Olympic Rowing

This past weekend I went to watch the world’s biggest rowing regatta: the Head of the Charles (held in Boston!). While I was there I saw 8 of the world’s best single scullers who all competed in the Olympics in single sculling boats this past summer, including Kim Brennan (Australian who won gold) and Dr. Gevvie Stone (American who won silver). For some context, single scullers row boats by themselves with two oars (shown below). Ever since we talked about torque in class I thought about how it applied the rowing stroke, also diagrammed below:

When rowing, you pulling on the handle moves your boat forward by rotation around a center of rotation, which is the location at which the blade enters the water. If you’ve ever rowed before it’s intuitive to think that the water is moving just as much as the boat, but in fact when the blade enters the water the boat itself is propelled forward but the water doesn’t move. Thus, if we think of the wrench example we talked about in class, the oar is the wrench and the bolt it’s rotating around is the point where the blade meets the water.

I decided to calculate the torque exerted by Gevvie Stone per stroke during her race at the Head of the Charles in her single. The course is 3 miles (4.8km) long, and she finished in a time of 19:15.6 min. I first converted into hours and calculated her average speed:

19 min x 60s  =  1,140 s + 15.6s = 1,155.6s

                1 min

1,155.6 x 1 hr  = .321 hrs

               3600s

4.8 km = 14.7 km/hr

.321 hr

To calculate torque, we need F_{perpendicular} and the radius. The force can be calculated by simply F = ma, where the mass of Gevvie and her boat combined are 85.3 kg.

F = (85.3kg)(Δv/Δt)

It takes approximately 1 second for the handle to travel 1.5m from the catch to the finish (see diagram above), thus the acceleration is 1.5 m/s².

F = (85.3kg)(1.5m/s²) = 128.0 N

The length of the oar (approximately 285 cm) serves as the radius, so we can create a triangle from the aerial view:

The average length of a sculling oar is 2.85m, and the approximate horizontal length from the blade to the boat is 2.9m, thus we have the hypotenuse and adjacent side lengths, and can calculate the angle to be 11.5 degrees. Now we can calculate the torque:

T = Fcos x * r

T = (128.0)(cos11.5)(2.8)

T = 351.2 Nm

And since she has two oars (one in each hand) her total torque each stroke is 2(351.2) = 702.4 Nm

References

https://www.vespoli.com/boats/singles.php

https://www.bostonglobe.com/sports/2016/10/22/head/qVP4hCL80MlXbTmdQOz39K/picture.html?p1=Article_Gallery

http://www.concept2.com/service/oars/setting-inboard

Baseball Bat and Moment of Inertia

In the midst of the World Series, I thought it would be interesting to examine some aspects of physics in the sport.  Many baseball players (including most of those on the Indians and Cubs I would assume) work hard in practice to improve their swing speed in order to hit the ball a greater distance.  These players lift weights to strengthen their muscles in order to exert greater forces.  Bats of different weights are also known to help players improve their swing speeds.  I thought it would be interesting to see how the concentration of weight within the bat could contribute to swing speed.  If a player has two bats, that are equal in size and weight, but differ in where the mass is concentrated, which bat can be swung the fastest when the same force is exerted?

"Axis"= location of hand grip on bat 

Bat 1 = long uniform rod 

Bat 2= long rod with mass concentrated at the end away from the hand grip

deltaKE= -deltaPE + W(NC) 

1/2Iw^2 + 1/2mv^2 = -(0J-mgh)

1/2I(v/r)^2 + 1/2mv^2 = mgh 

1/2v^2(m+I/(r^2)) = mgh

If mgh is equal for the two bats, 1/2v^2(m+I/(r^2)) must be constant.   The bat with the larger I (moment of inertia) will have the slower speed. Therefore, the player should choose Bat 1, with the smaller I, in order to swing faster. 

Fall from a horse

This September I fell from my horse while jumping. I decided to look into the physics behind it to find out how bad the fall was on paper compared to the actual outcome. The fall happened because my horse and I left the ground too far away from the jump and crashed through the top rails. We started with the velocity (in the x-direction) of 4.0 m/s and left from 2.0 m away to a 1.20 jump. At the top of the jump we experienced an inelastic collision with the stationary rails.

