using this mass the Inertia can be found and thus the kinetic energy of Hurricane Sandy can be found.

KE= (1/2)Iw2= (1/2)(1/2)(M)(R2)(voutside/R)2= (1/4) (M)(v2outside)

      KE=(1/4) (4.16x1015kg)((120km/h)((1km/h)/(3.6m/s)))2=1.16x1018J

Hurricane Sandy had an energy of  about 1.16x1018J. This was the rotational kinetic energy of the storm and in reality was probably even greater with the added energy contributed by the storm surge of water it created when it hit my coastal hometown in New Jersey. 

Physics of a Fouetté Turn in Ballet

Physics of a Fouetté Turn

      In ballet, there are several types of turns a dancer can do. One of the most difficult types is a Fouetté. The video below describes the motions involved in such a turn. 

As the dancer turns, she brings her leg inward which decreases her moment of inertia. This in turn increases her angular velocity. I decided to calculate what the dancer's velocity would be right after she brings her leg in. To do this, I timed how long it took the dancer to turn with her leg extended. She completed one turn in 0.8 seconds, which is about 2.5*pi rad/s. Additionally, I considered the dancer with her leg close to the body as a uniform cylinder, giving a moment of inertia as I=1/2MR^2 (Even though the dancer does not bring her leg all the way in, I did this to simplify my calculations). I considered her leg to be a uniform rod rotating about its end. 

Here are my knowns:

Radius of body= 0.1524 meters
Mass of person= 60 kg
Mass of leg= 6 kg
Length of leg= 0.762 meters
w2=7.85 rad/s

To solve for the angular velocity of the dancer with her leg brought in:


(1/2)MR^2(w1) = (1/2MR^2 + 1/3mL^2)(w2)
1/2(60)(0.1524^2)(w1)= (1/2(60)(0.1524^2)+1/3(6)(0.762^2))(7.85)

w1= 20.9 rad/s

In linear terms, this is about 3.2 m/s which is pretty fast. 

An observation I made was that when a dancer does a fouetté turn en pointe, they spin faster. This is because the dancer decreases the radius that they are spinning on. This is turn decreases their moment of inertia, which would later increase their angular velocity. This is a video of fouetté turns en pointe:

Thursday, November 29, 2012

The Physics of Swimming

By Clara Slight

I have swum all my life. Since we learned recently about drag force, I decided to calculate the drag force that is encountered while swimming. I decided to calculate this for two different aspects: for your hand while swimming freestyle and then for your entire body. When you take a stroke while swimming, your hand is moving faster than your entire body. I made some estimates, and used a velocity of 3 m/s, a cross-sectional area of .10 m, a viscosity of water of 8.9*10-4, and a coefficient of drag of a half sphere or .42 ( The drag force that a hand encounters while swimming freestyle is Fd = 1/2pCdv2A = ½(8.9*10-4)(.42)(32)(.1) = 1.68*10-4 N. For the entire body, the calculations change. I also made some estimates here, using a velocity of 1 m/s, a cross-sectional area of .4 m, the same viscosity of water, a coefficient of drag of a streamlined half body or .09. The drag force that a body encounters while swimming freestyle is Fd = 1/2pCdv2A = ½(8.9*10-4)(.09)(12)(.4) = 1.60*10-5. It is interesting to note that when a swimmer is completely submerged, the drag coefficient decreases to .04 and the subsequent drag force is less. The calculated value is Fd = 1/2pCdv2A = ½(8.9*10-4)(.04)(12)(.4) = 7.12*10-6. It is also interesting to observe that as the velocity increases, so does the drag force, which makes sense. 

How to win big at the carnival

By Sam Wopperer

I saw this video a few months ago, and it made me think back to when I played these carnival side games, though I was never as successful as this guy.

If you’ve ever been to a carnival or amusement park, you’ve probably noticed hundreds of these side games along the paths. For me, they use to be huge sinkholes for my time and money. Well, not anymore. Armed with the physics we’ve learned so far, you can figure out how to beat the odds, win the big prizes, and impress your friends (and understanding the different physics concepts behind each game is good review for the final). So, step right up, and give it a shot.

***Note: There are few actual calculations here. I mainly just used physics equations and the theory behind them to model how to most effectively play these games. When you next go to an amusement park, you probably won’t calculate the exact velocity or trajectory needed to throw a ball or ring in order to win the game, but I think as long as you can understand the physics behind the games, you’ll greatly increase your chances of winning. Also, you should never cheat, but in the name of winning big, fluffy stuffed animals, it’s okay to bend the rules a little bit here.

