5:45 Sat, Feb 11 < Classical Dynamics of Particles and Systems - Inst 9-0 (x₂-4/₂) x+ If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum point of the cycloid at (2a,0) the parametric equations are x = a (1 + cos 0) y = a(0+ sin 0) Since the particle starts from rest at the point (x₁, y₁), the velocity at any elevation x is [cf. Eq. 6.19] v = √28(x-x₁) Then, the time required to reach the point (x₂,₂) is [cf. Eq. 6.20] t= t= (₁₁) V8 Using (1) and the derivatives obtained therefrom, (3) can be written as 1+ cos 0 cos-cos, 1+y¹² 2g(x-x₁) t== 歐 V8⁰ 0 2 0₂-0L Now, using the trigonometric identity, 1+ cos 0 = 2 cos² 0/2, we have 0 COS de 2 11 dx cos2 Cos do 0, 2 (1) (2) (3) (4) O < 57%
5:45 Sat, Feb 11 < Classical Dynamics of Particles and Systems - Inst 9-0 (x₂-4/₂) x+ If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum point of the cycloid at (2a,0) the parametric equations are x = a (1 + cos 0) y = a(0+ sin 0) Since the particle starts from rest at the point (x₁, y₁), the velocity at any elevation x is [cf. Eq. 6.19] v = √28(x-x₁) Then, the time required to reach the point (x₂,₂) is [cf. Eq. 6.20] t= t= (₁₁) V8 Using (1) and the derivatives obtained therefrom, (3) can be written as 1+ cos 0 cos-cos, 1+y¹² 2g(x-x₁) t== 歐 V8⁰ 0 2 0₂-0L Now, using the trigonometric identity, 1+ cos 0 = 2 cos² 0/2, we have 0 COS de 2 11 dx cos2 Cos do 0, 2 (1) (2) (3) (4) O < 57%
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
How equation 4 of the
![5:45 Sat, Feb 11
< Classical Dynamics of Particles and Systems - Inst
2a
X
If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum
point of the cycloid at (2ª,0) the parametric equations are
x = a (1+cos €)
y = a(0+ sin 0)
Since the particle starts from rest at the point (x₁, y₁), the velocity at any elevation x is [cf. Eq.
6.19]
t =
+
0=0
(x2,Y₂)
= √28(x-x₁)
Then, the time required to reach the point (x₂,y₂) is [cf. Eq. 6.20]
x2
t = Ĵ
x1
a
v=
V8
Using (1) and the derivatives obtained therefrom, (3) can be written as
11/2
O
(x₁, y₁)
1+y₁²
2g(x-x₁)
0₂₁
Ĵ
₂=OL
a
t=
-A
Now, using the trigonometric identity, 1+ cos 0 = 2 cos² 0/2, we have
0
11
1+ cos 0
cos-cos₁
COS
0 cos²2
11/2
0
2
dx
- dᎾ
2
2 0₁
2
do
Cos²
□ O
(1)
(2)
(3)
|||
O
B
57%
Q :
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Transcribed Image Text:5:45 Sat, Feb 11
< Classical Dynamics of Particles and Systems - Inst
2a
X
If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum
point of the cycloid at (2ª,0) the parametric equations are
x = a (1+cos €)
y = a(0+ sin 0)
Since the particle starts from rest at the point (x₁, y₁), the velocity at any elevation x is [cf. Eq.
6.19]
t =
+
0=0
(x2,Y₂)
= √28(x-x₁)
Then, the time required to reach the point (x₂,y₂) is [cf. Eq. 6.20]
x2
t = Ĵ
x1
a
v=
V8
Using (1) and the derivatives obtained therefrom, (3) can be written as
11/2
O
(x₁, y₁)
1+y₁²
2g(x-x₁)
0₂₁
Ĵ
₂=OL
a
t=
-A
Now, using the trigonometric identity, 1+ cos 0 = 2 cos² 0/2, we have
0
11
1+ cos 0
cos-cos₁
COS
0 cos²2
11/2
0
2
dx
- dᎾ
2
2 0₁
2
do
Cos²
□ O
(1)
(2)
(3)
|||
O
B
57%
Q :
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