## Thursday, October 31, 2013

### A Planet similar to earth

Scientists have found a new planet very close to the earth. Named Kepler-78b, this planet has 1.8 times the mass as that of the earth and is 400 light-years from the earth. It takes Kepler-78b 8.5 hrs to complete an orbit around its sun-like start in a radius of 900,000 miles. It does not rotate around its own axis. So I thought it will be interesting to compare the angular and linear speed of this new planet to earth. In order to calculate that, I assume it orbits in a circular path.

The orbit radius=900,000 miles*1609 m/1 mile=1.45*10^9 m.

time=8.5 hrs=30600 s.

w=2pi/30600 s=0.00021 rad/s

v=rw=0.00021 rad/s*1.45*10^9 m=3.0 *10^5 m/s.

The angular and linear speed of the plant can be compared to that of the earth, which has a linear speed

of 29780 m/s, which equals an angular speed of 29780 m/1.50*10^11 m=2.0*10^-7 rad/s. The planet rotates much faster than the earth around the star.

Sources:

http://en.wikipedia.org/wiki/Earth

http://news.discovery.com/space/alien-life-exoplanets/kepler-78b-mystery-exoplanet-shouldnt-even-exist-131030.htm

## Monday, October 28, 2013

### Chicago Marathon Winner and His Apparent Lack of Work

The Chicago Marathon was run on October 13, the first major marathon in the United States since the Boston Marathon. Its male winner was Dennis Kimetto, and he finished the marathon in 2 hours, 3 minutes, and 45 seconds, beating the record of 2:04:38 set last year. He was just a farmer a few years ago, and now is setting records in international marathons.

I wanted to calculate his average speed as well as his total work (just for the marathon not training) and see how much energy he had to expend to run the marathon in record time. To calculate his average speed (assuming a flat course) I found the total amount of time he was running (7425 seconds) and divided the total distance (26.1 miles or 42155.8 m) and found his average velocity to be 5.6 m/s. Assuming he accelerated to this speed in the first 10 meters of the race, I found his acceleration by using vf=vo+2ax. I found his acceleration to his top speed to be 0.28 m/s^2. And then assumed his acceleration for the remainder of the race to be 0 m/s^2, assuming he held a constant speed for the remaining 42145.8 m. So his average acceleration would be 6.64x10^-5 m/s^2. I also assumed his mass to be approximately 50 kg, and calculated his total work by using W=Fd, finding his total work to be 140J.

This seems absurdly low compared to the amount of energy expended by Usain Bolt in his record setting run (28500 J), and even the energy expended (1968 J) by the world high jump record holder (Javier Sotomayor) in his world record jump of 2.45m. This comparison is because our current work equations do not take into account the force necessary to maintain one's speed over a period of time, merely looking at the amount of acceleration. Though Kimetto was assumed to be running at a constant speed for over two hours, his acceleration was almost 0, meaning that with our current equations, this seems as if he was inputting very little work for his two hours of running at record setting pace. Obviously, Kimetto was putting a lot of work into his running, but by our numbers, it was much easier for him to run 26.1 miles in two hours than for Usain Bolt to run 100m in 9.58s.

Sources:

http://www.nytimes.com/2013/10/14/sports/amid-heavier-security-course-record-is-set-in-chicago.html?_r=0

http://en.wikipedia.org/wiki/Javier_Sotomayor

http://colgatephys111.blogspot.com/2013/10/the-physics-behind-worlds-fastest-man.html

I wanted to calculate his average speed as well as his total work (just for the marathon not training) and see how much energy he had to expend to run the marathon in record time. To calculate his average speed (assuming a flat course) I found the total amount of time he was running (7425 seconds) and divided the total distance (26.1 miles or 42155.8 m) and found his average velocity to be 5.6 m/s. Assuming he accelerated to this speed in the first 10 meters of the race, I found his acceleration by using vf=vo+2ax. I found his acceleration to his top speed to be 0.28 m/s^2. And then assumed his acceleration for the remainder of the race to be 0 m/s^2, assuming he held a constant speed for the remaining 42145.8 m. So his average acceleration would be 6.64x10^-5 m/s^2. I also assumed his mass to be approximately 50 kg, and calculated his total work by using W=Fd, finding his total work to be 140J.

