physics behind roller coaster

PHYSICS BEHIND ROLLER COASTER

REPORT OF DISCUSSION

Presented to fulfill the duty of Kapita Selekta Fisika Sekolah

that has been teached by Mr. Sutopo

By Group VI:

1. Miftachul Choiriyah 109321417091 2. Aurora Kumala Cahyaningtyas 109321417099 3. Mukhammad Adharul M. 10932141709 4. Sinta Nurrisa Karonsih 109321417113

STATE UNIVERSITY OF MALANG

FACULTY OF MATHEMATICS AND NATURAL SCIENCES

DEPARTMENT OF PHYSICS

November, 2011

PROBLEM A

Position of Cars where the Passengers feel zero and 8g

Apparent Weight Respectively

Some people excited to play roller coaster. They usually play this

amusement-park ride because they think that this is fun and they can enjoy the

sensation when they ride it. One of the causes of the sensation that felt by the

passenger of roller coaster is because the rapid changes in passengers apparent

weight comes from the normal force that we (as passengers) felt from the floor,

seat, etc. The magnitudes of apparent weights (normal force) that felt by the

passenger of roller coaster are different at each position on the paths of roller

coaster.

For example if we want to discuss about the magnitude of apparent weight

that felt by the passenger of roller coaster in the case at the given problem.

Consider the friction force at this case is zero and WA= WB = WC = WD = WE = WF

= W= mg. We can divide the path of roller coaster at given problem to several

paths in order to make us easy. For example path A-B is linear path before the

valley, path B-C-D is first circular paths with radius R that make a valley at the

bottom of this circular path, then path D-E is linear path after the valley, and the

last is path E-F-G is path of second circular path with radius R that make a

mountain (or hill) from the arc of second circle that has height h from the

ground that shown by this figure:

Figure1. The profile of roller coaster being designed

Path A-B ( Point 1)

Based on figure 2:

PATH B-C-D

There are different sensations of our apparent weight that we felt when we

passed this path. This path is an arc of a circle with radius R that fits the bottom of

Figure2. Forces that acted at car on

the roller coaster at path A-B.

Figure was taken from Physics for Scientists and Engineers (2004,

p.134)

If we consider that the angle of straight

line tracks with horizontal line is and 0 <

< 90o, we can find the components of force

that acted roller coaster cars (symbolized

with a box) as shown at figure 2. If we also

consider there is no friction force during the

motion of the cars, f at figure 2 is equal to

zero. Apparent weight that felt by passengers

is proportional to the normal force that acted

on cars (shown with n at figure 2). We can

write n = N at this case to shown normal

force that acted on the cars.

yy maF0.cos1 mmgN

cos1 mgN

11 mgN

So at this point,

cos1 gg

And the apparent weight that felt by

passengers at this point is:

cos1 mgN

valley as figure in figure3. Roller coaster cars were symbolized with a point at this

figure. At this path, the magnitude of the angle between radius R to the normal

line at the bottom of valley are change from to 0 at path B-C and change from 0

to at path C-D.

Point 2 that moved from B-C

When we observe point 2 that moved at path B-C, there were some

components of force that acted at that point at that position along the path B-C.

The magnitude of the angle between radius R to the normal line at the bottom of

valley is change from to 0 and the height of cars at this path from the ground is

also change from hB (h ) to hC (0 m). Based on second Newtons law,

R

vgmN

R

vmmgN

R

vmWN

R

vmWN

maF

B

B

yy

2

22

2

22

2

22

2

22

cos

cos

cos

cos

R

vgg

R

vgg

so

mgN

B

2

2

22

22

cos

cos

As we know that

y

x

The height of the car was changed from h to hC (0 m), based on conservation Law

of mechanical energy;

Point 3 at Bottom of the Valley (C)

When the car passed point C at the bottom of the valley, there were some

components of force that acted at that point at that position. There were no angle

that formed between radius R to the normal line at this position ( = 0). In this

position, we usually feel that our apparent weight is very heavy. Based on second

Newtons law,



R

hhggmN

R

vgmN

R

vgmN

2cos

cos

cos

2

2

2

2

22

R

vgmN

R

vmmgN

R

vmWN

R

vmWN

maF

C

C

yy

2

3

3

2

3

3

2

33

2

3

3

R

vgg

so

mgN

2

3

3

33

As we know that

hhgv

mvhhmg

mvmghmgh

mghmvmghm

mghmvmghmv

EMEM

BBAA

BA

2

2

21

21)0(

21

21

21

2

2

2

22

22

y

x

Based on conservation Law of mechanical energy;

Point 4 that moved from C-D

When the car moved at path C-D, the magnitude of the angle between

radius R to the normal line at the bottom of valley is change from 0 to and the

height of cars at this path from the ground is also change from hC (0 m) to hD (h).

