The document discusses the physics of a particle's motion involving angular velocities and accelerations in a radial coordinate system. It includes formulas for calculating the particle's acceleration and motion based on radius and angles, incorporating gravitational effects. Various integrals and specific values are presented to demonstrate the findings in a practical context.

1. 1.26
Datos del problema
 Ƴ = tan−1 ℎ
2𝜋𝑟
 Θ = Ƴ
 Rθ’ = G.senƳ
Tenemosentoncesque el radioesconstante,porlotanto:
 r’ = 0
 r’’ = 0
hallandolaaceleraciónde laparticula
a = (r’’- rθ’2
)ur + (rθ’’+2r’θ’)uθ +z’k
a = (-rθ’2
)ur + ( rθ’’)uθ + zk
calculando θ’
 θ’ = Ƴ’
 Ƴ’ = arctan’(
ℎ
2𝜋𝑟
) =
ℎ
2𝜋
(
1
𝑟2
)
(4𝜋2+
ℎ2
4𝜋2 𝑟2
Ahora
rθ’’ = gsenƳ
θ’’ =
gsenƳ
𝑟
a = -r(
ℎ
2𝜋
(
1
𝑟
)
(4𝜋2+
ℎ2
4𝜋2 𝑟2
)2
ur +(r.
gsenƳ
𝑟
)uθ + zk
a =
−𝑟(
ℎ2
4𝜋2(
1
𝑟4
))
(4𝜋2+
ℎ2
4𝜋2 𝑟2) 2ur + (gsenƳ) uθ + z’k

2. calculandoz’
hθ = 2𝜋𝑧
z =
ℎ𝜃
2𝜋
→ z’ =
ℎ𝜃’
2𝜋
z’ =
−ℎ
16𝜋4+ℎ2
 a=(
−ℎ
8𝜋3 𝑟3+
ℎ2
2𝜋
)ur + (gsenƳ)uθ + (
−ℎ
16𝜋4 𝑟2+ℎ2
)k

3. 2.28
∑ 𝐹𝜃 = 𝑚. 𝑎θ
0 = m [𝑟𝜃’’ + 2𝑟’𝜃’ ] = m[
1
𝑟
𝑑(𝑟2 𝜃)
𝑑𝑥
]
De donde:
d(r2
θ) = 0
integrando :
para ro = 0.5 y r = 0.25
se tiene :
(0.5)2
( 1) = c = (0.25)2
θ’
Para r = 0.25 : θ = 4 rads/s
Como jalamos a velocidad de 0.2 m/s
Entonces: r = -0.2m/s : r = 0
ar = r’’ - rθ’2
= 0-0.25(4)2
= -4m/s
sea la 2da ley en la dirección radial
∑ 𝐹𝑟 = 𝑚𝑎r : -T = 2(-4)
T = 8