People
are wild about amusement parks. Each day, we flock by the millions to the
nearest park, paying a sizable hunk of money to wait in long lines for a short
60-second ride on our favourite roller coaster. The thought prompts one to
consider what is it about a roller coaster ride that provides such widespread
excitement among so many of us and such dreadful fear in the rest? Is our
excitement about coasters due to their high speeds? Absolutely not! In fact, it
would be foolish to spend so much time and money to ride a selection of roller
coasters if it were for reasons of speed. It is more than likely that most of
us sustain higher speeds on our ride along the interstate highway on the way to
the amusement park than we do once we enter the park. The thrill of roller
coasters is not due to their speed, but rather due to their accelerations and
to the feelings of weightlessness and weightiness that they produce. Roller
coasters thrill us because of their ability to accelerate us downward one
moment and upwards the next; leftwards one moment and rightwards the next.
Roller coasters are about acceleration; that's what makes them thrilling. And
in this part of Lesson 2, we will focus on the centripetal acceleration
experienced by riders within the circular-shaped sections of a roller coaster
track. These sections include the clothoid loops (that we will approximate as a
circle), the sharp
180-degree banked turns, and the small dips
and hills found
along otherwise straight sections of the track.
The
most obvious section on a roller coaster where centripetal acceleration occurs
is within the so-called clothoid loops. Roller
coaster loops assume a tear-dropped shape that is geometrically referred to as
a clothoid. A clothoid is a section of a spiral in which the radius is
constantly changing. Unlike a circular loop in which the radius is a constant
value, the radius at the bottom of a clothoid loop is much larger than the
radius at the top of the clothoid loop. A mere inspection of a clothoid reveals
that the amount of curvature at the bottom of the loop is less than the amount
of curvature at
the top of the loop. To simplify our analysis of the physics of clothoid loops,
we will approximate a clothoid loop as being a series of overlapping or adjoining
circular sections. The radius of these circular sections is decreasing as one
approaches the top of the loop. Furthermore, we will limit our analysis to two
points on the clothoid loop - the top of the loop and the bottom of the loop.
For this reason, our analysis will focus on the two circles that can be matched
to the curvature of these two sections of the clothoid. The diagram at the
right shows a clothoid loop with two circles of different radius inscribed into
the top and the bottom of the loop. Note that the radius at the bottom of the
loop is significantly larger than the radius at the top of the loop.
As
a roller coaster rider travels through a clothoid loop, she experiences an
acceleration due to both a change in speed and a change in direction. A
rightward moving rider gradually becomes an upward moving rider, then a
leftward moving rider, then a downward moving rider, before finally becoming a
rightward-moving rider once again. There is a continuous change in the
direction of the rider as she moves through the clothoid loop. a change in direction is one characteristic of an
accelerating object. In addition to changing directions, the rider also changes
speed. As the rider begins to ascend (climb upward) the loop, she begins to
slow down. As energy
principles would
suggest, an increase in height (and in turn an increase
in potential energy) results in a decrease in kinetic energy and speed. And
conversely, a decrease in height (and in turn a decrease in potential energy)
results in an increase in kinetic energy and speed. So the rider experiences
the greatest speeds at the bottom of the loop - both upon entering and leaving
the loop - and the lowest speeds at the top of the loop.
This
change in speed as the rider moves through the loop is the second aspect of the
acceleration that a rider experiences. For a rider moving through a circular
loop with a constant speed, the acceleration can be described as being
centripetal or towards the center of the circle. In the case of a rider moving
through a noncircular loop at non-constant speed, the acceleration of the rider
has two components. There is a component that is directed towards the center of
the circle (ac) and attributes itself to the direction
change; and there is a component that is directed tangent (at) to the
track (either in the opposite or in the same direction as the car's direction
of motion) and attributes itself to the car's change in speed. This tangential
component would be directed opposite the direction of the car's motion as its
speed decreases (on the ascent towards the top) and in the same direction as
the car's motion as its speed increases (on the descent from the top). At the
very top and the very bottom of the loop, the acceleration is primarily
directed towards the center of the circle. At the top, this would be in the
downward direction and at the bottom of the loop it would be in the upward
direction.