Game 1: Basket toss

Objective: Throw the ball into the basket
Sounds pretty easy right? Well, if you’ve tried this one before, you’ll know it’s not quite that simple, so here’s the trick. Toss the ball along the sides of the basket.

The physics behind it: When I was thinking about this, I thought of conservation of momentum and conservation of energy. When you toss the ball, it has a momentum equal to p (p=mv=Ft). When the ball hits the backboard, the basket really doesn’t move, so the only way momentum can be conserved is if the balls retains its initial velocity, but in the opposite direction. (This isn’t exactly the case due to air resistance, but the work of this is extremely small and doesn’t change the final velocity that drastically.) By retaining its initial velocity in the x-direction, the ball will go into the basket, hit the back wall, and then come out of the basket at the same speed, so you can’t just toss the ball into the basket. So, when you throw your ball, aim along the inside edges of the basket. Upon hitting the inside edge, the y-component of the velocity will be converted into an inward component (see diagram below). This will help direct the ball towards a more angled collision with the back wall as opposed to the head on collision it would have experienced if the ball had just been tossed straight in. Additionally, the increased contact between the ball and the basket increases the work friction does on the ball and acts to reduce its velocity. This will reduce the final velocity of the ball and help it stay in the basket.

You can draw tangent lines along the curvature of the basket, which illustrate the new direction the ball travels in as each infinitesimally small collision between the ball and basket take place.

Game 2: Test your strength

Objective: Hit the detector with enough force to ring the bell
Well, this game isn’t so difficult if you hang out at the gym all day. But even if you don’t do that, you’re in luck. There’s a secret to this game too. If you increase the velocity with which you spin the hammer, make sure you hit the detector exactly head on, and hold the hammer at the very end, you’ll be ringing the bell and winning prizes in no time.

The physics behind it: When I was thinking about this game, I thought of how can I maximize the force being imparted onto the detector, and I thought of three ways to do this. First off, use what we know about torque (t=Frsin(theta)) to get the most out of your swing. Hold onto the hammer at the very end. This will increase the radius term in the torque equation. Additionally, the angle that you must hit the detector must be as close to 180 as possible. This will allow all of your force to be imparted unto the detector and will prevent any component of the force being applied to something else (see diagram below). You can also increase the momentum of your hammer by increasing the velocity with which you move the hammer. vfinal=sqrt(2ad) (from kinematics where vinitial=0), and by increasing the distance term by moving the hammer outward and then over your head, you generate a larger arc than you would have otherwise by just lifting the hammer straight over your head and then straight down onto the target.

Game 3: Ring toss

Objective: Toss a ring on the bottle
This is probably the most difficult of the three games I’ve analyzed. From what I can think of, you must toss the ring along a row of bottles with more x-velocity than y-velocity, so the ring sort of skims over the row of bottles. Additionally, you must toss the ring in a way so that it is slightly angled away from the tops of the bottles (see diagrams below).

The physics behind it: When you first toss the ring, the force of gravity will act to pull the ring down onto a bottle. However, if you can toss the ring horizontally, its chance of landing on a bottle in the row is increased because if it misses one bottle, it may hit any of the ones after it in line. Tossing the ring just straight up in the air reduces the odds of landing it on a bottle because the area of the ring can only encircle one bottle top (see diagram below). Additionally, if the ring is tossed at an angle slightly away from the bottle top, when it finally does fall onto a bottle, the ring will lasso the bottle top and stick on the bottle top. This is more easily seen in the diagram below.

Physics behind Fear Factor –The Rotating Platform Escape

By Siyun Zou

                  Fear factor is an American reality game show where a bunch of contestants compete against one another performing dangerous stunts for $50,000. The rotating platform escape is one of the most interesting stunts.

Total moment of Inertia:
Total mass= 60kg
L=2.74 m
W= 1.22 m
I= 1/12 M (L2+W2)
I= 1/12 (60kg)(2.742 +1.222)
I= 45 kgm2
Estimated rotational Velocity of the spinning platform from video: 1 revolution per 16s
f = 0.0625 rev/sec
ω = 2πf= 2π(0.0625rev/sec)= 0.393 rad/s
The contestant took approximately 40s to set herself free from the rotating platform
ωf= ωi + αt2
0.393 rad/s= 0 + α (40s)2
α = 2.26E-4 rad/s2
Calculation of her buoyant force when she is completely submerged:
Average swimming pool volume= 375m3
Volume of the platform = (2.74m*1.22m*0.508) = 1.70 m3
FB = ρVg = 1.70 m3 (9.8m/s2) (1000 kg/m3)
FB = 16660 N
∑F= FB-Fg
∑F= Vg (ρf – ρi) -ma
∑F=16660 N- (60kg*9.8m/s2)
∑F=1.6E4 N 