(702 kg)(4.0 m/s) +0=(702+18)V’

V’=3.9 m/s (x-direction)

After the crash I separated from my horse and fell at an estimated angle of 80 degrees. I was able to find the linear velocity at that angle to be 4.5 m/s and landed at a distance of 0.211 m from the jump. In addition to the linear motion I also rotated during the fall which contributed to my injuries. To find the F of the ground on me, I found KE energy of the linear and rotational motion and then used the work equation to find my acceleration. I then converted from m/s² to g’s and found that I was in the range to have obtained a concussion.

KE=(.5)(50.4)(5.82)²+(.5)(54)(4.5)²

KE=1399.68 J

W=∆KE               W=FdCosϴ       -1399.68=F(.211)Cos 80            F=38201

F=ma                38201=(54)a     a=707 m/s²                   72.2 g’s (70 to get a concussion)

For this problem I had to estimate quite a bit on the distances and velocity as well as the rotational numbers since the fall wasn’t recorded and I do not remember what happened. I also ignored air resistance and friction with the poles.

Yoga in Space?

An article was recently published bringing to light the long-term back problems of astronauts that have spent extended time periods in space.  Due to the reduction in gravity, astronauts experience less force on their bodies, and are actually known to "grow" two inches while in space.  While this may seem like a good thing to reduce the forces on our bodies, this loss of gravity can length the torso, weaken the spine, and lead to a host of physiological problems.  The article brings up possible future (nearly month long) missions to Mars that could come with the risk of more back pain and muscular atrophy.

Mars has a gravitational pull 38% of Earth's surface gravity, meaning that it would be 3.7 m/s^2.  A 70 kg person on Earth would feel a force due to gravity of 686 N while the same person would feel a force due to gravity of only 259 N on Mars.

(70 kg)(9.8 m/s^2) = 686 N
(70 kg)(3.7 m/s^2) = 259 N

The article went on to talk about the benefits of astronauts doing strength exercises and yoga to alleviate the negative effects of loss of gravity.  However, yoga depends on using the force of gravity to stretch muscles.  In addition, many strength training exercises only work because the force of tension of the muscles are opposing the force of gravity (concentric contractions) or because gravity is greater than the opposing muscle contraction (eccentric contractions).  If the force due to gravity is significantly lower in space or on another planet, then these exercises will be less effective.  If the Fg is lower, then the Ft of the muscles will be lower to overcome or oppose it.  Better exercises need to be developed and implemented in order to avoid these long term problems in the future.

Read the article here: http://www.cnn.com/2016/10/26/health/astronaut-back-pain-spine-health-space/index.html

How to Hit a Home Run

One could guess that hitting a home run depends on the pitch and the force of the bat on the ball. However, not many people realize the importance of "follow through." When a pitcher throws a ball toward the home plate, it has a certain momentum due to its mass and velocity. Most pitches thrown in the major leagues are around 100 mph or 160 km/hr. In order to hit a home run, the batter must cause a significant change in momentum because because the batter wants the velocity of the ball to be in the opposite direction and at the greatest speed possible. 

Now consider the momentum change of the ball in terms of impulse. Impulse is equal to the change is momentum which is the sum of the forces multiplied by time. To get the greatest impulse, meaning the greatest change in momentum, you want your force and time to be as big at possible. The batter can only apply so much force, so in order to get the greatest impulse, the batter should increase the time of impact between the baseball and bat, known as "follow through". This can also be applied to tennis or golf. 

P=mv

Change in momentum = Impulse = F x t
The greater the time, the greater the impulse, meaning the greater the change in momentum.
Because the mass of the ball is not changing, only the velocity changes. To hit a home run, the batter wants the greatest change in velocity possible, where the ball changes direction and has a high magnitude. Thus, a greater impact time from the follow through will lead to a greater change in velocity, and inevitably a home run!


Read more: http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Momentum-Real-life-applications.html#ixzz4OCfw2OTd

Slap Shot! Can you stop it?