Based on second Newtons law,

Because the height of cars at D is same with the height of cars at B, so the velocity

of cars at D is same too with the velocity of cars at B (based on conservation law

of mechanical energy), so:



R

ghgmN

R

vgmN

23

2

3

3

ghv

mvmgh

mvmgh

mgmvmghm

mghmvmghmv

EMEM

CCAA

CA

2

2

21

02

1)0(2

1

21

21

2

3

2

3

2

3

2

3

2

22

R

vmWN

R

vmWN

maF

D

D

yy

2

44

2

44

R

vgmN

R

vmmgN

D

2

4

2

44

hhgvvv 222

4

2

2

x

y

And the apparent weight that felt by passengers at this point is:

Path D-E (Point 5)

The components of force of roller coaster cars at this path is same with the

components of force of roller coaster cars at path A-B because the magnitude of

the angle of this path (as a straight line) with horizontal line is same with the

magnitude of angle of path A-B with the horizontal line, that are . The different

of the cars motion between path A-B and D-E is the cars are moving down at

path A-B (the cars are speeding up), but at path D-E the cars are moving up so we

(as passenger) feel that the cars are slowing down. The components of force that

acted on the roller coaster cars at this path can be seen in this figure below, with

the roller coaster cars symbolized as a point like in this figure:

R

hhggmN

R

vgmN

R

vgmN

2cos

cos

cos

4

2

4

2

44

x

y

Same with point A at path A-B,

And the magnitude of apparent weight that felt by passengers at this path is same

with the magnitude of apparent weight that felt by passengers at path A-B:

PATH E-F-G

Figure8. Path E-F-G at roller coaster cars track

After passed a straight line track at path D-E, the roller coaster cars will

pass a hill (or mountain) as the arc of the circle track that has same radius R

with the arc of the first circle as shown as figure 8 above. The roller coaster cars

are symbolized with a point. Consider that point 6 represent the roller coaster cars

that moved from path E-F and point 7 is represent the roller coaster cars at the top

of the arc of the second circle (located at h from the ground at F).

coscos

0cos5

mgWN

mWN

maF

EE

E

yy

As we know that

15

55

cos ggg

so

mgN

cos51 mgNN

Point 6 (The roller coaster cars moved at path E-F)

Figure9. The forces that acted on the roller coaster cars at E

Consider that point 6 represent the roller coaster cars that moved from

path E-F that formed as an arc of the circle as shown as figure9. The magnitude of

the angle between the radius of this circle at E and the normal force at the peak of

arch of the circle as the track of motion is . To distinguish this angle with the first

angle at the previous circle that make valley on roller coaster track, we can

symbolized as , and has the same magnitude. The components of force that

acted on point 6 is same with the components force on point 5 because point 5 and

6 was observed at the same position, that is E. But we must realize that the path of

cars was different, from E until G the track of the motion is in circular path.

R

vgmN

R

vmmgN

R

vmWN

R

vmNW

maF

E

E

yy

2

6

6

2

6

6

2

66

2

6

6

cos

cos

cos

cos

As we know that

R

vgg

R

vgg

so

mgN

2

6

2

6

6

66

cos

cos

x

y


Point 7 at the Peak of Mountain (F)

Figure10. The components of force that acted on the roller coaster cars at the

peak (top) of the second circle (F position)

As a passenger, we usually feel that our apparent weight is very small

(even we feel that we can fly if we dont use safety belt) when we reach the top of

mountain at the track of roller coaster. The top of roller coaster is shown at F in

figure10. The components of force that acted on the roller coaster cars at this

position are:

R

hhggmN

R

vgmN

R

vgmN

2cos

cos

cos

6

2

6

2

6

6

hhgv

mvhhmg

mvmghmgh

mghmvmghm

mghmvmghmv

EMEM

EEAA

EA

2

2

21

21)0(

21

21

21

2

2

2

22

22

So the apparent weight that felt by


R

vmWN

R

vmNW

maF

F

yy

2

7

77

2

7

7

R

vgmN

R

vmmgN

2

7

7

2

7

7

As we know that so


From the given problem, we know that the roller coaster provide

passengers with the sensation of apparent weight that varies from zero to 8g (our

normal weight is 1g WNormal = mg=1g so m=1).