This seems absurdly low compared to the amount of energy expended by Usain Bolt in his record setting run (28500 J), and even the energy expended (1968 J) by the world high jump record holder (Javier Sotomayor) in his world record jump of 2.45m. This comparison is because our current work equations do not take into account the force necessary to maintain one's speed over a period of time, merely looking at the amount of acceleration. Though Kimetto was assumed to be running at a constant speed for over two hours, his acceleration was almost 0, meaning that with our current equations, this seems as if he was inputting very little work for his two hours of running at record setting pace. Obviously, Kimetto was putting a lot of work into his running, but by our numbers, it was much easier for him to run 26.1 miles in two hours than for Usain Bolt to run 100m in 9.58s.

Sources:

http://www.nytimes.com/2013/10/14/sports/amid-heavier-security-course-record-is-set-in-chicago.html?_r=0

http://en.wikipedia.org/wiki/Javier_Sotomayor

http://colgatephys111.blogspot.com/2013/10/the-physics-behind-worlds-fastest-man.html

### Cell phones as bullet proof vests?

In Florida this morning around 4:30 AM, a robber shot at a gas station clerk who was unable to open the safe. Miraculously, the clerk walked away from the shooting with only mild chest pain. In fact, the clerk didn't even realize he was hit until he took out his cell phone and found the bullet lodged in it.

I thought it was pretty amazing that the cell phone was able to stop the bullet. I was curious about how much force the cell phone exerted on the bullet and the deceleration experienced by the bullet.

In order to determine this, I made several assumptions. I found the average

**mass**of a bullet to be 200grains or**.01kg**. I also determined that the average muzzle**velocity**is 850ft/s or**259m/s**. The article stated that the robber shot at the clerk on his way out of the store, so I assumed that the bullet traveled 5m before hitting the clerk. A bullet traveling at 259m/s would travel 5m in**.02s.**
So, we have the following information:

m = .01kg

v0 = 259m/s

vF = 0m/s

t = .02s

We now have enough information to calculate the net force exerted on the bullet by the cell phone.

Force = change in momentum / time

change in momentum of bullet = mFvF - m0v0

=(.01kg)(0m/s) - (.01kg)(259m/s)

=-2.59kg m/s

Fcell phone = (-2.59kg m/s) / .02s

= -129.5N

Now that have calculated the net force, we can figure out the deceleration experienced by the bullet using:

F = ma

-129.5N = (.01kg)a

a = -12950m/s2

That's incredible!

source: http://www.nydailynews.com/news/national/clerk-cell-phone-stops-crook-bullet-article-1.1499126

## Saturday, October 19, 2013

### Physics of Shaolin Soccer by Rehman Momin

Physics of Shaolin Soccer

So my friend showed me this video ( http://www.youtube.com/watch?v=uV-dqhGd-YQ ) and at around 2:19, one player kicks the ball incredibly fast into the chest of an opposing player and the momentum of the ball pushes the player into the goal. So I wanted to figure out just how fast the player had to kick the ball to get both the ball and the opposing player to fly into the goal.

Seeing as how the man who got the ball kicked into his chest was a rather large fellow, I assumed his weight to be 100 kg. I found that the mass of a FIFA recognized level 5 ball is 0.450 kg. I also assumed that the chubby man was standing at the edge of the penalty box (16.5 m from the goal) and when hit by the ball, he and the ball both took 2 seconds (as recorded by the timeing in the video) to get into the goal.