We
learned that the inwards
acceleration of an
object is caused by an inwards net
force. Circular motion (or merely motion along a curved path)
requires an inwards component of net force. If all the forces that act upon the
object were added together as vectors, then the net force would be directed
inwards. Neglecting friction and air resistance, a roller coaster car will
experience two forces: the force of
gravity (Fgrav)
and the normal force (Fnorm). The normal force is
directed in a direction perpendicular to the track
and the gravitational force is always directed downwards. We will concern
ourselves with the relative magnitude and direction of these two forces for the
top and the bottom of the loop. At the bottom of the loop, the track pushes
upwards upon the car with a normal force. However, at the top of the loop the
normal force is directed downwards; since the track (the supplier of the normal
force) is above the car, it pushes downwards upon the car. The free-body
diagrams for these two positions are shown in the diagrams at the right.
The
magnitude of the force of gravity acting upon the passenger (or car) can easily
be found using the equation Fgrav = m•g where g =
acceleration of gravity (9.8 m/s2). The magnitude of the normal
force depends on two factors - the speed of the car, the radius of the loop and
the mass of the rider. As depicted in the free body diagram, the magnitude of Fnorm is always greater at the bottom of the loop
than it is at the top. The normal force must always be of the appropriate size
to combine with the Fgrav in such a
way to produce the required inward or centripetal net force. At the bottom of
the loop, the Fgrav points
outwards away from the center of the loop. The normal force must be
sufficiently large to overcome this Fgrav and supply some excess force to result in a
net inward force. In a sense, Fgrav and Fnorm are in a tug-of-war; and Fnorm must win by an amount equal to the net force.
At the top of the loop, both Fgrav and Fnorm are directed inwards. The Fgrav is found in the usual way (using the equation
Fgrav = m•g).
Once more the Fnorm must
provide sufficient force to produce the required inward or centripetal net
force.
Earlier
TOPIC, the use of Newton's second law and free-body diagrams to solve
circular motion diagrams was illustrated. It was emphasized at that time that
any given physical situation could be analyzed in terms of the individual
forces that are acting upon an object. These individual forces must add up as
vectors to the net force. Furthermore, the net force must be equal to the mass
times the acceleration. The process of conducting a force analysis of a
physical situation was EXPLAIND. Now we will investigate the use of these
fundamental principles in the analysis of situations involving the motion of
objects in circles. We will utilize the basic problem-solving approach that was introduced earlier in Lesson 2. This
approach can be summarized as follows.
Suggested Method of Solving
Circular Motion Problems 1. From the verbal description of the
physical situation, construct a free-body diagram. Represent each force by a
vector arrow and label the forces according to type. 2. Identify the given and the unknown
information (express in terms of variables such as m= ,a= , v= ,
etc.). 3. If any of the individual forces are
directed at angles to the horizontal and vertical, then use vector
principles to resolve such
forces into horizontal and vertical components. 4. Determine the magnitude of any known
forces and label on the free-body diagram. 5. Use circular
motion equations to determine
any unknown information. 6. Use the remaining information to solve
for the requested information. · If the problem requests the value of an
individual force, then use the kinematic information (R, T and v) to
determine the acceleration and the Fnet; then use the free-body
diagram to solve for the individual force value. · If the problem requests the value of
the speed or radius, then use the values of the individual forces to
determine the net force and acceleration; then use the acceleration to
determine the value of the speed or radius. |
Combine
a force analysis with the above method to solve the following roller coaster
problem.
Sample Roller Coaster Problem Anna
Litical is riding on The Demon at Great America. Anna experiences a downward
acceleration of 15.6 m/s2 at the top of the loop and an upward
acceleration of 26.3 m/s2 at the bottom of the loop. Use Newton's
second law to determine the normal force acting upon Anna's 864 kg roller
coaster car. |
Steps
1 and 2 involve the construction of a free body diagram and the identification
of known and unknown quantities. This is shown in below.
