Exercicios Resolvidos do Halliday sobre Rotação- Cap11- Exercicio 67

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67. (a) We use conservation of mechanical energy to find an expression for ω2 as a function of the angle
θ that the chimney makes with the vertical. The potential energy of the chimney is given by
U = Mgh, where M is its mass and h is the altitude of its center of mass above the ground. When
the chimney makes the angle θ with the vertical, h = (H/2) cos θ. Initially the potential energy is
Ui = Mg(H/2) and the kinetic energy is zero. The kinetic energy is 12Iω
2 when the chimney makes
the angle θ with the vertical, where I is its rotational inertia about its bottom edge. Conservation
of energy then leads to
MgH/2 = Mg(H/2) cos θ +
1
2
Iω2 =⇒ ω2 = (MgH/I)(1− cos θ) .
The rotational inertia of the chimney about its base is I = MH2/3 (found using Table 11-2(e) with
the parallel axis theorem). Thus
ω =
√
3g
H
(1− cos θ) .
(b) The radial component of the acceleration of the chimney top is given by ar = Hω2, so ar =
3g(1− cos θ).
(c) The tangential component of the acceleration of the chimney top is given by at = Hα, where α is
the angular acceleration. We are unable to use Table 11-1 since the acceleration is not uniform.
Hence, we differentiate ω2 = (3g/H)(1− cos θ) with respect to time, replacing dω/dt with α, and
dθ/dt with ω, and obtain
dω2
dt
= 2ωα = (3g/H)ω sin θ =⇒ α = (3g/2H) sin θ .
Consequently, at = Hα = 3g2 sin θ.
(d) The angle θ at which at = g is the solution to 3g2 sin θ = g. Thus, sin θ = 2/3 and we obtain
θ = 41.8◦.
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