Thus, we have the following:

Mass of man: 100 kg

Mass of ball: 0.450 kg

Distance and time for ball and man travel into the goal: 16.5 m in 2 seconds

-therefore, velocity of man and ball is 8.25 m/s

Initial velocity of man: 0 m/s

Using the conservation of momentum formula for an inelastic collision (because once the ball hits the man, the ball and man move together):

Initial momentum = Final momentum

m(ball)v(ball) + m(man)v(man) = [m(ball) + m(man)]v(ball + man)

(0.450 kg)(V) + (100 kg)(0 m/s) = [0.450 kg + 100 kg] (8.25 m/s)

V is of the kicked ball is roughly 1,841 m/s or 4119 mph

The fastest kick ever recorded in game was 114 mph (50.96 m/s) by David Hirst.

Gotta give it up to Shaolin Soccer!

So my friend showed me this video ( http://www.youtube.com/watch?v=uV-dqhGd-YQ ) and at around 2:19, one player kicks the ball incredibly fast into the chest of an opposing player and the momentum of the ball pushes the player into the goal. So I wanted to figure out just how fast the player had to kick the ball to get both the ball and the opposing player to fly into the goal.

Seeing as how the man who got the ball kicked into his chest was a rather large fellow, I assumed his weight to be 100 kg. I found that the mass of a FIFA recognized level 5 ball is 0.450 kg. I also assumed that the chubby man was standing at the edge of the penalty box (16.5 m from the goal) and when hit by the ball, he and the ball both took 2 seconds (as recorded by the timeing in the video) to get into the goal.

Thus, we have the following:

Mass of man: 100 kg

Mass of ball: 0.450 kg

Distance and time for ball and man travel into the goal: 16.5 m in 2 seconds

-therefore, velocity of man and ball is 8.25 m/s

Initial velocity of man: 0 m/s

Using the conservation of momentum formula for an inelastic collision (because once the ball hits the man, the ball and man move together):

Initial momentum = Final momentum

m(ball)v(ball) + m(man)v(man) = [m(ball) + m(man)]v(ball + man)

(0.450 kg)(V) + (100 kg)(0 m/s) = [0.450 kg + 100 kg] (8.25 m/s)

V is of the kicked ball is roughly 1,841 m/s or 4119 mph

The fastest kick ever recorded in game was 114 mph (50.96 m/s) by David Hirst.

Gotta give it up to Shaolin Soccer!

## Tuesday, October 15, 2013

### Woman gets Stuck on Drawbridge

This week in Florida, a 55 year old woman got stuck on a drawbridge while participating in a breast cancer walk. The woman held on to the bridge, while it was fully vertical, for a full 20 minutes at a height of 22 feet! The woman was eventually rescued and is now safe.

I was interested in this article and how it related to physics in regards to potential energy and the woman's final velocity if she were to hit the water.

Calculation of Potential Energy:

PE=mgh

•Assume mass to be: 70kg

•g= 9.8 m/s^2

•h= 22ft = 6.7m

PEg= 4,596.2 J

Calculation of Final Velocity:

•We
know that all the PE is converted to KE at the bottom of here fall (due to conservation of energy)

•KE=1/2mv2

•
4,596.2J= ½ (70kg) v2

v=
11.46 m/s

Sources: http://i.dailymail.co.uk/i/pix/2013/10/13/article-2456893-18B4F52400000578-991_634x844.jpg, http://www.huffingtonpost.com/2013/10/12/woman-stuck-on-bridge_n_4091061.html

## Sunday, October 13, 2013

I was curious to see how effective the crumple zone of my car (Honda Accord Coupe) was and how it compared to a car that I've always considered to be relatively unsafe (Smart Car).

----------------------------------------------------------------------------------------------------------------------------------

BACKGROUND

First thing first, the crumple zone of a car was introduced by Mercedes Benz as a safety feature in the 1950s. Over the years, more research has produce a more effective crumple zone. The Crumple Zone of a car is usually located in the front and the back of a car. Focusing primarily on the front end, the crumple zone is created by purposely making the front end weaker in some areas so that the car "crumples" (kind of like an accordion). The physics behind the crumple zone entails the use of the equation for Impulse.

Impulse = change in momentum = All forces * change in time

The idea behind crumple zones is to increase the time it takes for a car to decelerate, thereby making the force acting on the passengers at any point during the collision to be less. In other words, the longer the collision (or longer it takes to decelerate, the lower the magnitude of the force of the collision on the passenger).

