In honor of the end of my volleyball career…

A volleyball player’s approach and arm swing when they hit is very intentional and can be quite intricate. The angle of approach, depth of wind up, shoulder rotation, and hand contact and length of contact with the ball are just a few technicalities that ultimately affect the ball’s direction and velocity. But let’s try to simplify it in order to calculate the force I can exert on the ball.

If I hit a 0.28kg volleyball at 80mph or about 36m/s (close to the top male Olympic speeds, but let's just pretend!), that means the ball had a momentum of 10.1 kgm/s because p=mv.

We will assume the ball started at rest. This is decently accurate because when the ball leaves the setter’s hands it is traveling parallel to the net. We will assume I am hitting ‘down the line’, which means I am manipulating the ball with my arm swing and wrist to cross the net perpendicularly. So the ball was only moving in the x-direction before I hit it, and afterwards it moves in the y-direction.

We know that F= Δp/Δt. Contact time between my hand and the ball depends on what I want the ball to do. If I want the ball to have top-spin and drop quickly to the floor my fingers wrap around the ball and I remain in contact with it longer. If I want the ball to ‘float’ I make very brief contact with the ball with mostly my palm.

Let’s say I want the ball to float and that t=0.008 seconds. That leads us to: F= 10.1kg/s * 0.008s. F= 1262.5 N. So a volleyball player’s arm is really quite the weapon! 

P.S. Notice how the figure's knees are buckling inward before he jumps? That reduces the amount of force they can put into their jump. BAD FORM 

Vampire Bat Incisors

Halloween is far gone, but vampire bats are always fascinating to learn about. A common misconception to many is that vampire bats are violent, threatening creatures that can kill. However, vampire bats are relatively small and rarely kill their prey, even being quite gentle in their ways. Instead of using the long canines (the fangs) on the sides, vampire bats normally use their tiny incisors on the front to cut into the skin. On top of that, the victim rarely notices the incision because of how sharp these incisors are. 

Obviously, the main issue we need to discuss here is the physics behind it. So, how much force is a vampire bat's incisor actually exerting onto the prey's skin, and how does it compare with a human incisor? Based on an equation we learned in class, I noticed that the force is related to the pressure and area:

P = F/A

F = PA (where in this case, A is the area of contact between the skin and the incisor)

I found that bat incisors are typically 2 mm in width, and 0.2 mm in thickness. The total area must therefore be:

A = 2 mm x 0.2 mm

A = 0.4 mm^2

Thus, the total force required would be:

F = PA

F = 0.4P

If we look at a typical human incisor, which is about 9 mm wide and 1 mm thick, the total area is:

A = 9 mm x 1 mm

A = 9 mm^2

Comparing the bat's incisor to the the human's incisor, we can see that the force needed to apply the same pressure will be significantly greater:

F = PA

F = 9P

The force required by human incisors is almost 23 times more than that of bat incisors. So if, for any reason, you decide you want to become a humanoid vampire, perhaps sharpening your teeth or working on that jaw muscle might be a good idea.

Sources:

http://www.arkive.org/common-vampire-bat/desmodus-rotundus/

http://www.science.smith.edu/msi/pdf/642_Pteropus_vampyrus.pdf

http://www.nytimes.com/2016/11/01/science/fangs.html?rref=collection%2Fsectioncollection%2Fscience&action=click&contentCollection=science&region=stream&module=stream_unit&version=latest&contentPlacement=120&pgtype=sectionfront

Splash-Less Bleach

While grocery shopping I noticed bleach for sale advertising a "No Splash Formula". I was immediately reminded of physics and the videos we watched in class to demonstrate viscosity, that had different liquids being poured into a container. I searched the product online, and was able to find a product description promoting a thick formula to prevent splashes.

