Monday, December 9, 2013

Thanksgiving Day Balloons

This Thanksgiving Day my family went to watch the Thanksgiving Day Parade in New York City.  It was particularly windy and cold at 32 degrees Fahrenheit.  They said that the workers had to hold the parade balloons particularly low.  This made me wonder how much force the workers must apply to the balloons to hold them down.  If the balloon is not moving at all in the Y direction, with up being positive, the sum of the forces equals the buoyant force, the force of the balloon and the force of the people holding it down.

Fy=pair*g*Vairdisplaced + phelium*g*Vballoon - Fpeople=0 N
The buoyant force and the force of the balloon are acting in the positive direction because the buoyant force acts up and the force of the ballon is the force of the helium, whose density is less than that of air so it pulls the balloon upward.  The force of the people is then the only force in the downward direction.

The largest balloon ever was a Shamu balloon with 18000ft^3 or 509.7m^3.  How much force was needed to hold this balloon down?
Fpeople(total) = 6340 N

If there are approximately 40 people holding down each balloon then each person is applying about 159N of force.  This is only the force in the Y direction.  With the wind there were likely other forces in the X direction which would have put more work on the balloon holders to hold it in place.


Sunday, December 8, 2013

How does the length of a bridge change with the seasons?

Harnessing my inner 10 year old, I think bridges are completely rad. The fact that we can build such huge structures across great distances safely is pretty awesome So, given my fascination with bridges, I figured I’d take a closer look at the physics that bridge builders have to consider when planning and putting together these remarkable pieces of engineering. For the sake of ease of calculations, I’m considering only the effects of temperature, not any other weather factors. I’m also assuming that the bridge will respond in a linear fashion to the changes in temperature.

The bridge I’m considering is essentially a solid slab of steel spanning a river. I’m going to take a look both at winter and summer effects of temperature on the bridge.
ΔT = +30oC
α = 12x10-6 oC-1
Lo = 200m
ΔL = ?

ΔL = αLoΔT
ΔL = (12x10-6 oC-1)(200m)(30oC)
ΔL = 0.072m = 7.2cm

ΔT = -20 oC
α = 12x10-6 oC-1
Lo = 200m
ΔL = ?

ΔL = αLoΔT
ΔL = (12x10-6 oC-1)(200m)(-20oC)
ΔL = -0.048m = -4.8cm

As a means of comparison, what if it was an aluminum bridge?
ΔT = +30 oC
α = 25x10-6 oC-1
Lo = 200m
ΔL = ?

ΔL = αLoΔT
ΔL = (25x10-6 oC-1)(200m)(30oC)
ΔL = 0.15m = 15cm

ΔT = -20 oC
α = 25x10-6 oC-1
Lo = 200m
ΔL = ?

ΔL = αLoΔT
ΔL = (25x10-6 oC-1)(200m)(-20oC)
ΔL = -0.10m = -10cm

Guess it’s a good thing we make bridges out of steel and not aluminum! 

Physics of Fly systems

So I was working in a theater over the past week setting up lighting equipment for a show, and I needed to raise what is called a batten up using what is called a loft block. A batten is a 80 gauge steel pipe that weighs around 111.97 kg and a loft block is a winch that has a radius of 150mm or .150m. In addition to the wight of the pipe itself, there were 10 lighting instruments on the batten. There were four "source 4"s three 4" fresnels and 3 8" fresnels which weigh 6.4kg, 3.3kg, and 6.5kg respectively. This brings the weight of the batten to 166.97kg in total which represents a force of 1636.306 N pulling on an aircraft cable that suspends the pipe. So I wanted to find out what torque was required to lift the batten:

166.97kg* 9.8 m/s^2 = 1636.306 N

Fg = Ft which means that the force due to gravity is the same as the force of the torque required to move the batten

This means that the torque required to move the batten is:
T= rF (because the sin() is going to equal 1)
T = .150 m * (1636.306 N) = 245.4 N/m

In addition, I wanted to know how much energy would be required to move the batten using the loft block. Assuming that only one cable was attached to batten, I found that the kinetic energy equaled:

I = 1/3 ( 166.97kg) (9.144m)^2 = 4653.6 kgm^2
KE = 1/2 (I) (w^2)
w = v/r assuming that the batten was raised at a speed of 1 m/s
w = 1.0 m/s / .150m = 6.67 rads/s
KE = .5 * 4653.6 kgm^2 * (6.67)^2
KE = 207033.8 J = 207.033 kJ

Though this may seem like a large amount of energy needed to raise a batten, there are multiple lift lines on the pipe which would lessen the amount of energy needed by lessening the moment of inertia. But even without multiple lift lines, it is possible to lift a batten using a hand crank. However, most theaters use a mechanic crank that pulls much more effectively with little to no effort from a stage hand.