Wednesday, November 28, 2012

Turkey Diffusion

By Meredith Barthold

Since we just had Thanksgiving and we ate a rather large turkey in my house, I was wondering how much mass the turkey loses as it is baked in the oven. I know the initial weight, and I could have just weighed the turkey after being cooked, but instead I decided to use diffusion to calculate the change in mass after the turkey is cooked. So I measured the turkeys dimensions before putting it in the oven, kept track of the time it took to cook, and I used Fick’s Law of Diffusion.

Most variables were obvious, but the concentration of water inside the turkey is an estimation based on the pressure required to penetrate human skin. It takes 7 bar to penetrate the skin with a needle (http://www.ncbi.nlm.nih.gob/pmc/articles/PMC2291478), and to convert bar to kg/m3, I multiplied 7 bar times the density of air in an estimated 400°F oven, which is 0.5243 ( This gives an estimation of 3.6701 kg/m3 inside of the turkey.

For the concentration of water vapor outside the turkey I used the same value as was used in class for the water vapor outside the sunflower.

All variables are as follows:
Length of turkey: 35.5 cmà0.355 m
Cross sectional Area of turkey: (0.29 m)*(0.18 m)= 0.0522 m2
Concentration of water vapor outside the turkey: 0.011 kg/m3
Concentration of water inside the turkey: 3.6701 kg/m3
Time in the oven to cook: 5 hrs. 33 mins. à 333 minutes à19980 seconds
Diffusion constant, water vapor in air: 2.4 x 10-5 m2/s

Using the following equation:
Rate of diffusion = Δm/Δt = DA [(C2-C1)/L]
Δm = (2.4 x 10-5 m2/s)(0.0522 m2)[(3.6701-0.011)/(0.355 m)](19980 seconds)
Δm= 0.258 kg = 0.569 lbs = 0.6 lbs

So even though we bought a turkey that weighed 22.6 pounds initially, due to diffusion, the turkey weighed only 22.0 pounds when it was put on our dining room table.

By Meghan Eisold

Recently, concussions have been a major concern among soccer players. When I was in high-school, some girls wore protective head gear, pictured below, to prevent concussions. I decided to determine whether the headgear can significantly lower the force that the ball exerts on a soccer player while heading the ball.
Mass of Soccer ball: 0.43 kg
Mass of soccer player’s: 75 kg
Horizontal Velocity of a kicked soccer ball(v1): 70 km/h (19.44 m/s)
I assumed that the Vertical Velocity of the ball is negligible 
Horizontal velocity of soccer player (v2): -5 km/h (-1.39 m/s) (negative=direction soccer player is traveling)
Vertical Velocity of Soccer Player is assumed to be 0 (player is not jumping)

First I determined the velocity of the ball and the player after the collision:

Elastic collisions
Equation 1: m1v1 + m2v2 = m1v1+m2v2
(0.43 kg)(19.44 m/s)+(75 kg)(-1.39 m/s)=(0.43 kg)v1+(75 kg)v2
Equation 2: v1+v1= v2 + v2
(19.44 m/s) +v1= (0.43 kg) + v2
v1= v2 – 19.01
Combine equations: (0.43 kg)(19.44 m/s)+(75 kg)(-1.39 m/s)=(0.43 kg)(v2 – 19.01
m/s) +(75 kg)v2
-95.89= -8.1743 +0.43v2+(75 kg)v2’            v2=-1.16 m/s
(19.44 m/s) +v1= (1.39 m/s) + (1.60 m/s)    v1=-19.21 m/s

I then used the final velocities to determine the force that the ball would exert on the player during the appropriate collision time. I estimated that the headgear would extend the collision time by 5 ms.

Collision without headgear: t=25 ms =0.0025s
FΔ t = Δ p  
Fb on p(0.0025s)=(75 kg)(-1.16 m/s)-( 75 kg)(-1.39 m/s)
Fb on p=6900 N

Collision with Collision with headgear: t=30 ms =0.0030s
FΔ t = Δ p  
Fb on p(0.0030s)=(75 kg)(-1.16 m/s)-( 75 kg)(-1.39 m/s)
Fb on p=5750 N

From these results I can conclude that the headgear does lower the force that is exerted on the player, but it is not a significant difference. 

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.

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.