 

I recognized that this product must have a higher coefficient of velocity, η. The higher the coefficient is the more viscous the fluid is, and the more resistant to flow it is. Knowing that water splashes, I hypothesized that the η of the bleach must be higher than that of water, which is given as about 1 Millipascal seconds (mPas). I also figured that it would likely have an η lower than that of a fluid that pours poorly, such as honey, which as an η equal to about 80 mPas. I found that the main ingredient in the bleach, Sodium Hypocholorite has η=1 mPas. This would explain why most bleach splashes, as water also splashes. However, this product also contains Sodium Hydroxide, which has a much higher η of 10 mPas. It is therefore considerable that the higher η of the ingredient Sodium Hydroxide is this products claim to fame. 

References:

http://www.engineeringtoolbox.com/viscosity-converter-d_413.html

http://www.engineeringtoolbox.com/kinematic-viscosity-d_397.html

https://www.walmart.com/ip/Clorox-Splash-less-Regular-Bleach-Concentrated-116-Fluid-Ounces/21633504#about-item

Maximum Speed of a Helicopter

Growing up as a kid my neighbor was a helicopter pilot for the local medical center, he would airlift patients who needed urgent care to the hospital as fast as possible to increase their chances of surviving whatever accident they had experienced.  I had always wondered just how fast he could get someone to the hospital, so I did a little research on the maximum speed of helicopters.

The topic ended up being much more complex than I had initially anticipated, and it has to do with the basic design of a helicopter's rotor blade.

If we consider a helicopter going 100m/s in the forward direction, and the blades are rotating at about 400m/s.  If we consider the one direction motion of the blades, then we can see that the blades will not be traveling at the same speed in when they are considered in the same direction as the  helicopter.  This causes our imagined helicopter above to have two blade speeds:

Traveling in the same direction of the copter (advancing blade) - 400m/s(blade speed) + 100m/s(copter speed)=500m/s
Traveling in the opposite direction of the copter (retreating blade) - 400m/s(blade speed) + 100m/s(copter speed)=300m/s

This disparity of speed causes the force of lift that the blades are experiencing to be unequal, providing different lift for each side of the copter.  To balance these forces, the blades are put on an angle so that the retreating blade can cause for force and the advancing blade can cause less force, putting them equal to one another.  But as we approach higher and higher speeds, the angle of the blades can only make up for so much of a difference in blade speed.  So, through much more advanced math than this class has covered, the aspects of unequal speed of the blade limit the helicopter to about 250mph, which covers a mile in a little more than 14 seconds!

The Physics Behind Hitting a Golf Ball Far

Last week I was hitting golf balls at the driving range and wanted to learn how to hit the ball farther. After doing some research, I came up Jamie Sadlowski. Jamie Sadlowski won back-to-back World Long Drive Championships in 2008 and 2009. Unlike the typical long driver competitor, Jamie is only 5 feet, 11 inches tall and weighs a mere 168 pounds. Many golf professionals suggest that Jamie’s power stems from his background in junior ice hockey and ability to generate a massive degree of torque with his hips and shoulders. Sadlowski farthest drive is 445 yards, an impressive distance considering the average PGA tour drive is only around 288 yards. Not only is Jamie able to hit the ball so far because of his physical capabilities, but also his equipment.

Although a multitude of factors go into a driver built for long drive competition, one of the contributing factors is the long length of the driver shaft. Jamie Sadlowski uses a 48 inch (1.2192m) driver for long drive competitions. This makes reaching over 150 mph swing speeds slightly less difficult than with a standard driver, which is only about 45 inches (1.143m) in length. Using the equations for angular velocity, one is able to determine the clubhead speed that a long drive competitor, like Jamie Sadlowski, gains by using a longer driver shaft.

Assuming Jamie Sadlowski swings has a club head speed of 150 mph (67.056 m/s) and the golf club is swung in a circular arc, his angular club head speed would be:

Assuming the angular velocity of Sadlowski’s club head remains equal with the 45 inch driver, one can calculate how much translational club head speed Sadlowski would lose using a PGA tour driver.