Given Info: m = 864 kg atop = 15.6 m/s2 , down bottom =
26.3 m/s2 , up Find: Fnorm at top and bottom |
Step
3 of the
suggested method would not apply to this problem since there are no forces
directed "at angles" (that is, all the forces are either horizontally
or vertically directed). Step 4 of the suggested method involves the
determination of any known forces. In this case, the force of gravity can be
determined from the equation Fgrav = m • g. Using a g value of
9.8 m/s2, the force of gravity acting upon the 864-kg car is
approximately 8467 N. Step 5 of the suggested method would be used if the
acceleration were not given. In this instance, the acceleration is known. If
the acceleration were not known, then it would have to be calculated from speed
and radius information.
Step
6 of the
suggested method involves the determination of an individual force - the normal
force. This will involve a two-step process: first the net force (magnitude and
direction) must be determined; then the net force must be used with the free body
diagram to determine the normal force. This two-step process is shown below for
the top and the bottom of the loop.
Bottom of Loop Fnet = m * a Fnet = (864 kg) * (26.3 m/s2, up) Fnet = 22 723 N, up From FBD:
Fnorm must be greater than the Fgrav by 22723 N in order to supply a net
upwards force of 22723 N. Thus, Fnorm = Fgrav + Fnet Fnorm = 31190 N |
Top of Loop Fnet = m * a Fnet = (864 kg) * (15.6 m/s2,
down) Fnet = 13478 N, down From FBD:
Fnorm and Fgrav together must combine together (i.e.,
add up) to supply the required inwards net force of 13478 N. Thus, Fnorm = Fnet - Fgrav Fnorm = 5011 N |
Observe
that the normal force is greater at the bottom of the loop than it is at the
top of the loop. This becomes a reasonable fact when circular motion principles
are considered. At all points along the loop - which we will refer to as
circular in shape - there must be some inward component of net force. When at
the top of the loop, the gravitational force is directed inwards (down) and so
there is less of a need for a normal force in order to meet the net centripetal
force requirement. When at the bottom of the loop, the gravitational force is
directed outwards (down) and so now there is a need for a large upwards normal
force in order to meet the centripetal force requirement. This principle is
often demonstrated in a physics class using a bucket of water tied to a string.
The water is spun in a vertical circle. As the water traces out its circular
path, the tension in the string is continuously changing. The tension force in
this demonstration is analogous to the normal force for a roller coaster rider.
At the top of the vertical circle, the tension force is very small; and at the
bottom of the vertical circle, the tension force is very large. (You might try
this activity yourself outside with a small plastic bucket half-filled with
water. Give extra caution to stay clear of all people, windows, trees and
overhead power lines. Repeat enough cycles to observe the noticeable difference
in tension force when the bucket is at the top and the bottom of the circle.)
If
you have ever been on a roller coaster ride and traveled through a loop, then
you have likely experienced this small normal force at the top of the loop
and the large normal force at the bottom of the loop. The normal force provides
a feel for a person's weight., we
can never feel our weight; we can only feel other forces that act as a result
of contact with other objects.) The more you weigh, the more normal force that
you will experience when at rest in your seat. But if you board a roller
coaster ride and accelerate through circles (or clothoid loops), then you will
feel a normal force that is constantly changing and different from that which
you are accustomed to. This normal force provides a sensation or feeling of
weightlessness or weightiness. When at the top of the loop, a rider will feel partially weightless if the normal forces
become less than the person's weight. And at the bottom of the loop, a rider
will feel very "weighty" due to the increased normal forces. It is
important to realize that the force of gravity and the weight of your body are
not changing. Only the magnitude of the supporting normal force is changing!
(The phenomenon of weightlessness will be discussed in much more detail NEXT TOPIC.)
There
is some interesting history (and physics) behind the gradual usage of clothoid
loops in roller coaster rides. In the early days of roller coaster loops,
circular loops were used. There were a variety of problems, some of which
resulted in fatalities, as the result of the use of these circular loops.
Coaster cars entering circular loops at high speeds encountered excessive
normal forces that were capable of causing whiplash and broken bones. Efforts
to correct the problem by lowering entry speeds resulted in the inability of
cars to make it around the entire loop without falling out of the loop when
reaching the top. The decrease in speeds as the cars ascended the large
circular loop resulted in coaster cars turning into projectile cars (a
situation known to be not good for business). The solution to the problem
involved using low entry speeds and a loop with a sharper curvature at the top
than at the bottom. Since clothoid loops have a continually changing radius,
the radius is large at the bottom of the loop and shortened at the top of the
loop. The result is that coaster cars can enter the loops at high speeds; yet
due to the large radius, the normal forces do not exceed 3.5 G's. At the top of the loop, the
radius is small thus allowing a lower speed car to still maintain contact with
the track and successfully make it through the loop. The clothoid loop is a
testimony to an engineer's application of the centripetal acceleration equation
- a = v2/R. Now that's physics for better living!