On Saturday, I was watching the Boston Bruins play and began to think about the force needed to accelerate the puck during a slap shot. So, I went and did some research on slap shots. According to the Guinness World Records the fastest slap shot that was recorded in a competition was done by Denis Kulyash of Avangard Omsk in the Continental Hockey League's All-Star skills competition in St. Petersburg, Russia, on 5 February 2011. This shot was clocked to be traveling at 110.3 mph or 49.3 m/s. Although this is the record, I am sort of biased towards the second-place holder Zdeno Chara of the Boston Bruins, who set the record during the National Hockey League's All-Star skills competition in Montreal on 25 January 2009 with a shot clocked at 105.4 mph.  Either way both shots produce a hockey puck that is moving incredibly fast. To find the acceleration that puck experiences I started by describing the interaction as an elastic collision.  The average mass of a hockey puck is 170 g or .17 kg. The average mass of a new composite hockey stick is around 450 g or .45kg. The final velocity of the hockey stick is 3m/s. Using the conservation of momentum equation, you calculate the velocity of the hockey stick in the collision.

(Mp) (Vpuck intial) + (Ms) (Vstick intial) = (Mp) (Vpuck final) + (Ms) (Vstick final)

0 + (.45) (Vsi) = (.17) (49.3) + (.45) (3)

Vsi = 21.62 m/s

Then by finding the impulse you can solve for the force. The collision lasted for .001 seconds.

Change in momentum = sum of forces x change in time

(.17) (49.3)/.001 = 8381N

Then if you take that force and divide it by the mass of the puck you can find the acceleration that is experienced by the puck

8381N/.17kg = 49300 m/s² or 5030.6 times gravitational force (g)

Overall, there is both a crazy amount to force and acceleration.

*In this situation, I am not considering the flex of the stick.

1.http://www.guinnessworldrecords.com/world-records/fastest-ice-hockey-shot/

2.http://blog.purehockey.com/composite-hockey-stick-weigh-in/

BELIEVELAND heads to the World Series!

In light of recent events (Go Cleveland Indians!), I started thinking about the physics of baseball. The more I thought about it, it became clear that there are many physics examples in this sport. The article I read is titled, "The Physics of October Baseball in Wrigley". In this article, the author analyzes how the wind might affect the trajectory of a ball after it is hit as well as how it affects the pitch by looking at the average wind speed in the summer vs. the average wind speed in the fall. (I think the author is trying to make excuses for the Chicago Cubs...but I'll try to look past that.) He provides a comprehensive analysis of the different factors that might affect the pitch (i.e. speed of the ball; distance traveled; the rotational motion of the ball; etc.). I included two images of the statistics he recorded. The bottom image includes a "key" above it, which describes what the numbers between the slash lines signify.

The author includes the speed, distance, horizontal angle, vertical angle, and rotational motion for three different pitch types by Jake Arrieta.

The author analyzes different types of hits by assessing the distance traveled, the time in the air, and how much the ball drifts (in degrees).

Using his theoretical data, the author concludes that the shift in winds in Chicago might have an effect on the outcome of a pitch or a hit between the summer and fall. He provides data about the ball's velocity, distance traveled, and angle, all of which can be used in kinematics and energy equations to find additional information. Moreover, he discusses the rotational motion of the baseball, which, again, can be used in angular kinematic equations (given constant angular acceleration) or can be converted to linear speed. 

Prior to reading this article, I was going to discuss the topics of momentum and energy in baseball. Using the data from Jake Arrieta's pitches (from the article) and the mass of the ball (let's call it 1kg for simplicity purposes), we can determine the momentum of the ball. For example, the momentum of the ball in a sinker pitch would be: p=mv, so p=1kg(95.0mph) = p=1kg(42.5 m/s). So, p=42.5 kgm/s for a ball in a sinker pitch. Furthermore, we can analyze how work and energy are related using the equation [m(vf2 - vi2) = -mg(h2-h1) + Wnonconservative]. Since we know the speed of the ball (42.5 m/s), we could determine what force the baseball glove/catcher's hand needs to exert in order to bring the ball to a stop (vf=0). 

To conclude, there is obviously a lot of physics going on in a baseball game! The article "The Physics of October Baseball in Wrigley" discusses several topics in physics that we have covered thus far and definitely got me thinking about future topics, specifically regarding rotational motion. 