----------------------------------------------------------------------------------------------------------------------------------

So now looking at the two cars I am interested, Honda Accord Coupe and the Smart Car, I needed to collect some information. Going on to Youtube to find timed crash tests of both the cars, I found that the crumple zone makes the length of the collision 0.092 seconds (for the Honda Accord Coupe) and 0.061 seconds (for the Smart Car). These times were collected from the moment of contact with the crash wall to the moment just before rebound. I will assume that without the crumple zone, the car would stop in 0.01 seconds.

Now to see how much force my body (65 kg) would feel hitting a stationary wall at 100 km/hr (27.8 m/s), I calculated the following:

Me in a car with NO CRUMPLE ZONE

Impulse = change in momentum = All forces * change in time

mv-mv = F * t

(65 kg)(0 m/s)-(65 kg)(27.8 m/s) = F * (0.01 sec)

F = -181,000 N

Me in the HONDA ACCORD COUPE with Crumple Zone

Impulse = change in momentum = All forces * change in time

mv-mv = F * t

(65 kg)(0 m/s)-(65 kg)(27.8 m/s) = F * (0.092 sec)

F = -19,600 N

Me in the SMART CAR with Crumple Zone

Impulse = change in momentum = All forces * change in time

mv-mv = F * t

(65 kg)(0 m/s)-(65 kg)(27.8 m/s) = F * (0.061 sec)

F = -29,600 N

Thus, in looking at the values, my car is much safer than the Smart Car, but the Smart Car's crumple zone (as small as it may be), is still incredibly effective at decreasing the magnitude of the force.

While 19,600 N of force is still a lot, one must not forget the other safety features (seat belt, air bags, etc) that follow the same principle of trying to elongate the time of force.

RESOURCES

http://www.youtube.com/watch?v=aGf57UUhO7w

http://www.youtube.com/watch?v=1z0EhZ0-EWo

## Friday, October 11, 2013

### When A Thrill Gets Too Thrilling

By Liz Flory

While
at the gym Thursday morning, a report of the Rip Ride Rocket Roller Coaster
malfunction at Universal Studios Orlando came on the news. This 1,200 meter
long coaster reaches maximum speeds of 105 km/hr, and has an initial 51 meter
vertical climb, followed by a series of inversions. On Wednesday night, a
coaster cart filled with 12 thrill seeking individuals ascended the vertical
climb and, after beginning its descent, was suddenly stopped by safety features
that detect when there is a technical glitch. The thrill seekers were stuck for
over two hours and were eventually rescued: the coaster car was pushed back up
to the top of the vertical climb, and firefighters escorted the riders out of
the cart and down emergency ladders.

While
watching the news I could only imagine how strong the safety features and
breaks must have been to stop the cart and hold it in place for two and a half
hours. The work done by the crew to move the cart and rescue the thrill seekers
must have been a significant effort as well! Because work is only done when
there is displacement, I set out to find:

1. The force exerted by the
safety features to stop the cart from accelerating down the incline to reach
its maximum speed

2. The force exerted by the
safety features to hold the cart in place for two and a half hours

3. The work done by the crew
to move the cart and rescue the riders.

To answer these questions I had to make a few (or more!)
assumptions. I assumed that cart went over the top of the ride at a speed of
1.0 m/s and the cart proceeded to get stuck on a linear incline with an angle
of inclination of 45°. I also assumed that the cart was stopped 10 m below the
top of the coaster, based on the image in the news. This means that the
distance the cart traveled was (10m)/sin45=14m. The Rip Ride Rocket is a steel
coaster, and modern coasters have polyurethane wheels. I researched the kinetic
coefficient of friction between steel and polyuretheane Âµ