The
above discussion and force analysis applies to the circular-like motion of a
roller coaster car in a clothoid loop. The second section along a roller
coaster track where circular motion is experienced is along the small dips and
hills. These sections of track are often found near the end of a roller coaster
ride and involve a series of small hills followed by a sharp drop. Riders often
feel heavy as they ascend the hill (along regions A and E in the diagram
below). Then near the crest of the hill (regions B and F), their upward motion
makes them feel as though they will fly out of the car; often times, it is only
the safety belt that prevents such a mishap. As the car begins to descend the
sharp drop, riders are momentarily in a state of free fall (along regions C and
G in the diagram below). And finally as they reach the bottom of the sharp dip
(regions D and H), there is a large upwards force that slows their downward
motion. The cycle is often repeated mercilessly, churning the riders' stomachs
and mixing the afternoon's cotton candy into a slurry of
... . These small dips and hills combine the physics of circular
motion with the physics of projectiles in order to produce the ultimate thrill
of acceleration - rapidly changing magnitudes and directions of acceleration.
The diagram below shows the various directions of accelerations that riders
would experience along these hills and dips.
At
various locations along these hills and dips, riders are momentarily traveling
along a circular shaped arc. The arc is part of a circle - these circles have
been inscribed on the above diagram in blue. In each of these regions there is
an inward component of acceleration (as depicted by the black arrows). This
inward acceleration demands that there also be a force directed towards the
center of the circle. In region A, the centripetal force is supplied by the
track pushing normal to the track surface. Along region B, the centripetal
force is supplied by the force of gravity and possibly even the safety
mechanism/bar. At especially high speeds, a safety bar must supply even extra
downward force in order to pull the riders downward and supply the remaining
centripetal force required for circular motion. There are also wheels on the
car that are usually tucked under the track and pulled downward by the track.
Along region D, the centripetal force is once more supplied by the normal force
of the track pushing upwards upon the car.
The magnitude of the normal forces along these various regions
is dependent upon how sharply the track is curved along that region (the radius
of the circle) and the speed of the car. These two variables affect the
acceleration according to the equation
a = v2 /
R
and in turn affect the net force. As suggested by the
equation, a large speed results in a large acceleration and thus increases the
demand for a large net force. And a large radius (gradually curved) results in
a small acceleration and thus lessens the demand for a large net force. The
relationship between speed, radius, acceleration,
mass and net force can be used to determine the magnitude of the seat force (i.e., normal force) upon a roller coaster
rider at various sections of the track. The sample problem below illustrates
these relationships. In the process of solving the problem, the same problem-solving strategy enumerated above will be utilized.
Sample Roller Coaster Problem Anna
Litical is riding on The American Eagle at Great America. Anna is moving at
18.9 m/s over the top of a hill that has a radius of curvature of 12.7 m. Use
Newton's second law to determine the magnitude of the applied force of the
track pulling down upon Anna's 621 kg roller coaster car. |
Steps
1 and 2 involve the construction of a free body diagram and the identification
of known and unknown quantities. This is shown in below.
|
Given Info:
Find: Fapp at top of hill |
Step
3 of the
suggested method would not apply to this problem since there are no forces
directed "at angles" (that is, all the forces are either horizontally
or vertically directed). Step 4 of the suggested method involves the
determination of any known forces. In this case, the force of gravity can be
determined from the equation Fgrav = m * g. So the force of gravity acting
upon the 621-kg car is approximately 6086 N. Step 5 of the
suggested method involves the calculation of the acceleration from the given
values of the speed and the radius. Using the equation given in Lesson 1, the
acceleration can be calculated as follows
a = v2 /
R
a = (18.9 m/s)2 / (12.7 m)
a = 28.1 m/s2
Step
6 of the
suggested method involves the determination of an individual force - the
applied force. This will involve a two-step process: first the net force
(magnitude and direction) must be determined; then the net force must be used
with the free body diagram to determine the applied force. This two-step
process is shown below.