But for now, let's hope the Cleveland Indians can maintain their successful momentum throughout the World Series and clinch a Championship title!

sources:

http://www.hardballtimes.com/the-physics-of-october-baseball-in-wrigley/

http://www.nytimes.com/2016/10/20/sports/baseball/cleveland-indians-toronto-blue-jays-alcs.html?_r=0

Can Water Bounce?

http://phys.org/news/2016-10-lessons-unearthly-behavior-enormous-droplets.html

With the earth becoming more settled and globalization constantly increasing, the sights of scientists and explorers have turned to the stars and the bottoms of the oceans. In our attempts to push further into space, there have been incredible advancements in technology, bio-organics, and the preparation that goes into space travel. In order to further our ability to travel and potentially live in space and on other planets, we need to create better systems that can exist and sustain life.

To do so, scientists at Portland State University designed a setup where droplets of water were dropped down a tower, while placed on a surface. The free-fall of the surface lead to the sensation of "no gravity" for the water, and it therefore bounced off the surface and became a droplet. From this event, the observers could look at a host of concepts, including hydrophobic interactions, fluid mechanics, droplet velocity and ejection as a function of drop volume.
These advancements will be crucial to better travel and life sustainability as bubbles and droplets often become problematic when in space. Knowing more about their function and mechanics in a free-fall state allow us to create better water processing systems, coolants, and sustainable habitats.

Looking at the experiment itself, there are hydrophobic interactions, which are therefore a force to eject the puddle off of the surface to become a droplet. Therefore there is an inelastic collision occurring, between the water and the surface, a force creating that collision depending on its hydrophobic interactions. Also, once the droplet is formed and off the surface, it reaches free-fall, meaning no gravity. Therefore the puddle condenses to a droplet based on surface tension and the adhesion of the water to itself. A better understanding of how water acts in space would greatly enhance our ability to explore and potentially start new life out of the Earth's environment.

Look out a deer!!!

Every Fall students who travel to Colgate for school are usually baffled by two things. First, they are usually surprised about how cold it gets and when it starts snowing. Secondly and arguably more unique to Hamilton, they are startled by the sheer number of deer that roam unabashed by human presence on this campus.  The town of Hamilton itself has created deer management task force to deal with the issue that there were 49 white deer per mi².  This number per the task force proposal needs to be between 7-10 deer per mi² to be able to not damage the local ecosystem. So why deer on a physics blog? Due to the increased number of deer, the community deals with more collision with deer than normal.  Also, a deer ran into the SOMAC ambulance the other day, which got me thinking about the energy and momentum changes of a deer on car or vice versa collision.  Let’s do some physics!!

A white-tailed deer weighs on average about 150 lbs. or 68kg and an ambulance weights 2,200 lbs or 997kg. In this situation, the ambulance was at traveling down the road and deer came out of woods and ran right in front of the ambulance and was hit. Let’s say the ambulance was moving at 30mph or 13m/s. For this situation the deer is hit then rolls up the hood and then trapped on the box of the ambulance, making this an inelastic collision where only momentum is conserved. Without any breaking, I calculated the velocities of the ambulance and the deer after the collision.

 M(ambulance)V(Ambulance) + m(deer)V(deer) = (M+m)V(deer/ambulance together)

              997(13) + 68(0) = 1065 (V)

              (V) = 12.17m/s

For this problem, you will notice that for the velocity of the deer I used zero. As the deer, it moving perpendicular to the ambulance, the velocity is not contributing to the change in momentum. The collision alone will decrease the speed of the ambulance by .83m/s. Then if you do the change in momentum of the ambulance and know that the collision occurred over 0.0001s.

              1065(12.17) – 997(13)= 12,961.05 - 12961=  .05/0.0001 =  500 N

A good sized force was enacted on the ambulance by the deer. You can also find the deceleration of the ambulance.

              500/997kg =  .502m/s² 

Overall, it is not a fun experience for either party. Thus, the deer management task force exists.

*This can also be modeled as an elastic collision.

Sources:

1.       http://hamilton-ny.gov/village-of-hamilton/deer-culling

2.       https://en.wikipedia.org/wiki/White-tailed_deer

3.       http://www.emt-resources.com/Ambulance-types.html

Centrifugation - Separate particles with rotational motion

Centrifugation, in simple terms, is the process of separating particles using rotational motion around an axis. At the end of centrifugation, particles are sorted out by a density gradient inside the centrifuge tube, with denser (heavier) particles at the bottom and lighter particles on top.