_{K}=0.25. Based on the difference in kinetic and static friction between other materials and the fact that on a microscopic level the ridges of unmoving surfaces fit into each other, I assumed the static coefficient of friction would be 0.2 more, thus Âµ_{S}=0.45. Finally, based on research of the mass of other coaster carts with available data, I assumed that the mass of the cart, m_{cart}=550 kg. Using data from the CDC, the average weight of an American adult is 82.5 kg. Thus, the total mass held by the safety features of the coaster was (550 kg + 12(82.5 kg))= 1540 kg. With this information, I could begin answering my questions.
1. In order to find

**the force exerted by the safety features to stop the cart mid-descent**, I used Newton’s second law F=ma. To find the acceleration I needed to know the speed of the cart at its stopping point. Thus, I used the law of conservation of energy: ∆KE=-∆PE+W_{NC. }Because the cart had not descended far, I assumed that only frictional force between the wheels of the cart and the coaster, and not air resistance were contributing to W_{NC. }This frictional force is kinetic.
W

_{NC}=(F_{fr})(d)cosÎ¸=(-Âµ_{K}mgsinÎ¸)(d)cosÎ¸
1/2mv

_{2}^{2}-1/2mv_{1}^{2}=-(mgh_{2}-mgh_{1}) + W_{NC}
1/2(1540kg)(v

_{2}^{2})-1/2(1540kg)(1 m/s)^{2}=-((1540kg)(9.81m/s^{2})(41m-51m)) -(0.25)(1540kg)(9.81m/s^{2})(sin45)(14m)(cos45)
This gives a final velocity of 15 m/s. Using the kinematic
equation v

_{f}^{2}=v_{i}^{2}-2a∆d, acceleration is ((15 m/s)^{2}-(1 m/s)^{2})/(-2*-14m)=-8.2 m/s^{2}.
Thus, the force to stop the
cart is (1540kg)(-8.2 m/s

^{2}) =**13000N**.
2. Because gravity opposes
the force of the safety features and friction works in the same direction as
the force of the safety features, the

**force of the safety features to hold the cart in place**will equal the x component of gravity minus the force of static friction of the wheels against the steel.
F = mgcosÎ¸-Âµ

_{K}mgsinÎ¸ = (1540kg*9.81 m/s^{2})((cos45)-(0.45)(sin45)) =**5900N**—this is significantly less force than the initial force needed to stop the moving cart.
3. The cart must be moved by
the crew from its resting position 41m above the ground to a resting position at
the top of the coaster, 51m above the ground. Thus at the initial and final
points there is only PE

_{G }at play and friction works over the distance of movement. Using the Law of Conservation of Energy:
W

_{crew}= ∆KE=-∆PE+W_{NC}
Because the cart is at rest
in its initial and final positions, the cart has no change in kinetic energy, and
the W

_{crew}= 0. However, this answer is unsatisfying because we want to give the crew more credit!
If the cart were still
moving slightly at the top of the incline, right before stopping then:

W

_{crew}=-∆PE+W_{NC}
-((1540kg)(9.81m/s

^{2})(51m-41m))-(0.25)(1540kg)(9.81m/s^{2})(sin45)(14m)(cos45)=**-190,000 J**
Thank goodness for physics, safety
features and helpful firemen!

## Wednesday, October 9, 2013

By Laura Aseltine

Though I usually take the cruiser up the hill for class in the morning, lately I’ve been brainstorming some better ways to ride in style. One option I’ve been considering is to have my kind classmate Melissa Barnard pull me up the hill in a wagon. It can’t be that much work can it? I decided to use my nifty new physics calculate.

