Download Roller Coasters

From http://electron4.phys.utk.edu/141/oct1/October1.html
Roller Coasters
In an amusement park the most thrilling rides let the rider experience the largest fictitious forces.
Fictitious forces are experienced in accelerating reference frames. The seat or cart in which the rider is
sitting is accelerating, because it is speeding up, slowing down, or turning rapidly. A ride that is just
moving you along with constant velocity will probably not excite you. You might as well be in your car
on the freeway with the cruise control engaged.
Roller coasters offer thrilling rides. Most roller coaster rides start with a steep "vertical" drop. The cart
descends on a track down a very steep slope.
Assume the top of the hill is 50 m above the ground. The gravitational potential energy of a object on
top of the hill is M g h. Neglecting friction, this potential energy has been converted into kinetic energy
when the object reaches the bottom of the hill.
K = (1/2) M v2 = M g h. v2 = 2 g h = 2 × 9.8 m/s2 × 50 m = 980 (m/s)2.
v = 31.1 m/s = 70 mph.
At the bottom of the hill the object (the cart, you, …) moves with a speed of 70 mph.
The forces on the object as it descends are the force of gravity and the support from the track. The
component of the gravitational force perpendicular to the track is canceled by the support from the track,
and the component of the gravitational force tangential to the track accelerate the object. However if you
are sitting in the accelerating cart, then you are experiencing a fictitious force in the direction of
theopposite the velocity in addition to the other forces. This force cancels the tangential component of
gravity. You are left with an apparent weight only due to the perpendicular component of the force of
gravity, much less than your normal weight.
If the track pointed straight down, the perpendicular component of the gravitational force would be zero,
and your apparent weight would be zero. You would feel weightless. A friend on the ground, however,
observes you accelerating with acceleration g because of your weight. The weight of the cart also
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accelerates the cart at 9.8 m/s2. Your velocity and the velocity of the cart change at the same rate, your
relative velocity stays zero. You are both in free fall, not subject to any forces other than gravity, just as
a skydiver, who stepped out of an airplane.
Since the cart of a roller coaster is attached to a track, it is possible, using motors, to produce
accelerations in the direction of the velocity greater than g.
Many roller coasters have a vertical loop, 20 m high. Assume a cart enters the loop at a speed of 25 m/s
or 56 mph. As the cart moves through the loop, it is moving in a circle and therefore it is accelerating.
An observer on the ground concludes that the forces on the cart are the force of gravity and the support
from the track. The support from the track has to cancel the component of the weight perpendicular to
the track. But if it just did that, the cart would keep on moving tangential to the track, since there would
be no force pointing towards the center of the circle, providing the centripetal acceleration. So the force
that the track exerts on the cart must be the sum of the perpendicular component of mg and the
centripetal force m (v2 / r).
A rider in the accelerating cart again experience a fictitious force in addition to the real forces. At the
bottom of the loop the fictitious centrifugal force has magnitude mass × (25 m/s)2 / 10 m = mass × 62.5
m / s2 = mass × 6.4 g in the downward direction. This force is added to the rider's weight to yield an
apparent weight of mass × 7.4 g.
On the top of the loop the some of the carts kinetic energy has been converted into potential energy.
(1/2) m vtop2 = (1/2) m vbottom2 - m g h
vtop2 = vbottom2 - 2 g h = (25 m/s)2 - 2 × 9.8 m/s2 × 20 m = 233 (m/s)2
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v = 15.3 m/s
The fictitious centripetal force the rider experience is now directed upward and has magnitude m (v2 / r)
= mass × (15.3 m/s)2 / 10 m = mass × 23.4 m / s2 = mass × 2.4 g.
Subtracting the real weight of the rider, which points in the downward direction, we obtain an apparent
weight = mass × 1.4 g in the upward direction. Things seem to "fall" upward, towards the bottom of the
inverted car.
Halfway up the loop the weight points straight down, but the fictitious centrifugal force points
horizontally toward the outside of the loop. The apparent weight is therefore directed downward and
outward.
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