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_{K}mgsinθ)(d)cosθ
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 = mgcosθ-µ

_{K}mgsinθ = (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!

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