To
calculate the work I decided to split it up into W

_{horizontal }and W_{vertical}. Since W=F_{//}d I first needed to consider the forces in the horizontal direction, which would be F_{melissa }and F_{fr.}(I’m assuming that the only nonconservative force is F_{fr})._{ }I decided that we would be accelerating at 0.05 m/s^{2}and Melissa would be pulling me at a 15° angle. The force that she is pulling at is F=ma so I would need the mass. I added the mass of the wagon (13 kg), the mass of me (80kg) and the mass of my backpack (10kg) to get a total mass of 103 kg. F= 103kg * 0.05 m/s^{2}= 5.15 N. To determine the horizontal component of F_{melissa }I calculated cos(15°)*5.15= 4.97 N. I decided that the force of friction would be 0.5 N.
So to calculate the W

_{horizontal}I took the sum of the forces in the parallel direction (4.92N-0.5N) and multiplied it by the displacement. I used mapmyrun.com to map the displacement from my townhouse to the Ho. It was 0.8 miles, which converts to 1300 m. Therefore W_{horizontal }= (4.97N-0.5N)*1300= 5811J.
I then calculated the W

_{vertical.}The only force in the vertical direction is the y component of F_{Melissa}. I calculated this to be sin(15°)*5.14=1.33N. I multiplied this by the displacement in the y direction. Again I used mapmyrun.com to get the change in elevation, which was 197 ft, which converts to 60. m. Therefore W_{vertical}= 1.33N*60.m=79.8J.
To get the W

_{net}, I added these two works together. W_{net}= W_{horizontal}+ W_{vertical}= 5811J + 79.8J = 5889.9 J. Maybe this will convince her! Anyone have a wagon?## Sunday, October 6, 2013

### Hollywood Physics

I was watching Pacific Rim this weekend and thought of an amusing calculation to do. At one point in the film, a Kaiju (giant sea monster) and a Jaeger (giant robot designed to fight Kaiju's) were fighting at the edge of the atmosphere. The Kaiju is slain and the Jaeger, named Gypsy Danger, was falling back to earth. Ignoring all other forces, I was curious as to the gravitational potential energy of the Jaeger. I looked up how much Gypsy danger weighed, and from a fan site, found that it weighs 1,980 tons. Converted into kilograms, that's 1.8x10^6 kg. Assuming it was near the edge of Earth's atmosphere when it began falling, that would mean it was 80km (80000m) above the surface of the Earth (edge of the mesosphere). The formula for gravitational potential energy is PEg=mgh. That means the gravitational potential energy is (1.8x10^6)*(10)*(80,000) = 1.44x10^12 J. That is an enormous amount of energy. Put into perspective, it's recommended that average person takes in 2,000calories a day. Converted into joules, that's 8368J. You would need 172,084,130 people to consume that much energy.

http://pacificrim.wikia.com/wiki/Gipsy_Danger

http://www.srh.noaa.gov/jetstream/atmos/layers.htm

## Thursday, October 3, 2013

### The Physics Behind The World's Fastest Man

As I was checking out the recent sports
news, I saw an article about the world's fastest man, Usain Bolt of Jamaica,
and his contract extension with Puma. I knew that he was incredibly fast, but I
wondered what made Puma finally make this deal with him. Upon looking up the
time it took for Bolt to achieve his World Record the 100 m World Championships
in Berlin, I found that it only took him 9.58 seconds to complete the race. I
wanted to see how fast Bolt was running at different points during the race.

I plugged these values in to the
equation, v

_{avg}= (x_{f}– x_{o})/t, and found that Bolt’s average speed was 10.4 m/s. Although this value is very large, it is simply his average speed for the entire race. I looked up values of his time at different distances to determine his average speeds for each section of the race. From 0-20 meters, his average speed was 6.92 m/s. From 20-40 meters, his average speed was 11.4 m/s. From 40-60 meters, his average speed was 12.0 m/s. From 60-80 meters, his average speed was 12.4 m/s. From 80-100 meters, his average speed was 12.0 m/s. It is interesting to see that he is accelerating until he is at least 80 meters into the race, and that he only slightly slows down at the finish of the race.
After finding out how fast Usain Bolt
ran during his World Record race, I was curious about how much work he did
during the race. I found his average acceleration during the race to be about
3.00 m/s from an online source. I also found his mass to be about 95 kg. I used
the equation F = ma to determine the force that he exerted during the race. This
force came out to be 285 N. Using this force, the distance of the race, and the
equation, W = Fd, I determined Bolt’s work to be 28500 J. That's a lot of work
for only 9.77 seconds.

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