Therefore, Jamie Sadlowski gains about 10 mph on his clubhead speed by using a 48 inch driver instead of the standard PGA tour length 45 inch driver. Looks like I need to buy a longer driver!

Sources:
http://www.jamiesadlowski.net/about/

https://en.wikipedia.org/wiki/Jamie_Sadlowski

https://www.youtube.com/watch?v=AG-PkLfWLRY (Video from 2014 World Long Drive Championship)

Image:

http://alchetron.com/Jamie-Sadlowski-383210-W

Physics of Stop Ramps

While driving on the highway, I noticed that there were periodic ramps of sand off to the side of the road where trucks could roll into to stop if their breaks failed while going down the hill. The ramps seemed rather short, though, which made me start to wonder how the builders know how long to make the ramps so that they are effective at stopping the trucks traveling at highway speed down a hill. I decided to try to figure it out myself…
If an eighteen wheeler has a mass of 36287 kg and is traveling at a speed of 70mph , how long would a sand ramp have to be to stop the truck given its breaks don’t work?  The coefficient of friction of sand is 0.60.

F= ma

F_fr = µF_N = µmg

F_N = mg

µmg = ma

(0.60)(36287kg)(9.8) = (36287kg)a

a = 5.88 m/s²

a = Δv

      Δt

70mph = 31.3 m/s

5.88 m/s² = 31.3 – 0

                              Δt

Δt = 5.32 s

x = v₀t + ½ at²

x = (31.3m/s)(5.32s) + ½ (5.88m/s²)(5.32s)²

x = 250 m

The ramp must be at least 250 m long to stop a truck traveling at 70mph.

Velocity of a Bowling Bal

Over Thanksgiving break a few friends and I went bowling. While watching the bowling ball travel down the lane I realized that I approximate the velocity at which a ball was traveling. Important aspects of this scenario to note are I'm assuming the alley is friction-less and the ball is rolling without slipping. The length of a standard bowling alley lane is around 19 meters and the ball in question has a mass of 5kg.

The Potential energy of the ball equals the mass of the ball times g times the change in height. With the change in height being a little lower than my height when I bring my arm back before releasing the ball. A little less than my height of 1.76m, the approximate height I bring the ball up should be around 1.7m.

PE=mg(h1-h2)=5kg*9.8m/s2 *(1.7m-0m)=83.3 Joules

The total Kinetic Energy of the ball would be the sum of the KE(rot) and KE(trans)

KE(tot)=KE(trans) + KE(rot)                     I(sphere)=(2/5)mr2      w=(v/r)

KE(tot)=(1/2)mv2+(1/2)Iw=(1/2)mv2 + [(1/2)((2/5)mr2](v/r)2)=(7/10)mv^2

Assuming no friction, the entire PE is converted into KE as the ball travels 

KE=PE         (7/10)mv2   = 83.3 J               (7/10)(5 kg)v^2  = 83.3 J    v=4.9m/s

Solving for velocity a value of 4.9 m/s is obtained

Honey in my Coffee

I've alway used honey in my coffee rather than sugar, it's healthier and all around more tasty. Here at colgate the honey is stored in these large squirt bottle- like things and it takes a while to come out. However once I add coffee into the mix, the honey becomes easier to stir and seemingly melts into the coffee. However as we learned today what is actually happening is the heat from the coffee is increasing the viscosity of the honey by increasing the kinetic energy of the particles thus making the honey flow faster.

Basically as heat rises so does viscosity,

Heaviest Vehicle Pulled 100 ft. by a Pair

Jacob and Matthew Fast have just broken the record for the "Heaviest Vehicle Pulled Over 100 ft by a Male Pair". They are the sons of multiple record-breaking strongman Reverend Kevin Fast. They followed in their father's footsteps by pulling three fire trucks over 100 ft on flat ground. These fire trucks combined to weigh 86,350 kg or 190,400 lb. 

After hearing that these brothers pulled this crazy amount of weight I wondered how much work they had done.