World Cup Physics

In light of the recent World Cup drawing, I wanted to look at the physics behind arguably one of the best players in the world, Cristiano Ronaldo.  He has been recorded to kick at soccer ball at a rate of 60 miles per hour which 96.5 km/h. 96.5 km/h is approximately 26.8 meters/second.  The weight of a standard FIFA soccer ball is anywhere between 420-445 g.  The radius of a standard soccer ball is 11cm which is 0.11m. 

The rotational kinetic energy that a soccer ball has when kicked by Ronaldo:

The angular velocity is w=v/r
                                    w= 26.8 m/s / 0.11 m = 243.63 radians/sec

Minimum KErot = ½ I w2
                        = ½ (0.4*0.42 kg*(0.11m)2)  * (243.63 rad/s)2
                        = 60.32 J
Maximum KErot = ½ I w2
                        = ½ (0.4*0.445 kg*(0.11m)2) * (243.63 rad/s)2
                        = 63.92 J

Jack Could Have Lived

I analyzed the scene in the Titanic where Rose is on a piece of wood and jack is in the water. I wanted to determine whether Jack and Rose could both fit onto the piece of wood because I think Rose was being selfish. I estimated that the piece of wood to  be a rectangle with a length of 1.69 m and a width of .75m connected to a triangle with the same height and width. I estimated the thickness of the board to be .2 meters.


Volume of Board
V=1.90125 m2 X 0.2 m
V=.38025 m3

Jack is 73 kg
Rose is 63 kg

Buoyant Force of the Wood
Fb=(1025 kg/m3)(.38025 m3)(9.8 m/s2)
Fb=3819.61125 N

ΣF=Fb+Mgwood+ Mg (rose+jack)
ΣF=(3819.61125 N) - 228 kg(9.8m/s2) - (73+63kg)(9.8m/s2)
ΣF=252.41125 N

The total force is still pointing upward so Rose and Jack could have both fit on the board and been saved.

Saturday, December 7, 2013

The Tension Force in Spider Man’s Web

By Kody Lyng
This past week, a new trailer for the upcoming sequel to The Amazing Spider Man series was released. In one of the clips, Spider-Man falls from the top of the Empire State Building in order to pursue a criminal. I would like to take a closer look at the tension force in Spider-Man’s web when he is at the bottom of his fall. We must first make a few assumptions:

Spider-Man shoots his web at the top of the Empire State Building
At the bottom of his fall, he is 1 meter above the ground
His web does not stretch at all
Spider-Man weighs 62 kg
Ignore air resistance

In order to find his velocity at the bottom of his swing, we must set his kinetic energy at the bottom of the swing equal to his initial potential energy.

1/2mv2 = mgh
v = sqrt(2gh) = sqrt[2*9.8 m/s2*(443m-1m)] = 93.1 m/s

In order to find the tension force of the web, we must set the net force equal to the tension force subtracted by the force of gravity. The net force will be the centripetal force or
Fc = m*v2/r


m*v2/r = FT – mg  or  FT = mv2/r + mg
FT = (62 kg)*(93.1 m/s)2/(443 m-1 m) + (62 kg)*(9.8 m/s2)
FT = 1820 Newtons

Home Alone Paint Cans

I saw a commercial for the classic holiday movie "Home Alone" and wondered about the physics that I saw.  The scene I'm referring to is the one where Macaulay Culkin throws paint cans tied to a rope off of the banister that swing and hit burglars in the head.

I was interested in the force with which the paint cans hit the robbers.

  • I estimated the length of the rope and the mass of the paint can
  • L=3m
  • mass of paint can=4.5kg
  • I found this equation online for the velocity of a pendulum at the bottom of it's swing (aka the point of contact with the robbers' heads)
  • v = √{2gL[1-cos(a)]}
  • I assumed that the rope didn't get taught until ~45degrees
  • v=(2)(9.8m/s^2)(3m)[1-cos(45)]=5.3m/s
  • a(radial)=v^2/r
  • a(radial)=5.3^2/3m=9.3m/s^2
  • F=ma
  • F=(4.5kg)(9.3m/s^2)=42N

Physics of Plank Circles

During practice this week, we did a new exercise that I'm going to call plank circles. This exercise was simply getting into plank/push-up position and rotating in a 360 degree circle by moving just your hands, keeping your feet in place. I wanted to see how a difference in height could change the radial acceleration value.

