You are an architect designing a roller coaster for a theme park. You plan to construct a track with a loop radius of 75 m. What is the minimum cart speed so that the cart does not fall off the track at the top of the loop?
You are an architect designing a roller coaster for a theme park. You plan to construct a track with a loop radius of 75 m. What is the minimum cart speed so that the cart does not fall off the track at the top of the loop?
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter15: Oscillatory Motion
Section: Chapter Questions
Problem 15.12OQ: For a simple harmonic oscillator, answer yes or no to the following questions, (a) Can the...
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Kindly provide the solution to the following question using the GRASS method. Uniform and Circular Motion (Dynamics Unit):
You are an architect designing a roller coaster for a theme park. You plan to construct a track with a loop radius of 75 m. What is the minimum cart speed so that the cart does not fall off the track at the top of the loop?
Image attached is a guide on how to use the GRASS method on a question and the other image are the formulas. Please make sure to show all your work.
Transcribed Image Text:Motion
Ad = Ad₁ + Ad₂ + ...
b
sin B
a
sin A
V₂ V
t
VAB = VAC + VCB
a av
=
=
Forces
āc
F = ma
=
C
sin C
c²=a² + b² sin
d
t
āav ( t)²
2
F = mã,
Uniform Circular Motion
1²
F = ma
y=-
d = v₁ t+
VXY =
√xx
ā =
v=
x =
F
m
F₂
Unit 1 Dynamics
=
à
t
H
cos
-b+√b²-4ac
2a
F = mg
GMm
ā=
à = (v₁ + √₂) t
2
H
V
7.
t
tan 0 =
1.
A
A==bh
v² = ₁² +2αav
F₁ = FN
v=2-r
V=
T
c²=a² + b² - 2ab cos 0
Y₂ Y₁
X₂ X₁
āav( t)²
2
ā
F
A = lw
s,max
N
Slope=
d=v₂ t
a = 4 ² rf² T
r=}
V=
FN
GM
r
![tiver:
Ad Ad
Required:
Ad Ad Ad
Y
Anclysis:
Let mooth & west be positive tv
4d² = Ad + Ad
2
Sektion:
N
(20.0km) (sin 25°)
(20.0xmXcos 25°)
(45.0 kyksin 40) X
WE
(45.0km) (cos 40.00)
40.0⁰
Ad Ad + Ad
Bd² = (-20.0kn) (co₁ 250) + (45.0 kn)(sin 40.0°)
bota
= 10.799 km
Ad = A + Act
Ady = (20.0 km.) (sin 250) + (45.0 kn X(cos 40.0")
Ady = [42.924 km]
| AZ²₂1 = √(40₂)² + (Ad4₂) ²
142, 1 =√ √/(10.799 kn)² + (42.924 kn)²
1Ad1=44.262 km
Scheton (contid):
-
O + tan" (Ads)
ở bán 42.921 km
-1
10.799 Kn
0 = 75.9°
Statene ti
Then fore, the displacement of
the whole is 4.43 x 10¹ kn [W76″N](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b519efd-481f-44c9-b15a-1bd08236bdce%2Fc496acf6-81b1-4610-883a-1db47bd17364%2Fh8chn7_processed.png&w=3840&q=75)
Transcribed Image Text:tiver:
Ad Ad
Required:
Ad Ad Ad
Y
Anclysis:
Let mooth & west be positive tv
4d² = Ad + Ad
2
Sektion:
N
(20.0km) (sin 25°)
(20.0xmXcos 25°)
(45.0 kyksin 40) X
WE
(45.0km) (cos 40.00)
40.0⁰
Ad Ad + Ad
Bd² = (-20.0kn) (co₁ 250) + (45.0 kn)(sin 40.0°)
bota
= 10.799 km
Ad = A + Act
Ady = (20.0 km.) (sin 250) + (45.0 kn X(cos 40.0")
Ady = [42.924 km]
| AZ²₂1 = √(40₂)² + (Ad4₂) ²
142, 1 =√ √/(10.799 kn)² + (42.924 kn)²
1Ad1=44.262 km
Scheton (contid):
-
O + tan" (Ads)
ở bán 42.921 km
-1
10.799 Kn
0 = 75.9°
Statene ti
Then fore, the displacement of
the whole is 4.43 x 10¹ kn [W76″N
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