I initially looked up both the static and kinetic coefficients of friction. I then realized that overcoming the static friction occurs over such a small distance that the work done would essentially be zero. The kinetic coefficient friction = 0.8.

https://youtu.be/lujbMZ_VJ-E

           

For work, only the Forces in the x-direction need to be considered. After the initial movement, I will assume that there is no acceleration, so the forces sum to zero. 

                     Fapplied = Sum of FFriction

                               = FN(Coefficient of Kinetic Friction)

                               =m*g*uk 

                                    = 86,350 kg * 9.8 m/s^2 * 0.8

                               =676,984 N

So, assuming each brother applied the same force, Jacob and Matthew each applied a force of 338,492 N.

To find the work each brother did, we simply need to account for the distance (100 ft or 30.48 m)

                  W = F*d

                      = 338,492 N * 30.48 m

                      = 10,317,236 N*m

So each brother did 10,317,236 N*m of work. 

Let's put this in terms that we can understand.

The brothers applied this work over 38 seconds. 

So the Power of each brother was about 271,506 watts. 

This is equal to about 364 horsepower!

Hot Air Balloon Fiesta

As I was looking through my old pictures from my semester off-campus in Santa Fe, New Mexico last fall, I came across shots of the Albuquerque Hot Air Balloon Festival that takes place every October. Hundreds of balloons take flight each day for two weeks, often with unique and fun shapes. How do big balloons filled with air take flight? Using buoyant force! This force occurs when a fluid exerts a force that opposes gravity. Normally, the buoyant force of the air displaced is not big enough to cause a balloon to rise. However, we can increase this force by decreasing the density of the air inside the balloon by heating it. As heat is added, the air molecule have more energy and therefore move more, taking up more space. Because air pressure is kept the same, the balloon will not be crushed by the air outside. Since more dense fluids sink and less dense fluids float, the less dense air inside the balloon will create a force upward towards air with the same density.

F buoyant - F gravity = Negative = Net force towards ground = sink or not leave ground
F buoyant - F gravity = 0 No movement in y direction
F buoyant - F gravity = Positive = Net force towards sky = float

Average mass of balloon = 600. kg
Average mass of heated air inside balloon = 2500 kg
Average volume of balloon =2800 m cubed
Density of air unheated = 1.225 kg/m cubed

F gravity = (600.+2500)(9.81)=30,400 N
F buoyant needed to generate lift = 30,500 N = (rho)(V)(g) = (rho)(2800)(9.81)
Density of air needed to generate lift = 1.11 kg/m cubed

Static Equilibrium Could Win You One Million Dollars

CBS’s hit show Survivor first aired on May 31, 2000 and has continued, currently in its 33^rd season! The show features individuals who must survive harsh conditions, play a strong social game, and compete in challenges on their way to winning a grand prize of one million dollars. Some of these challenges have extremely high stakes, with the winner gaining an “immunity idol” that prevents them from being eliminated from the game.

As I watched a recent episode, I noticed that the immunity challenge relied almost entirely on the idea of static equilibrium in rotational motion. The challenge involved contestants balanced on a thin beam, while having to keep a small statue balanced using only a long bamboo pole (see the image below)

In order to keep the statue balanced, contestants would have to use their arm muscles to ensure that the bamboo pole maintained static equilibrium. The bamboo pole is ten feet long, or 3.05 meters. For this post, we will treat the bamboo pole as a uniform rod with a mass of one kilogram, as bamboo is an extremely light wood. Assuming the contestants back arm is the pivot point, and the front arm is applying a force of 30 degrees above horizontal, one meter from the back arm, we can calculate the force needed to maintain static equilibrium.

For static equilibrium,

 Using the free body diagram below, these three conditions will allow us to solve for both the magnitude of force exerted by the front arm, as well as the magnitude of force exerted by the back arm and the angle at which it acts.

To view the challenge this post is based on, visit: http://www.cbs.com/shows/survivor/video/642C2E55-51D6-0014-C840-8D39705ADE4F/survivor-reward-challenge-ferryman/