Physics of The Crumple Zone

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).

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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).

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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

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 µ_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 µ_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)cosθ=(-µ_Kmgsinθ)(d)cosθ

1/2mv₂²-1/2mv₁²=-(mgh₂-mgh₁) + W_NC

1/2(1540kg)(v₂²)-1/2(1540kg)(1 m/s)²=-((1540kg)(9.81m/s²)(41m-51m)) -(0.25)(1540kg)(9.81m/s²)(sin45)(14m)(cos45)

This gives a final velocity of 15 m/s. Using the kinematic equation v_f²=v_i²-2a∆d, acceleration is ((15 m/s)²-(1 m/s)²)/(-2*-14m)=-8.2 m/s².

Thus, the force to stop the cart is (1540kg)(-8.2 m/s²) = 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 = mgcosθ-µ_Kmgsinθ = (1540kg*9.81 m/s²)((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/s²)(51m-41m))-(0.25)(1540kg)(9.81m/s²)(sin45)(14m)(cos45)= -190,000 J

Thank goodness for physics, safety features and helpful firemen!

           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/s² 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/s² = 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?

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

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.

How hard does an NFL defender hit?

As an avid football fan, I spend every weekend watching hours of NFL action. Any football fan will tell you they love to see the big hits. I’m no different. In watching this weekend’s games, I found myself wondering exactly how hard do they hit and how fast are these athletes moving when they deliver these devastating blows? I did some research on average weight and speed stats for various positions for athletes performing at the NFL combine and used these in our kinematic and Newtonian equations to find out for myself. (Note, for ease of calculation, I’ve ignored wind resistance and assumed constant acceleration. All calculations are completed assuming 40yards of running distance.)

Cornerbacks: While not the biggest guys on the field, corners make up for their lack in size with blazing speed. Average 40yd times for cornerbacks in the past 5 years at the combine clock in at 4.55seconds. Using the kinematic equation Δx=V_°t+½at² (again assuming constant acceleration) I found that the average NFL corner can accelerate at 3.56m/s². Using the equation V=V_°+at I found that at the end of the 40 yards, corners are running at an average speed of 16.2m/s or 36.2mi/hr. With an average mass of 87.5kg, these corners can deliver a force of impact of 312N (after a 40yard running start, that is) (using the equation F=ma).

Linebackers: These big boys rely on their size to punish running backs and receivers unlucky enough to find themselves in the way of the oncoming pain train. Average 40yd times for linebackers in the past 5 years at the combine clock in at 4.77seconds. Using the same equations as above, I found that the average NFL linebacker can accelerate at 3.25m/s². At the end of the 40yds, ‘backers are traveling at an average speed of 15.5m/s or 34.7mi/hr. With an average mass of 109kg, ‘backers plow into the offense with a force of 354N.

Safeties: The last line of defense. These gifted athletes averaged a 40yd time of 4.62seconds over the past 5 years at the combine. Using this time, I found that safeties can accelerate at 3.46m/s². At the end of the 40yds, safeties are traveling an average speed of 16.0m/s or 35.8mi/hr. With an average mass of 94.8kg, safeties run over the competition with a force of 328N.

Needless to say, I don’t think I’d want to be on the wrong end of an NFL defender moving at full speed.

Static vs. Kinetic Friction

I was watching Top Gear this weekend (obviously the UK version) and thought about static versus kinetic friction in terms of a drag race and why static friction coefficients are greater than kinetic friction coefficients.  The hosts often test to see how fast a car can go in a given distance.  When they first step down on the gas peddle, there is a significant amount of static friction preventing the car from moving immediately.  Enough force must be applied to get the car to move to overcome the molecular interactions of the tires and ground "locking" together.  Once the car is moving, these forces are easier to overcome because the car already moving and has acceleration, forces being applied, etc.  This means there is less kinetic friction to overcome because the molecular interactions aren't as "locked" together.  So once the car is moving, it speeds up really quickly.  So the coefficient of kinetic friction is less than the coefficient of static friction.