To spin 360 degrees, it took me approximately 8 seconds. Thus, to find the average angular acceleration,
w avg=(delta theta)/(delta t)
w avg= (360 degrees (pi/180 degrees))/(8 s)
w avg= .785 rad/s

Next, I find linear velocity. I used 5 feet as the radius (my hand placement to my feet), assuming that from my shoulder height to my head is about 1 foot (for a total of 6 feet tall).
v=(5 ft (.3048 m/1 ft))(.785 rad/s)
v=1.20 m/s

Finally, to find radial acceleration,
a=(1.20 m/s)^2/(5 ft (.3048 m/1 ft))
a=.945 m/s^2

Now, for someone who is 6 inches shorter than me and also took 8 seconds to spin 360 degrees,
w avg=(delta theta)/(delta t)
w avg=(360 degrees (pi/180 degrees))/(8s)
w avg= .785 rad/s

They are moving at the same average angular velocity as me. To find linear velocity,
v=(4.5 ft (.3048 m/1 ft))(.785 rad/s)
v=1.08 m/s

They are moving at a slower linear velocity, however. To find radial acceleration,
a=(1.08 m/s)^2/(4.5 ft (.3048 m/1 ft))
a= .850 m/s^2

Since they are moving at a faster linear velocity, they are also accelerating faster radially.

Thermal Expansion and Violins

I used to play the violin when I was little. I remember that my violin would undergo changes and require more tuning depending on the temperature. I decided to do some calculations to determine how much a violin changes when the temperature changes, comparing winter to spring. In my hometown, Binghamton, NY, the average winter temperature is -9 degrees C in January and 8 degrees C in May ( I am going to calculate the change in length of a full size violin and the change in volume.

Change in Length=(alpha)(L0)(change in temp.)
Change in temp.= 17 degrees C
L0 of a full size violin= 590 mm= 0.590 m
Most violins are made with spruce and maple for the back but I am just going to use the alpha and beta values for sitka spruce. For alpha I am looking at the longitudinal change in length (not radial or tangential)
Change in L=(1.74*10^-6 degrees C^-1)(0.590m)(17 degrees C)
Change in L=1.75*10^-5 m

Now I want to see how the volume of the hollow part of a violin changes from winter to spring temperatures:

Change in volume= (beta)(V0)(change in temp.)

To calculate the volume of the violin, I am using the length of the corpus of the violin to be 356 mm, and since the width of a violin changes I am going to approximate the violin as a rectangle and I will average the widths of the three parts of the body (upper: 168 mm, middle: 112 mm, and lower: 208 mm) so 162.7 mm or 0.1627 m, and the height of the ribs on a violin are about 30 mm or 0.03 m.
V=0.00174 m^3

Change in volume= (82.61*10^-6 degrees C^-1)(0.00174m^3)(17 degrees C)
Change in volume= 2.44*10^-6 m^3

Although the changes in length and volume may not seem that large, since violins are very delicate instruments any slight changes in the wood has an impact on the sound of the violin. Now it makes sense why violinists are so careful to control the humidity and temperature of their instruments since the size of the instrument changes with changes in temperature.

Violin measurements from:

Violin wood information from:

Alpha value for wood from:

Beta value for wood from:

Violins and humidity:

Friday, December 6, 2013

The Rockefeller Christmas Tree

Rotational Dynamics:

What would be the moment of inertia of this pulley as it raises the star for the Rockefeller Christmas tree? (Pulley rotates about its center, initially at rest)

            Pulley radius = .381 meters
            Frictional Torque = tfr = 2.50 mN
            Force Applied to cord = 2000 N
            Angular speed = 15.0 rad/s2

St = RFT - tfr
      = (.381 m)(200.0 N) – (5.50 mN) = 756.5 mN

I = St/a
  = (73.7 mN) / (15.0 rad/s2) = 50.43 kg m2

Using the moment of inertia we can now determine the angular acceleration of the pulley, and finally the linear acceleration of the star

            Mass of star: 250 kg
a = (mgR - tfr) / (I + mR2)
  = (2000 N)(.381 m) – (5.50 mN)
      (50.43 kg m2) + (250 kg)(.381m2)
 = 5.19 rad/s2

Linear Acceleration of star:
 a = Ra
    = (0.381 m)(5.19 rad/s2)
    =  1.98 m/s2
            This answer makes sense, the set up crew would want to move this star very slowly as to prevent any swaying or damages to the crystal or its surroundings!

Stress and Fracture

This year’s tree is from Connecticut, with a trunk about 4 feet in diameter and 75 feet high. How much would the tree shorten once the 200 N compressive force of the star is exerted on the tree? (Assuming long uniform rod)

Length of tree = 75 feet = 22.9 meters
Cross sectional area = p(.61m)2 = 1.17 m2
Compressive Strength of wood parallel to grain = 35 x 106 N/m2

F/A = (2000N)/(1.17) = 1709 N/m2  
(which is less than the compressive strength so it will not break)
L = Lo  F
         E    A
 = [(22.9m) / (10 x 109N/m2)] x 1709 N/m2

=3.91 x 10-6 m

Therefore the tree will shrink negligibly when the star is mounted.