Table1. The apparent weight that felt by passengers at each position on the roller coaster track

Path Point The Apparent Weigh Gravitational acceleration (g)

A-B 1

B-C-D

2

3

4

D-E 5

E-F-G

6

R

hhgg

2cos

7

The magnitude of is same with the magnitude of , h higher than h and

h higher than h, but h is still higher than h. The maximum value of cos =

cos are 1 and the minimum value of cos = cos are 0, so the magnitude of

apparent weight that felt by passengers are change during the motion of the roller

coaster cars depend on the position of the car.

77 mgN

R

vgg

2

7

7

R

hhggmmgN

R

vgmmgN

'

77

2

7

77

2

cosmg

So the apparent weight that felt by


'27

2

7

'

2

7

'

'2

7

2

22

2

2

21

21)0(

21

21

21

hhgv

mvhhmg

mvmghmgh

mghmvmghm

mghmvmghmv

EMEM

FFAA

FA

cosg

R

hhggm

2cos

R

hhgg

2cos

R

ghgm

2

R

ghg

2

R

hhggm

2cos

R

hhgg

2cos

cosmg cosg

R

hhggm

2cos

R

hhggm

'2 R

hhgg

'2

Based on table1, the maximum magnitude of apparent weight that felt by

passengers is in point 3 that located at the bottom of the valley at path B-C-D at

the roller coaster track, that is . Because the maximum apparent

weight that felt by passengers are 8g and m =1 (our normal weight is 1g), so the

value of should be 7g in order to make the value of

The minimum magnitude of apparent weight that felt by passengers is in

point 7 that located at the top of the mountain at path E-F-G at the roller coaster

track, that is . Because the minimum apparent weight that felt

by passengers are 0 and m =1 (our normal weight is 1g), so the value of

should be g in order to make the value of

PROBLEM B

The Radius R of the Arc of the Circle that Fits

The Bottom of Valley

From the previous explanation, we know that the value of should be

7g, so . Then we also know that the value of should be g so

Based on Equation 1 and 2, we can conclude that:

R

ghgm

2

R

gh2 gR

ghgm 8

2

R

hhggm

'2

R

hhg '2

0

2 '

R

hhggm

hR

R

h

gR

gh

7

2

2

7

72

ggR

gh

gR

ghg

gR

gh

R

gh

gR

hhg

72

27

22

2

'

'

'

'

R

gh2

R

hhg '2

3

32

6

62

'

'

'

hR

R

h

gR

gh

Eq.1

Eq.

2

37

2 'hhR Eq.3

PROBLEM C

The Height of h and the Possibility of Roller Coaster Cars Movement that

Keep on the Track Without any Machine if the Track Joining Lower Circle

and Upper Circle is a Straight Line

The top of the next mountain is an arc of a second circle of the same radius

R with the radius of circle that make bottom of a valley before the mountain at the

roller coaster track. From problem A, we know that the magnitude of apparent

weight that felt by passengers of the roller coaster cars is zero (0g) at the top of

the mountain at F that located h from the ground.

We can define the magnitude of h using the concept of the conservation

law of mechanical energy that has been showed in the previous problem. Finally

from equation 3, we know that so

Its impossible to keep the roller coaster cars remain stay moved on the

roller coaster track without any machine if we consider that the track joining

lower circle and upper circle is a straight line track and the mountain on the

track was an arc of the circle that caused centripetal force that influence the

motion of roller coaster cars at this path (path E-F-G). Its caused by the negative

value of the apparent weight that felt by passengers at point 6 when the roller

coaster cars moved in the arc of the second circle that make mountain when the

cars moved from E to F along the track that formed as an arc of the second circle.