Centrifugation is much more efficient in creating a density gradient and separate particles by mass than just using gravity. The rotational motion of centrifugation creates strong force on the particles inside the microfuge tube. In particular, the particles will be impacted by the centripetal force, which is proportion to the centripetal components of the linear acceleration in rotational motion of the centrifuge.

For example, a minicentrifuge can reach maximum speed of 10,000 rpm (revolutions per minute). Assume sample is 0.08 m from the rotation axis of the centrifuge.

     10,000 rmp = 10,000*2π/60 =1047.2 rad/s

So: αR = ω2r = 1047.22 * 0.08 = 87729.8 rad/s2 = 8952.02 g’s

Using rotational motion, the minicentrifuge is able to get the sample to move at a centripetal acceleration that is about 9000 times larger than gravitational acceleration, thus the centripetal force is 9000 times larger than gravity for the same sample.

http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS8_8/PS8_8.htm

http://www.public.asu.edu/~laserweb/woodbury/classes/chm467/bioanalytical/centrifugation%20notes.html

Blue Origin In-Flight Escape Test

On October 5th, Blue Origin, a private spaceflight company started by Amazon founder Jeff Bezos, successfully completed an in-flight escape test. Blue Origin is attempting to develop reusable rockets first for space tourism in the suborbital atmosphere, but eventually to enable individuals to be able to live and work in space.

The safety feature that was tested ensures that future users of Blue Origin rockets will be safe even in the event of a malfunction. The escape feature allows the crew capsule to separate from the rest of the rocket in case of an emergency. The capsule is propelled away from the main body of the rocket and then carried safely to the ground using parachutes. This will ensure that all passengers are kept safe if their is any sort of problem with the main body of the rocket.





The separation of these two pieces of the spacecraft can be thought of as a perfectly inelastic collision. This means that we know m_Rv_R = m_Bv_B + m_Cv_C to be true where R is the full rocket, B is the booster which the capsule C is seperating from. Since this is perfectly inelastic a large amount of energy which comes in the form of the propulsion from the capsules rocket, is needed to allow this seperation to occur. After the separation the capsule experiences a large external force which once applied will mean that the system which we originally considered will not follow conservation of momentum. Only in the instant during which the seperation occurs can the system be modeled using our above equation. In order two seperate the two pieces a rocket on the bottom of the crew capsule fired for 1.8 seconds exerting 70,000 lbs or about 310,000 Newtons of force. Knowing this we can calculate the impulse or change in momentum of the capsule after it seperates.
Impulse = (310,000 N)(1.8 s) = 560,000 Ns.

Using this technology along with many of the other groundbreaking technologies being created by Blue Origin along with other private spaceflight companies it is possible that in the not too distant future space travel could be very safe, affordable, and commonplace.

http://www.nytimes.com/2016/10/06/science/space/blue-origin-rocket-crew-capsule.html?rref=collection%2Fsectioncollection%2Fscience

Why Running on the Curve of a Track is Slower

Nick Baglieri

Professor Metzler

Physics 111

15 October 2016

           

Why is Running on The Curve of the Track Slower?

            According to a study published in the Journal of Biology and to the testaments of most runners, sprinting on a curve is slower than sprinting on a straightaway. For example, if a 200 meter race was performed on a track with a 21 meter curve, it would be .12 seconds faster than a 15 m curve. 200 meter sprinters run .4 seconds slower on straight tracks than curved tracks.  “This decrease in maximum speed is related to the curvature of the track lane and can potentially result in one sprinter gaining an advantage of up to 0.12 s over a competitor in an adjacent inside lane.”

One attempted explanation has been by modeling a sprinter as a point mass moving in a circular path. When a runner runs straight, he/she exerts a vertical force on the ground, however on a curve, the runner exerts a lateral and vertical force that results in a resultant force vector. This lateral force generates a centripetal force for the runner. Centripetal force is required to change the angular momentum vector of the runner. The runner exerts the same resultant force regardless of the surface, so the vertical force exerted by the runner must be smaller on the curved surface. This increases the contact time with the ground, and decreases the impulse of the runner, making he/she run slower.

Source: http://jeb.biologists.org/content/210/6/971#F1