Fnet = m • a Fnet = (621 kg) • (28.1 m/s2,
down) Fnet = 17467 N, down As shown in FBD at right: Fapp = Fnet - Fgrav Fnorm = 11381 N |
Fapp and Fgrav must combine together (i.e., add up)
to supply the required downwards net force of 17467 N. |
This
same method could be applied for any region of the track in which roller
coaster riders momentarily experience circular motion.
.
1.
Anna Litical is riding on The Shock Wave at Great America. Anna experiences a
downward acceleration of 12.5 m/s2 at the top of the loop and an upward
acceleration of 24.0 m/s2 at the
bottom of the loop. Use Newton's second law to determine the normal force
acting upon Anna's 50-kg body at the top and at the bottom of the loop.
Fnorm = 135 N (top) and Fnorm =
1690 N (bottom)
Use a, m, and g (9.8 m/s/s) with Fnet = m
• a and Fgrav = m•g to find
Fnet and Fgrav.
Then use a free-body diagram to find Fnorm.
Top:
Fnet =
625 N, down and Fgrav = 490 N, down.
So Fnorm =
135 N, down.
Bottom:
Fnet =1200
N, up and Fgrav = 490 N,
So Fnorm =1690 N, up.
2.
Noah Formula is riding a roller coaster and encounters a loop. Noah is
traveling 6 m/s at the top of the loop and 18.0 m/s at the bottom of the loop.
The top of the loop has a radius of curvature of 3.2 m and the bottom of the
loop has a radius of curvature of 16.0 m. Use Newton's second law to determine
the normal force acting upon Noah's 80-kg body at the top and at the bottom of
the loop.
Fnorm = 116 N (top) and Fnorm =
2404 N (bottom)
Use a, m, and g (9.8 m/s/s) with Fnet = m
• a and Fgrav = m • g to
find Fnet and Fgrav.
Then use a free-body diagram to find Fnorm.
Top:
a = v2 /
R = (6 m/s)2 / (3.2 m) = 11.25 m/s2
Fnet = 900 N, down and Fgrav =
784 N, down.
So Fnorm =
116 N, down.
Bottom:
a = v2 /
R = (18 m/s)2 / (16 m) = 20.25 m/s2
Fnet = 1620 N, up and Fgrav =
784 N,
So Fnorm =
2404 N, up.
3.
Noah Formula is riding an old-fashioned roller coaster. Noah encounters a small
hill having a radius of curvature of 12.0 m. At the crest of the hill, Noah is
lifted off his seat and held in the car by the safety bar. If Noah is traveling
with a speed of 14.0 m/s, then use Newton's second law to determine the force
applied by the safety bar upon Noah's 80-kg body.
Fnorm = 523 N
Solution is as follows:
First, draw a free-body diagram and note that Fgrav =
784 N, down.
Second, calculate acceleration by
a = v2 /
R = (14 m/s)2 / (12 m) =
16.3 m/s/s.
Then note that Fnet = m • a = 1307 N down
(toward center).
Now Fgrav supplies 784 N of this downward force,
so the Fnorm must supply the rest. Therefore, Fnorm =
523 N.
4.
Anna Litical is riding a "woody" roller coaster. Anna encounters the
bottom of a small dip having a radius of curvature of 15.0 m. At the bottom of
this dip Anna is traveling with a speed of 16.0 m/s and experiencing a much
larger than usual normal force. Use Newton's second law to determine the normal
force acting upon Anna's 50-kg body.
Fnorm = 1343 N, up
Solution is as follows:
First, draw a free-body diagram and note that Fgrav =
490 N, down.
Second, calculate acceleration by
a = v2 /
R = (16 m/s)2 / (15 m) =
17.1 m/s/s, up
Then note that Fnet = m • a = 853 N, up (toward
center).
Now Fgrav supplies
490 N of downward force, so the Fnorm must overcome this down
force and still supply the sufficient Fnet in the up direction.
Therefore, Fnorm = 1343 N,up.