The magnitude of the apparent weight that felt by passengers of the roller

coaster cars that equivalent with the magnitude of normal force of that cars at this

path of track if we consider the track of the car is an arc of a circle is:

37

2 'hhR Rhh 3

7

6' Eq.4

hhmgRR

mgN

R

hhmgmgRN

R

hhmgmgN

R

hhggmN

2cos

2cos

2cos

2cos

Eq.5

Observe circle 2 that make mountain at the roller coaster track that

shown in figure below:

Figure11. The Valley and the Mountain of Roller Coaster

Based on equation 4, so

Then based equation 1, we know that so

Substitute equation 6 and 7 to the equation 5:

a

b

R

cos

cos

Ra

R

a

cos

cos

RRc

cRR

caR

cos'

'

'

RRhh

chh

chh

Rhh 37

6'

cos2

cos3

RRh

RRRh

Eq.6

hR7

2 Rh

2

7 Eq.7

cos2

2

72cos2cos RRRmgR

R

mghhmgR

R

mgN

The maximum value of cos is 1 so the maximum value of N is

It shown that the magnitude of the apparent weight that felt by passengers

at this position was negative because the value of (mg)2 bigger than mg. Its

indicated that the roller coaster cars will be out of the track although it still touch

the track (shown with the normal force at this path) if we consider the track of

roller coaster cars is an arc of circle at this path.

So to make the car can move on the track without any machine if the track

joining lower circle and upper circle is a straight line, we must consider that the

track after straight line D-E is parabolic track, so roller coaster cars still have

velocity at h, that is velocity to the horizontal direction (vx).

Based on the conservation law of mechanical energy, total energy at any

points on the track is always constant (the differences between one positions to

another is in the change of the value of kinetic and potential energy). At point C

(bottoms of the valley), roller coaster cars have maximum velocity (vx and vy).

After passed this position, velocity of cars decreases because the cars climbs the

mountain of roller coaster track. The magnitude of velocity at this path (path D-E)

depends on the magnitude of angle between the track with the horizontal line and

also the high of cars at this path. The higher positions of the cars, the velocity of

cars become smaller.

cos2

32cos

cos2

32cos

cos2

32cos

cos2

472cos

mgmgN

mgRRR

mgN

RR

mgRR

mgN

RRR

mgRR

mgN

21

2

121

12

321

mgmgmgmgN

mgmgN

mgmgN

After passed the straight line track (path D-E), roller coaster cars will pass

parabolic track (path E-F-G). At this path, velocity of cars is the resultant velocity

from velocity of cars at the horizontal and vertical direction. Follow the properties

of parabolic motion, the magnitude of velocity at horizontal direction (vx) are

always constant (motion along straight line with constant velocity [GLB]) at any

point in this path, but velocity of cars at vertical direction are change go to zero

(roller coaster cars were slowing down) until it reach the top of roller coaster

mountain at F. At this position, the vertical velocity of cars (vy) is zero but cars

still can move to G because it still has horizontal velocity (vx) that always constant

along this path. The car speeding up again after it passed the top of mountain. The

shorter the height of cars, the resultant velocity of cars will greater follow the

concept of conservation law of mechanical energy. Same with the previous case,

resultant velocity at path F-G also depend on the velocity at horizontal and

vertical direction.

PROBLEM D

The Changes of Apparent Weight that Felt by Passengers of the Roller

Coaster Cars during Their Journey

PROBLEM E

The Reason Why the First Hill in Roller Coaster Ride

Be the Highest One

We must make the first hill in roller coaster ride be the highest one in

order to make maximum value of potential energy of that system. If at that first

hill roller coaster cars have zero velocity, the magnitude of energy mechanic at

any position of cars at that roller coaster is same with the magnitude of potential

energy of cars at the first hill. The higher the height of the first hill, the

mechanical energy of that roller coaster car will also be greater. Energy mechanic

at any points at this system is remain constant (follow the conservation law of

mechanical energy), the difference between one position to others is at the

magnitude of kinetic and potential energy that always changes depend on the

velocity and the height of cars at any position.

So if roller coaster cars move through the track of roller coaster that has

the highest position at the first hill, roller coaster cars still can pass the next hill

(or mountain) that shortened than the first hill because it has enough energy to

pass it points. But if the next hill higher than the first hill, roller coaster cars cant

pass this point because it doesnt has enough energy to pass this point if we dont

use any machine to produce some force to pull roller coaster cars until it reach the

top of the next hill that has higher position than the first hill.

physics behind roller coaster

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